Решить 1 2 3: Калькулятор дробей онлайн — Таловская средняя школа

Калькулятор онлайн — Решение иррациональных уравнений и неравенств

Этот математический калькулятор онлайн поможет вам решить иррациональное уравнение или неравенство. Программа для решения иррациональных уравнений и неравенств не просто даёт ответ задачи, она приводит подробное решение с пояснениями, т.е. отображает процесс получения результата.

Данная программа может быть полезна учащимся старших классов общеобразовательных школ при подготовке к контрольным работам и экзаменам, при проверке знаний перед ЕГЭ, родителям для контроля решения многих задач по математике и алгебре. А может быть вам слишком накладно нанимать репетитора или покупать новые учебники? Или вы просто хотите как можно быстрее сделать домашнее задание по математике или алгебре? В этом случае вы также можете воспользоваться нашими программами с подробным решением.

Таким образом вы можете проводить своё собственное обучение и/или обучение своих младших братьев или сестёр, при этом уровень образования в области решаемых задач повышается.

Вы можете посмотреть теорию и общие методы решения иррациональных уравнений и неравенств.

Примеры подробного решения >>

sqrt(x) — квадратный корень x
x^(1/n) — корень степени n

Введите иррациональное уравнение или неравенство

Обнаружено что не загрузились некоторые скрипты, необходимые для решения этой задачи, и программа может не работать.
Возможно у вас включен AdBlock.
В этом случае отключите его и обновите страницу.

Сообщение отправлено. Спасибо.

Содержание

Решение иррациональных уравнений и неравенств

1. Иррациональные уравнения

Иррациональными называют уравнения, в которых переменная содержится под знаком радикала или под знаком возведения в дробную степень. Для таких уравнений ищут, как правило, только действительные корни.

Основной метод решения иррациональных уравнений — метод возведения обеих частей уравнения в одну и ту же степень. При этом следует иметь в виду, что возведение обеих частей уравнения в одну и ту же нечётную степень есть равносильное преобразование уравнения, а в чётную — НЕравносильное. Значит, основные принципиальные трудности связаны с возведением обеих частей уравнения в одну и ту же чётную степень, когда из-за неравносильности преобразования могут появиться посторонние корни, а потому обязательна проверка всех найденных корней.

ПРИМЕР 1.
$ \sqrt[\Large6\normalsize]{x^2-5x} = \sqrt[\Large6\normalsize]{2x-6} $

Возведя обе части уравнения в шестую степень, получим:
$ x^2-5x = 2x-6 \Rightarrow $
$ x^2-7x +6= 0 \Rightarrow $
$ x_1=1, \; x_2=6 $
Проверка. «Хорошие» корни можно проверить непосредственной подстановкой в исходное уравнение. При x = 1 заданное уравнение принимает вид $ \sqrt[\Large6\normalsize]{-4} = \sqrt[\Large6\normalsize]{-4} $, во множестве действительных чисел такое «равенство» не имеет смысла. Значит, 1 — посторонний корень, он появился по причине расширения ОДЗ уравнения после возведения в шестую степень. При х = 6 заданное уравнение принимает вид $ \sqrt[\Large6\normalsize]{6} = \sqrt[\Large6\normalsize]{6} $ — это верное равенство.
Итак, уравнение имеет единственный корень: х = 6.
Ответ: х = 6

ПРИМЕР 2.
$ \sqrt{x^2-x+2}+\sqrt{x^2-x+7} = \sqrt{2x^2-2x+21} $

Введя новую переменную $ u=x^2-x$, получим существенно более простое иррациональное уравнение:
$ \sqrt{u+2}+\sqrt{u+7} = \sqrt{2u+21} $.
Возведём обе части уравнения в квадрат:
$ (\sqrt{u+2}+\sqrt{u+7})^2 = (\sqrt{2u+21})^2 \Rightarrow $
$ u+2 +2\sqrt{u+2}\sqrt{u+7} +u+7 = 2u+21 \Rightarrow $
$ \sqrt{(u+2)(u+7)} = 6 \Rightarrow $
$ u^2+9u+14=36 \Rightarrow $
$ u^2+9u-22=0 \Rightarrow $
$ u_1=2, \; u_2=-11 $
Проверка найденных значений их подстановкой в уравнение $ \sqrt{u+2}+\sqrt{u+7} = \sqrt{2u+21} $ показывает, что $ u_1=2 $ — корень уравнения, а $ u_2=-11 $ — посторонний корень.
Возвращаясь к исходной переменной x, получаем уравнение $ x^2-x=2 \Rightarrow x^2-x-2=0 $, решив которое находим два корня: $ x_1=2, \; x_2=-1 $
Ответ: 2; -1.

ПРИМЕР 3.
$ x^2+3-\sqrt{2x^2-3x+2} = 1{,}5(x+4) $

Уединение корня и возведение обеих частей уравнения в квадрат привело бы к громоздкому уравнению. В то же время, если проявить некоторую наблюдательность, можно заметить, что уравнение легко сводится к квадратному. Действительно, умножим обе его части на 2:
$ 2x^2 +6 -2\sqrt{2x^2-3x+2} = 3x+12 \Rightarrow $
$ 2x^2 -3x +2 -2\sqrt{2x^2-3x+2} -8 = 0 \Rightarrow $

Введя новую переменную $ y=\sqrt{2x^2-3x+2} $, получим: $ y^2-2y-8=0 $, откуда $ y_1=4, \; y_2=-2 $. Значит, исходное уравнение равносильно следующей совокупности уравнений:
$ \left[\begin{array}{l} \sqrt{2x^2-3x+2} =4 \\ \sqrt{2x^2-3x+2} = -2 \end{array}\right. $

Из первого уравнения этой совокупности находим: $ x_1=3{,}5; \; x_2=-2 $. Второе уравнение корней не имеет.

Проверка. Так как совокупность уравнений равносильна исходному уравнению, причём второе уравнение этой совокупности корней не имеет, то найденные корни можно проверить подстановкой в уравнение $ \sqrt{2x^2-3x+2} =4$. Эта подстановка показывает, что оба найденных значения x являются корнями этого уравнения, а значит, и исходного уравнения.
Ответ: 3,5; -2.

ПРИМЕР 4.
$ 2x -5 +2\sqrt{x^2-5x} +2\sqrt{x-5} +2\sqrt{x}= 48 $

Областью определения уравнения является луч $ [5; \; +\infty) $. В этой области выражение $ \sqrt{x^2-5x} $ можно представить следующим образом: $ \sqrt{x^2-5x} = \sqrt{x}\sqrt{x-5} $. Теперь уравнение можно переписать так:
$ x+x -5 +2\sqrt{x}\sqrt{x-5} +2\sqrt{x-5} +2\sqrt{x} -48 = 0 \Rightarrow $ $ (\sqrt{x})^2 +2\sqrt{x}\sqrt{x-5} +(\sqrt{x-5})^2 +2(\sqrt{x-5}+\sqrt{x}) -48 = 0 \Rightarrow $ $ (\sqrt{x-5} +\sqrt{x})^2 +2(\sqrt{x-5}+\sqrt{x}) -48 = 0 $

Введя новую переменную $ y= \sqrt{x-5} +\sqrt{x} $, получим квадратное уравнение $ y^2+2y-48=0 $, из которого находим: $ y_1=6, \; y_2=-8 $. Таким образом, задача свелась к решению совокупности уравнений:
$ \left[\begin{array}{l} \sqrt{x-5} +\sqrt{x} =6 \\ \sqrt{x-5} +\sqrt{x} = -8 \end{array}\right. $
Из первого уравнения совокупности находим $ x= \left( \frac{41}{12} \right)^2 $, второе уравнение совокупности решений явно не имеет.

Проверка. Нетрудно проверить (подстановкой), что $ x= \left( \frac{41}{12} \right)^2 $ — является корнем уравнения $ \sqrt{x-5} +\sqrt{x} =6 $. Но это уравнение равносильно исходному уравнению, значит, $ x= \left( \frac{41}{12} \right)^2 $ — является корнем и исходного уравнения.
Ответ: $ x= \left( \frac{41}{12} \right)^2 $

Иногда при решении иррациональных уравнений оказывается удобным ввести две новые переменные.

ПРИМЕР 5.
$ \sqrt[\Large4\normalsize]{1-x} + \sqrt[\Large4\normalsize]{15+x} =2 $

Введём новые переменные: $ \left\{\begin{array}{l} u=\sqrt[\Large4\normalsize]{1-x} \\ v=\sqrt[\Large4\normalsize]{15+x} \end{array}\right. $

Тогда уравнение примет вид $u+v=2$. Но для нахождения значений двух новых переменных одного уравнения недостаточно. Возведя в четвёртую степень обе части каждого из уравнений системы, получим:
$ \left\{\begin{array}{l} u^4=1-x \\ v^4= 15+x \end{array}\right. $

Сложим уравнения последней системы: $u^4 +v^4 =16$. Таким образом, для нахождения u, v мы имеем следующую симметрическую систему уравнений:
$ \left\{\begin{array}{l} u+v=2 \\ u^4 +v^4 =16 \end{array}\right. $
Решив её, находим: $ \left\{\begin{array}{l} u_1=0 \\ v_1 =2; \end{array}\right. $ $ \left\{\begin{array}{l} u_2=2 \\ v_2 =0 \end{array}\right. $

Таким образом, исходное уравнение свелось к следующей совокупности систем уравнений: $ \left\{\begin{array}{l} \sqrt[\Large4\normalsize]{1-x} =0 \\ \sqrt[\Large4\normalsize]{15+x} =2; \end{array}\right. $ $ \left\{\begin{array}{l} \sqrt[\Large4\normalsize]{1-x} =2 \\ \sqrt[\Large4\normalsize]{15+x} =0 \end{array}\right. $

Решив эту совокупность, находим: $x_1=1, \; x_2=-15 $

Проверка. Проще всего проверить найденные корни непосредственной подстановкой в заданное уравнение. Проделав это, убеждаемся, что оба значения являются корнями исходного уравнения.
Ответ: 1; -15.

ПРИМЕР 6.
$ \sqrt[\Large3\normalsize]{2x+1} + \sqrt[\Large3\normalsize]{6x+1} = \sqrt[\Large3\normalsize]{2x-1} $

Возведём обе части уравнения в куб:
$ 2x+1 + 3\sqrt[\Large3\normalsize]{(2x+1)^2} \cdot \sqrt[\Large3\normalsize]{6x+1} + 3\sqrt[\Large3\normalsize]{2x+1} \cdot \sqrt[\Large3\normalsize]{(6x+1)^2} +6x+1 = 2x-1 \Rightarrow $ $ 3\sqrt[\Large3\normalsize]{2x+1} \cdot \sqrt[\Large3\normalsize]{6x+1} \cdot (3\sqrt[\Large3\normalsize]{2x+1} + \sqrt[\Large3\normalsize]{6x+1} ) = -6x-3 $

Воспользовавшись исходным уравнением, заменим сумму $ \sqrt[\Large3\normalsize]{2x+1} + \sqrt[\Large3\normalsize]{6x+1} $ на выражение $ \sqrt[\Large3\normalsize]{2x-1} $:
$ 3\sqrt[\Large3\normalsize]{2x+1} \cdot \sqrt[\Large3\normalsize]{6x+1} \cdot \sqrt[\Large3\normalsize]{2x-1} = -6x-3 \Rightarrow $
$ 3\sqrt[\Large3\normalsize]{ (2x+1)(6x+1)(2x-1) } = -2x-1 $
Возведём обе части в куб:
$ (2x+1)(6x+1)(2x-1) = -(2x+1)^3 \Rightarrow $
$ (2x+1)((6x+1)(2x-1) + (2x+1)^2) =0 \Rightarrow $
$ 16x^2(2x+1) =0 \Rightarrow $
$ x_1= -0{,}5; \; x_2=0 $

Проверка. Подстановкой найденных значений x в исходное уравнение убеждаемся, что его корнем является только x = -0,5.
Ответ: -0,5.

2. Иррациональные неравенства

Рассмотрим иррациональное неравенство вида $ \sqrt{f(x)} Ясно, что его решения должны удовлетворять условию \( f(x) \geq 0 $ и условию $ g(x) > 0 $. Осталось лишь заметить, что при одновременном выполнении указанных выше условий обе части заданного иррационального неравенства неотрицательны, а потому их возведение в квадрат представляет собой равносильное преобразование неравенства.

Таким образом, иррациональное неравенство \( \sqrt{f(x)} \( \left\{\begin{array}{l} f(x) \geq 0 \\ g(x) > 0 \\ f(x) ПРИМЕР 7.
\( \sqrt{x^2-x-12}

Данное неравенство равносильно системе неравенств:
$ \left\{\begin{array}{l} x^2-x-12 \geq 0 \\ x > 0 \\ x^2-x-12 0 \\ x > -12 \end{array}\right. $

Получаем: $ x \geq 4$

Ответ: $ x \geq 4$

Рассмотрим теперь неравенство вида $ \sqrt{f(x)} > g(x) $.

Ясно, во-первых, что его решения должны удовлетворять условию $ f(x) \geq 0 $.
Во-вторых, замечаем, что при $ g(x) g(x) $ не вызывает сомнений.
В-третьих, замечаем, что если $ g(x) \geq 0 $, то можно возвести в квадрат обе части заданного иррационального неравенства.

Таким образом, иррациональное неравенство $ \sqrt{f(x)} > g(x) $ равносильно совокупности систем неравенств:
$ \left\{\begin{array}{l} f(x) \geq 0 \\ g(x) (g(x))^2 \end{array}\right. $

Во второй системе первое неравенство является следствием третьего, его можно не писать.

ПРИМЕР 8.
$ \sqrt{x^2-x-12} \geq x $

Данное неравенство равносильно совокупности систем неравенств:
\( \left\{\begin{array}{l} x^2-x-12 \geq 0 \\ x

Имеем:
\( \left\{\begin{array}{l} (x-4)(x+3) \geq 0 \\ x

Из первой системы находим: $ x \leq -3$, вторая система не имеет решений.
Ответ: $ x \leq -3$

ПРИМЕР 9.
$ (x+5)(x-2) +3\sqrt{x(x+3)} >0 $

Преобразуем неравенство к виду $ x^2+3x-10 +3\sqrt{x^2+3x} >0 $ и введём новую переменную $ y= \sqrt{x^2+3x} $. Тогда последнее неравенство примет вид $ y^2+3y-10 >0 $, откуда находим, что либо $y 2$.

Таким образом, задача сводится к решению совокупности двух неравенств:
$ \left[\begin{array}{l} \sqrt{x^2+3x} 2 \end{array}\right. $

Первое неравенство не имеет решений, а из второго находим:
$ x^2+3x >4 \Rightarrow $
$ (x+4)(x-1) >0 \Rightarrow $
$ x1 $
Ответ: $ x1 $.

Решение предела функции · Калькулятор Онлайн

Введите функцию и точку, для которых надо вычислить предел

Сайт предоставляет ПОДРОБНОЕ решение по нахождению предела функции.

Займемся вычислением (решением) пределов функций в точке.
Дана функция f(x). Вычислим ее предел в точке x0.
Для примера, находит предел функции в нуле и предел на бесконечности.

Правила ввода выражений и функций

Выражения могут состоять из функций (обозначения даны в алфавитном порядке):

absolute(x): Абсолютное значение x
(модуль x или |x|)
arccos(x): Функция — арккосинус от x
arccosh(x): Арккосинус гиперболический от x
arcsin(x): Арксинус от x
arcsinh(x): Арксинус гиперболический от x
arctg(x): Функция — арктангенс от x
arctgh(x): Арктангенс гиперболический от x
e: e число, которое примерно равно 2.7
exp(x): Функция — экспонента от x (что и e^x)
log(x) or ln(x): Натуральный логарифм от x
(Чтобы получить log7(x), надо ввести log(x)/log(7) (или, например для log10(x)=log(x)/log(10))
pi: Число — «Пи», которое примерно равно 3.14
sin(x): Функция — Синус от x
cos(x): Функция — Косинус от x
sinh(x): Функция — Синус гиперболический от x
cosh(x): Функция — Косинус гиперболический от x
sqrt(x): Функция — квадратный корень из x
sqr(x) или x^2: Функция — Квадрат x
tg(x): Функция — Тангенс от x
tgh(x): Функция — Тангенс гиперболический от x
cbrt(x): Функция — кубический корень из x

В выражениях можно применять следующие операции:

Действительные числа: вводить в виде 7.5, не 7,5
2*x: — умножение
3/x: — деление
x^3: — возведение в степень
x + 7: — сложение
x — 6: — вычитание

Другие функции:

floor(x): Функция — округление x в меньшую сторону (пример floor(4.5)==4.0)
ceiling(x): Функция — округление x в большую сторону (пример ceiling(4.5)==5.0)
sign(x): Функция — Знак x
erf(x): Функция ошибок (или интеграл вероятности)
laplace(x): Функция Лапласа

Решите неравенство 5^(x+2)+5^(x+1)-5^x

Дано неравенство:
$$- 5^{x} + 5^{x + 1} + 5^{x + 2} \leq — 3^{\frac{x}{2} — 1} + — 3^{\frac{x}{2}} + 3^{\frac{x}{2} + 1}$$
Чтобы решить это нер-во — надо сначала решить соотвествующее ур-ние:
$$- 5^{x} + 5^{x + 1} + 5^{x + 2} = — 3^{\frac{x}{2} — 1} + — 3^{\frac{x}{2}} + 3^{\frac{x}{2} + 1}$$
Решаем:
$$x_{1} = \frac{- 2 \log{\left (5 \right )} + 2 \log{\left (87 \right )}}{- \log{\left (25 \right )} + \log{\left (3 \right )}}$$
$$x_{1} = \frac{- 2 \log{\left (5 \right )} + 2 \log{\left (87 \right )}}{- \log{\left (25 \right )} + \log{\left (3 \right )}}$$
Данные корни
$$x_{1} = \frac{- 2 \log{\left (5 \right )} + 2 \log{\left (87 \right )}}{- \log{\left (25 \right )} + \log{\left (3 \right )}}$$
являются точками смены знака неравенства в решениях.
Сначала определимся со знаком до крайней левой точки:
$$x_{0} \leq x_{1}$$
Возьмём например точку
$$x_{0} = x_{1} — \frac{1}{10}$$
=

2*(-log(5) + log(87))   1 
--------------------- - --
                    1   10
 (-log(25) + log(3))

=
$$\frac{- 2 \log{\left (5 \right )} + 2 \log{\left (87 \right )}}{- \log{\left (25 \right )} + \log{\left (3 \right )}} — \frac{1}{10}$$
подставляем в выражение
$$- 5^{x} + 5^{x + 1} + 5^{x + 2} \leq — 3^{\frac{x}{2} — 1} + — 3^{\frac{x}{2}} + 3^{\frac{x}{2} + 1}$$

                                                                                                    2*(-log(5) + log(87))   1         2*(-log(5) + log(87))   1     2*(-log(5) + log(87))   1     
                                                                                                    --------------------- - --        --------------------- - --    --------------------- - --    
 2*(-log(5) + log(87))   1         2*(-log(5) + log(87))   1         2*(-log(5) + log(87))   1                          1   10                            1   10                        1   10    
 --------------------- - -- + 2    --------------------- - -- + 1    --------------------- - --      (-log(25) + log(3))               (-log(25) + log(3))           (-log(25) + log(3))          
                     1   10                            1   10                            1   10     -------------------------- + 1    --------------------------    -------------------------- - 1
  (-log(25) + log(3))               (-log(25) + log(3))               (-log(25) + log(3))                       2                                 2                             2                 
5                               + 5                               - 5                            9    2*(-log(5) + log(87))    19   2*(-log(5) + log(87))      1    2*(-log(5) + log(87))     19   -log(5) + log(87)      21   -log(5) + log(87)      1    -log(5) + log(87)
 -- + ---------------------    -- + ---------------------    - -- + ---------------------     -- + -----------------    - -- + -----------------    - -- + -----------------
 10     -log(25) + log(3)      10     -log(25) + log(3)        10     -log(25) + log(3)   
значит решение неравенства будет при:
$$x \leq \frac{- 2 \log{\left (5 \right )} + 2 \log{\left (87 \right )}}{- \log{\left (25 \right )} + \log{\left (3 \right )}}$$ _____          
      \    
-------•-------
       x1

Решите неравенство x^2-3*x+1-(x^3+x^2+3*x-21)*1/x>

=3 (х в квадрате минус 3 умножить на х плюс 1 минус (х в кубе плюс х в квадрате плюс 3 умножить на х минус 21) умножить на 1 делить на х больше или равно 3) Дано неравенство:
$$x^{2} — 3 x + 1 — \frac{1}{x} \left(3 x + x^{3} + x^{2} — 21\right) \geq 3$$
Чтобы решить это нер-во — надо сначала решить соотвествующее ур-ние:
$$x^{2} — 3 x + 1 — \frac{1}{x} \left(3 x + x^{3} + x^{2} — 21\right) = 3$$
Решаем:
Дано уравнение:
$$x^{2} — 3 x + 1 — \frac{1}{x} \left(3 x + x^{3} + x^{2} — 21\right) = 3$$
преобразуем:
Вынесем общий множитель за скобки
$$- \frac{1}{x} \left(x + 3\right) \left(4 x — 7\right) = 0$$
знаменатель
$$x$$
тогда

x не равен 0

Т.к. правая часть ур-ния равна нулю, то решение у ур-ния будет, если хотя бы один из множителей в левой части ур-ния равен нулю.
Получим ур-ния
$$- x — 3 = 0$$
$$4 x — 7 = 0$$
решаем получившиеся ур-ния:
1.
$$- x — 3 = 0$$
Переносим свободные слагаемые (без x)
из левой части в правую, получим:

-x = 3

Разделим обе части ур-ния на -1

x = 3 / (-1)

Получим ответ: x1 = -3
3.
$$4 x — 7 = 0$$
Переносим свободные слагаемые (без x)
из левой части в правую, получим:
$$4 x = 7$$
Разделим обе части ур-ния на 4

x = 7 / (4)

Получим ответ: x2 = 7/4
но

x не равен 0

$$x_{1} = -3$$
$$x_{2} = \frac{7}{4}$$
$$x_{1} = -3$$
$$x_{2} = \frac{7}{4}$$
Данные корни
$$x_{1} = -3$$
$$x_{2} = \frac{7}{4}$$
являются точками смены знака неравенства в решениях.
Сначала определимся со знаком до крайней левой точки:
$$x_{0} \leq x_{1}$$
Возьмём например точку
$$x_{0} = x_{1} — \frac{1}{10}$$
=
$$- \frac{31}{10}$$
=
$$- \frac{31}{10}$$
подставляем в выражение
$$x^{2} — 3 x + 1 — \frac{1}{x} \left(3 x + x^{3} + x^{2} — 21\right) \geq 3$$

                              3         2                    
                        /-31 \    /-31 \    3*(-31)          
      2                 |----|  + |----|  + ------- - 21     
/-31 \    3*(-31)       \ 10 /    \ 10 /       10            
|----|  - ------- + 1 - -------------------------------- >= 3
\ 10 /       10                           1                  
                                    /-31 \                   
                                    |----|                   
                                    \ 10 /

562     
--- >= 3
155

значит одно из решений нашего неравенства будет при:
$$x \leq -3$$

 _____           _____          
      \         /
-------•-------•-------
       x1      x2

Другие решения неравенства будем получать переходом на следующий полюс
и т.д.
Ответ:
$$x \leq -3$$
$$x \geq \frac{7}{4}$$ 90000 Solve inequalities with Step-by-Step Math Problem Solver 90001 90002 90003 Enter an equation along with the variable you wish to solve it for and click the Solve button. 90004 90003 In this chapter, we will develop certain techniques that help solve problems stated in words. These techniques involve rewriting problems in the form of symbols. For example, the stated problem 90004 90003 «Find a number which, when added to 3, yields 7» 90004 90003 may be written as: 90004 90003 3 +? = 7, 3 + n = 7, 3 + x = 1 90004 90003 and so on, where the symbols?, N, and x represent the number we want to find.We call such shorthand versions of stated problems equations, or symbolic sentences. Equations such as x + 3 = 7 are first-degree equations, since the variable has an exponent of 1. The terms to the left of an equals sign make up the left-hand member of the equation; those to the right make up the right-hand member. Thus, in the equation x + 3 = 7, the left-hand member is x + 3 and the right-hand member is 7. 90004 90015 SOLVING EQUATIONS 90016 90003 Equations may be true or false, just as word sentences may be true or false.The equation: 90004 90003 3 + x = 7 90004 90003 will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this example) is called the solution of the equation. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable and determining the truth or falsity of the result. 90004 90003 Example 1 Determine if the value 3 is a solution of the equation 90004 90003 4x — 2 = 3x + 1 90004 90003 Solution We substitute the value 3 for x in the equation and see if the left-hand member equals the right-hand member.90004 90003 4 (3) — 2 = 3 (3) + 1 90004 90003 12 — 2 = 9 + 1 90004 90003 10 = 10 90004 90003 Ans. 3 is a solution. 90004 90003 The first-degree equations that we consider in this chapter have at most one solution. The solutions to many such equations can be determined by inspection. 90004 90003 Example 2 Find the solution of each equation by inspection. 90004 90003 a. x + 5 = 12 90042 b. 4 · x = -20 90004 90003 Solutions a. 7 is the solution since 7 + 5 = 12. 90042 b. -5 is the solution since 4 (-5) = -20.90004 90015 SOLVING EQUATIONS USING ADDITION AND SUBTRACTION PROPERTIES 90016 90003 In Section 3.1 we solved some simple first-degree equations by inspection. However, the solutions of most equations are not immediately evident by inspection. Hence, we need some mathematical «tools» for solving equations. 90004 90003 EQUIVALENT EQUATIONS 90004 90003 Equivalent equations are equations that have identical solutions. Thus, 90004 90003 3x + 3 = x + 13, 3x = x + 10, 2x = 10, and x = 5 90004 90003 are equivalent equations, because 5 is the only solution of each of them.Notice in the equation 3x + 3 = x + 13, the solution 5 is not evident by inspection but in the equation x = 5, the solution 5 is evident by inspection. In solving any equation, we transform a given equation whose solution may not be obvious to an equivalent equation whose solution is easily noted. 90004 90003 The following property, sometimes called the 90060 addition-subtraction property 90061, is one way that we can generate equivalent equations. 90004 90003 90060 If the same quantity is added to or subtracted from both members of an equation, the resulting equation is equivalent to the original equation.90061 90004 90003 In symbols, 90004 90003 a — b, a + c = b + c, and a — c = b — c 90004 90003 are equivalent equations. 90004 90003 Example 1 Write an equation equivalent to 90004 90003 x + 3 = 7 90004 90003 by subtracting 3 from each member. 90004 90003 Solution Subtracting 3 from each member yields 90004 90003 x + 3 — 3 = 7 — 3 90004 90003 or 90004 90003 x = 4 90004 90003 Notice that x + 3 = 7 and x = 4 are equivalent equations since the solution is the same for both, namely 4.The next example shows how we can generate equivalent equations by first simplifying one or both members of an equation. 90004 90003 Example 2 Write an equation equivalent to 90004 90003 4x- 2-3x = 4 + 6 90004 90003 by combining like terms and then by adding 2 to each member. 90004 90003 Combining like terms yields 90004 90003 x — 2 = 10 90004 90003 Adding 2 to each member yields 90004 90003 x-2 + 2 = 10 + 2 90004 90003 x = 12 90004 90003 To solve an equation, we use the addition-subtraction property to transform a given equation to an equivalent equation of the form x = a, from which we can find the solution by inspection.90004 90003 Example 3 Solve 2x + 1 = x — 2. 90004 90003 We want to obtain an equivalent equation in which all terms containing x are in one member and all terms not containing x are in the other. If we first add -1 to (or subtract 1 from) each member, we get 90004 90003 2x + 1 1 = x — 2 1 90004 90003 2x = x — 3 90004 90003 If we now add -x to (or subtract x from) each member, we get 90004 90003 2x-x = x — 3 — x 90004 90003 x = -3 90004 90003 where the solution -3 is obvious. 90004 90003 The solution of the original equation is the number -3; however, the answer is often displayed in the form of the equation x = -3.90004 90003 Since each equation obtained in the process is equivalent to the original equation, -3 is also a solution of 2x + 1 = x — 2. In the above example, we can check the solution by substituting — 3 for x in the original equation 90004 90003 2 (-3) + 1 = (-3) — 2 90004 90003 -5 = -5 90004 90003 The symmetric property of equality is also helpful in the solution of equations. This property states 90004 90003 If a = b then b = a 90004 90003 This enables us to interchange the members of an equation whenever we please without having to be concerned with any changes of sign.Thus, 90004 90003 If 4 = x + 2 then x + 2 = 4 90004 90003 If x + 3 = 2x — 5 then 2x — 5 = x + 3 90004 90003 If d = rt then rt = d 90004 90003 There may be several different ways to apply the addition property above. Sometimes one method is better than another, and in some cases, the symmetric property of equality is also helpful. 90004 90003 Example 4 Solve 2x = 3x — 9. (1) 90004 90003 Solution If we first add -3x to each member, we get 90004 90003 2x — 3x = 3x — 9 — 3x 90004 90003 -x = -9 90004 90003 where the variable has a negative coefficient.Although we can see by inspection that the solution is 9, because — (9) = -9, we can avoid the negative coefficient by adding -2x and +9 to each member of Equation (1). In this case, we get 90004 90003 2x-2x + 9 = 3x- 9-2x + 9 90004 90003 9 = x 90004 90003 from which the solution 9 is obvious. If we wish, we can write the last equation as x = 9 by the symmetric property of equality. 90004 90015 SOLVING EQUATIONS USING THE DIVISION PROPERTY 90016 90003 Consider the equation 90004 90003 3x = 12 90004 90003 The solution to this equation is 4.Also, note that if we divide each member of the equation by 3, we obtain the equations 90004 90003 90170 90004 90003 whose solution is also 4. In general, we have the following property, which is sometimes called the division property. 90004 90003 90060 If both members of an equation are divided by the same (nonzero) quantity, the resulting equation is equivalent to the original equation. 90061 90004 90003 In symbols, 90004 90003 90181 90004 90003 are equivalent equations. 90004 90003 Example 1 Write an equation equivalent to 90004 90003 -4x = 12 90004 90003 by dividing each member by -4.90004 90003 Solution Dividing both members by -4 yields 90004 90003 90194 90004 90003 In solving equations, we use the above property to produce equivalent equations in which the variable has a coefficient of 1. 90004 90003 Example 2 Solve 3y + 2y = 20. 90004 90003 We first combine like terms to get 90004 90003 5y = 20 90004 90003 Then, dividing each member by 5, we obtain 90004 90003 90207 90004 90003 In the next example, we use the addition-subtraction property and the division property to solve an equation.90004 90003 Example 3 Solve 4x + 7 = x — 2. 90004 90003 Solution First, we add -x and -7 to each member to get 90004 90003 4x + 7 — x — 7 = x — 2 — x — 1 90004 90003 Next, combining like terms yields 90004 90003 3x = -9 90004 90003 Last, we divide each member by 3 to obtain 90004 90003 90224 90004 90015 SOLVING EQUATIONS USING THE MULTIPLICATION PROPERTY 90016 90003 Consider the equation 90004 90003 90231 90004 90003 The solution to this equation is 12. Also, note that if we multiply each member of the equation by 4, we obtain the equations 90004 90003 90236 90004 90003 whose solution is also 12.In general, we have the following property, which is sometimes called the multiplication property. 90004 90003 90060 If both members of an equation are multiplied by the same nonzero quantity, the resulting equation Is equivalent to the original equation. 90061 90004 90003 In symbols, 90004 90003 a = b and a · c = b · c (c ≠ 0) 90004 90003 are equivalent equations. 90004 90003 Example 1 Write an equivalent equation to 90004 90003 90253 90004 90003 by multiplying each member by 6. 90004 90003 Solution Multiplying each member by 6 yields 90004 90003 90260 90004 90003 In solving equations, we use the above property to produce equivalent equations that are free of fractions.90004 90003 Example 2 Solve 90265 90004 90003 Solution First, multiply each member by 5 to get 90004 90003 90270 90004 90003 Now, divide each member by 3, 90004 90003 90275 90004 90003 Example 3 Solve 90278. 90004 90003 Solution First, simplify above the fraction bar to get 90004 90003 90283 90004 90003 Next, multiply each member by 3 to obtain 90004 90003 90288 90004 90003 Last, dividing each member by 5 yields 90004 90003 90293 90004 90015 FURTHER SOLUTIONS OF EQUATIONS 90016 90003 Now we know all the techniques needed to solve most first-degree equations.There is no specific order in which the properties should be applied. Any one or more of the following steps listed on page 102 may be appropriate. 90004 90003 Steps to solve first-degree equations: 90004 90301 90302 Combine like terms in each member of an equation. 90303 90302 Using the addition or subtraction property, write the equation with all terms containing the unknown in one member and all terms not containing the unknown in the other. 90303 90302 Combine like terms in each member.90303 90302 Use the multiplication property to remove fractions. 90303 90302 Use the division property to obtain a coefficient of 1 for the variable. 90303 90312 90003 Example 1 Solve 5x — 7 = 2x — 4x + 14. 90004 90003 Solution First, we combine like terms, 2x — 4x, to yield 90004 90003 5x — 7 = -2x + 14 90004 90003 Next, we add + 2x and +7 to each member and combine like terms to get 90004 90003 5x — 7 + 2x + 7 = -2x + 14 + 2x + 1 90004 90003 7x = 21 90004 90003 Finally, we divide each member by 7 to obtain 90004 90003 90328 90004 90003 In the next example, we simplify above the fraction bar before applying the properties that we have been studying.90004 90003 Example 2 Solve 90333 90004 90003 Solution First, we combine like terms, 4x — 2x, to get 90004 90003 90338 90004 90003 Then we add -3 to each member and simplify 90004 90003 90343 90004 90003 Next, we multiply each member by 3 to obtain 90004 90003 90348 90004 90003 Finally, we divide each member by 2 to get 90004 90003 90353 90004 90015 SOLVING FORMULAS 90016 90003 Equations that involve variables for the measures of two or more physical quantities are called formulas.We can solve for any one of the variables in a formula if the values of the other variables are known. We substitute the known values in the formula and solve for the unknown variable by the methods we used in the preceding sections. 90004 90003 Example 1 In the formula d = rt, find t if d = 24 and r = 3. 90004 90003 Solution We can solve for t by substituting 24 for d and 3 for r. That is, 90004 90003 d = rt 90004 90003 (24) = (3) t 90004 90003 8 = t 90004 90003 It is often necessary to solve formulas or equations in which there is more than one variable for one of the variables in terms of the others.We use the same methods demonstrated in the preceding sections. 90004 90003 Example 2 In the formula d = rt, solve for t in terms of r and d. 90004 90003 Solution We may solve for t in terms of r and d by dividing both members by r to yield 90004 90003 90376 90004 90003 from which, by the symmetric law, 90004 90003 90381 90004 90003 In the above example, we solved for t by applying the division property to generate an equivalent equation. Sometimes, it is necessary to apply more than one such property.90004 90003 Example 3 In the equation ax + b = c, solve for x in terms of a, b and c. 90004 90003 Solution We can solve for x by first adding -b to each member to get 90004 90003 90390 90004 90003 then dividing each member by a, we have 90004 90003 90395 90004 .90000 Solve inequalities with Step-by-Step Math Problem Solver 90001 90002 90003 Enter a polynomial inequality along with the variable to be solved for and click the Solve button. 90004 90003 In chapter 2 we established rules for solving equations using the numbers of arithmetic. Now that we have learned the operations on signed numbers, we will use those same rules to solve equations that involve negative numbers. We will also study techniques for solving and graphing inequalities having one unknown.90004 90007 SOLVING EQUATIONS INVOLVING SIGNED NUMBERS 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to solve equations involving signed numbers. 90004 90003 90014 Example 1 90015 Solve for x and check: x + 5 = 3 90004 90003 Solution 90004 90003 Using the same procedures learned in chapter 2, we subtract 5 from each side of the equation obtaining 90004 90003 90022 90004 90003 90014 Example 2 90015 Solve for x and check: — 3x = 12 90004 90003 Solution 90004 90003 Dividing each side by -3, we obtain 90004 90003 90033 90004 90035 90036 90037 Always check in the original equation.90038 90039 90040 90003 90042 90004 90035 90036 90037 Another way of solving the equation 90047 3x — 4 = 7x + 8 90047 would be to first subtract 3x from both sides obtaining 90047 -4 = 4x + 8, 90047 then subtract 8 from both sides and get 90047 -12 = 4x . 90047 Now divide both sides by 4 obtaining 90047 — 3 = x or x = — 3. 90038 90039 90040 90003 90058 90004 90035 90036 90037 First remove parentheses. Then follow the procedure learned in chapter 2. 90038 90039 90040 90007 LITERAL EQUATIONS 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to: 90004 90072 90073 Identify a literal equation.90074 90073 Apply previously learned rules to solve literal equations. 90074 90077 90003 An equation having more than one letter is sometimes called a 90014 literal equation 90015. It is occasionally necessary to solve such an equation for one of the letters in terms of the others. The step-by-step procedure discussed and used in chapter 2 is still valid after any grouping symbols are removed. 90004 90003 90014 Example 1 90015 Solve for c: 3 (x + c) — 4y = 2x — 5c 90004 90003 Solution 90004 90003 First remove parentheses.90004 90003 90091 90004 90003 At this point we note that since we are solving for c, we want to obtain c on one side and all other terms on the other side of the equation. Thus we obtain 90004 90003 90096 90004 90003 90099 90004 90035 90036 90037 Remember, abx is the same as 1abx. 90047 We divide by the coefficient of x, which in this case is ab. 90038 90039 90040 90003 90109 90004 90035 90036 90037 Solve the equation 2x + 2y — 9x + 9a by first subtracting 2.v from both sides. Compare the solution with that obtained in the example. 90038 90039 90040 90003 Sometimes the form of an answer can be changed. In this example we could multiply both numerator and denominator of the answer by (- l) (this does not change the value of the answer) and obtain 90004 90003 90120 90004 90003 The advantage of this last expression over the first is that there are not so many negative signs in the answer. 90004 90035 90036 90037 Multiplying numerator and denominator of a fraction by the same number is a use of the fundamental principle of fractions.90038 90039 90040 90130 90003 The most commonly used literal expressions are formulas from geometry, physics, business, electronics, and so forth. 90004 90003 90014 Example 4 90015 90136 is the formula for the area of a trapezoid. Solve for c. 90004 90003 90139 90004 90003 90142 90004 90035 90036 90037 A trapezoid has two parallel sides and two nonparallel sides. The parallel sides are called bases. 90047 Removing parentheses does not mean to merely erase them. We must multiply each term inside the parentheses by the factor preceding the parentheses.90047 Changing the form of an answer is not necessary, but you should be able to recognize when you have a correct answer even though the form is not the same. 90038 90039 90040 90003 90014 Example 5 90015 90155 is a formula giving interest (I) earned for a period of D days when the principal (p) and the yearly rate (r) are known. Find the yearly rate when the amount of interest, the principal, and the number of days are all known. 90004 90003 Solution 90004 90003 The problem requires solving 90155 for r.90004 90003 90163 90004 90003 Notice in this example that r was left on the right side and thus the computation was simpler. We can rewrite the answer another way if we wish. 90004 90003 90168 90004 90007 GRAPHING INEQUALITIES 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to: 90004 90072 90073 Use the inequality symbol to represent the relative positions of two numbers on the number line. 90074 90073 Graph inequalities on the number line.90074 90077 90003 We have already discussed the set of 90014 rational numbers 90015 as those that can be expressed as a ratio of two integers. There is also a set of numbers, called the 90014 irrational numbers, 90015, that can not be expressed as the ratio of integers. This set includes such numbers as 90187 and so on. The set composed of rational and irrational numbers is called the 90014 real numbers. 90015 90004 90003 Given any two real numbers a and b, it is always possible to state that 90192 Many times we are only interested in whether or not two numbers are equal, but there are situations where we also wish to represent the relative size of numbers that are not equal.90004 90003 The symbols are 90014 inequality symbols 90015 or 90014 order relations 90015 and are used to show the relative sizes of the values of two numbers. We usually read the symbol as «greater than.» For instance, a>

b is read as «a is greater than b.» Notice that we have stated that we usually read a 90004 90003 90014 a 90015 90004 90003 90205 90047 90004 90035 90036 90037 What positive number can be added to 2 to give 5? 90038 90039 90040 90003 90215 90047 90004 90003 In simpler words this definition states that a is less than b if we must add something to a to get b.Of course, the «something» must be positive. 90004 90003 If you think of the number line, you know that adding a positive number is equivalent to moving to the right on the number line. This gives rise to the following alternative definition, which may be easier to visualize. 90004 90003 90014 Example 1 90015 3 90004 90003 90227 90047 90004 90035 90036 90037 We could also write 6> 3. 90038 90039 90040 90003 90014 Example 2 90015 — 4 90004 90003 90241 90047 90004 90035 90036 90037 We could also write 0> — 4.90038 90039 90040 90003 90014 Example 3 90015 4> — 2, because 4 is to the right of -2 on the number line. 90004 90003 90255 90047 90004 90003 90014 Example 4 90015 — 6 90004 90003 90263 90047 90004 90003 The mathematical statement x 90004 90035 90036 90037 Do you see why finding the largest number less than 3 is impossible? 90038 90039 90040 90003 As a matter of fact, to name the number x that is the largest number less than 3 is an impossible task. It can be indicated on the number line, however.To do this we need a symbol to represent the meaning of a statement such as x 90004 90003 90014 The symbols (and) used on the number line indicate that the endpoint is not included in the set. 90015 90004 90003 90014 Example 5 90015 Graph x 90004 90003 Solution 90004 90003 90287 90047 90004 90003 Note that the graph has an arrow indicating that the line continues without end to the left. 90004 90035 90036 90037 This graph represents every real number less than 3. 90038 90039 90040 90003 90014 Example 6 90015 Graph x> 4 on the number line.90004 90003 Solution 90004 90003 90305 90047 90004 90035 90036 90037 This graph represents every real number greater than 4. 90038 90039 90040 90003 90014 Example 7 90015 Graph x> -5 on the number line. 90004 90003 Solution 90004 90003 90321 90047 90004 90035 90036 90037 This graph represents every real number greater than -5. 90038 90039 90040 90003 90014 Example 8 90015 Make a number line graph showing that x> — 1 and x 90004 90003 Solution 90004 90003 90337 90047 The statement x> — 1 and x 90004 90035 90036 90037 This graph represents all real numbers that are between — 1 and 5.90038 90039 90040 90003 90014 Example 9 90015 Graph — 3 90004 90003 Solution 90004 90003 90353 90004 90003 If we wish to include the endpoint in the set, we use a different symbol, 90356:. We read these symbols as «equal to or less than» and «equal to or greater than.» 90004 90003 Example 10 x>; 4 indicates the number 4 and all real numbers to the right of 4 on the number line. 90004 90003 90014 The symbols [and] used on the number line indicate that the endpoint is included in the set.90015 90004 90035 90036 90037 You will find this use of parentheses and brackets to be consistent with their use in future courses in mathematics. 90038 90039 90040 90003 90371 90004 90035 90036 90037 This graph represents the number 1 and all real numbers greater than 1. 90038 90039 90040 90003 90380 90004 90035 90036 90037 This graph represents the number 1 and all real numbers less than or equal to — 3. 90038 90039 90040 90003 90014 Example 13 90015 Write an algebraic statement represented by the following graph.90004 90003 90393 90004 90003 90014 Example 14 90015 Write an algebraic statement for the following graph. 90004 90003 90400 90004 90035 90036 90037 This graph represents all real numbers between -4 and 5 90405 including 90406 -4 and 5. 90038 90039 90040 90003 90014 Example 15 90015 Write an algebraic statement for the following graph. 90004 90003 90415 90004 90035 90036 90037 This graph includes 4 but 90405 not 90406 -2. 90038 90039 90040 90003 90014 Example 16 90015 Graph 90428 on the number line.90004 90003 Solution 90004 90003 This example presents a small problem. How can we indicate 90428 on the number line? If we estimate the point, then another person might misread the statement. Could you possibly tell if the point represents 90428 or maybe 90435? Since the purpose of a graph is to clarify, 90405 always 90406 label the endpoint. 90004 90003 90440 90004 90035 90036 90037 A graph is used to communicate a statement. You should always name the zero point to show direction and also the endpoint or points to be exact.90038 90039 90040 90007 SOLVING INEQUALITIES 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to solve inequalities involving one unknown. 90004 90003 The solutions for inequalities generally involve the same basic rules as equations. There is one exception, which we will soon discover. The first rule, however, is similar to that used in solving equations. 90004 90003 90014 90405 If the same quantity is added to each side of an inequality 90406, the results are unequal in the same order.90015 90004 90003 90014 Example 1 90015 If 5 90004 90003 90014 Example 2 90015 If 7 90004 90003 We can use this rule to solve certain inequalities. 90004 90003 90014 Example 3 90015 Solve for x: x + 6 90004 90003 Solution 90004 90003 If we add -6 to each side, we obtain 90004 90003 90481 90004 90003 Graphing this solution on the number line, we have 90004 90003 90486 90004 90003 90489 90004 90035 90036 90037 Note that the procedure is the same as in solving equations.90038 90039 90040 90003 We will now use the addition rule to illustrate an important concept concerning multiplication or division of inequalities. 90004 90003 Suppose x> a. 90004 90003 Now add — x to both sides by the addition rule. 90004 90003 90504 90004 90035 90036 90037 Remember, adding the same quantity to both sides of an inequality does not change its direction. 90038 90039 90040 90003 Now add -a to both sides. 90004 90003 90515 90004 90003 The last statement, — a> -x, can be rewritten as — x <-a.Therefore we can say, "If x> a, then — x 90004 90003 If an inequality is multiplied or divided by a 90014 negative 90015 number, the results will be unequal in the 90014 opposite 90015 order. 90004 90035 90036 90037 For example: If 5> 3 then -5 90038 90039 90040 90003 90014 Example 5 90015 Solve for x and graph the solution: -2x> 6 90004 90003 Solution 90004 90003 To obtain x on the left side we must divide each term by — 2. Notice that since we are dividing by a negative number, we must change the direction of the inequality.90004 90003 90540 90004 90035 90036 90037 Notice that as soon as we divide by a negative quantity, we must change the direction of the inequality. 90038 90039 90040 90003 Take special note of this fact. Each time you divide or multiply by a negative number, you must change the direction of the inequality symbol. This is the only difference between solving equations and solving inequalities. 90004 90035 90036 90037 When we multiply or divide by a positive number, there is no change.When we multiply or divide by a negative number, the direction of the inequality changes. Be careful-this is the source of many errors. 90038 90039 90040 90003 Once we have removed parentheses and have only individual terms in an expression, the procedure for finding a solution is almost like that in chapter 2. 90004 90003 Let us now review the step-by-step method from chapter 2 and note the difference when solving inequalities. 90004 90003 90014 First 90015 Eliminate fractions by multiplying all terms by the least common denominator of all fractions.(No change when we are multiplying by a positive number.) 90047 90014 Second 90015 Simplify by combining like terms on each side of the inequality. (No change) 90047 90014 Third 90015 Add or subtract quantities to obtain the unknown on one side and the numbers on the other. (No change) 90047 90014 Fourth 90015 Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed.(This is the important difference between equations and inequalities.) 90004 90035 90036 90037 The only possible difference is in the final step. 90038 90039 90040 90003 90580 90004 90035 90036 90037 What must be done when dividing by a negative number? 90038 90039 90040 90003 90589 90004 90035 90036 90037 Dont forget to label the endpoint. 90038 90039 90040 90007 SUMMARY 90008 90009 Key Words 90010 90601 90073 A 90014 literal equation 90015 is an equation involving more than one letter.90074 90073 The symbols are 90014 inequality symbols 90015 or 90014 order relations 90015. 90074 90073 a a is to the left of b on the real number line. 90074 90073 The double symbols 90356: indicate that the 90014 endpoints are included in the solution set 90015. 90074 90619 90009 Procedures 90010 90601 90073 To solve a literal equation for one letter in terms of the others follow the same steps as in chapter 2. 90074 90073 To solve an inequality use the following steps: 90047 90014 Step 1 90015 Eliminate fractions by multiplying all terms by the least common denominator of all fractions.90047 90014 Step 2 90015 Simplify by combining like terms on each side of the inequality. 90047 90014 Step 3 90015 Add or subtract quantities to obtain the unknown on one side and the numbers on the other. 90047 90014 Step 4 90015 Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed. 90047 90014 Step 5 90015 Check your answer. 90047 90074 90619 .90000 Ex 3.6, 1 (i) and (ii) 90001 90002 Last updated at May 2, 2020 by Teachoo 90003 90004 90002 Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check — Linear Equations in 2 Variables — Class 10 90003 90007 90002 90009 90003 90011 90012 90013 90002 90015 90016 90003 90002 90003 90002 90021 90022 90023 90024 90003 90002 90027 90016 90003 90030 90002 Transcript 90003 90033 Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (I) 1 / 2𝑥 + 1 / 3𝑦 = 2 1 / 3𝑥 + 1 / 2𝑦 = 13/6 1 / 2𝑥 + 1 / 3𝑦 = 2 1 / 3𝑥 + 1 / 2𝑦 = 13/6 Let 1 / 𝑥 = u 1 / 𝑦 = v So, our equations become 1/2 u + 1/3 v = 2 (3𝑢 + 2𝑣) / (2 × 3) = 2 3u + 2v = 12 1/3 u + 1/2 v = 13/6 (2𝑢 + 3𝑣) / (2 × 3) = 13/6 2u + 3v = 13 Our equations are 3u + 2v = 12 … (3) 2u + 3v = 13 … (4) From (3) 3u + 2v = 12 3u = 12 — 2v u = (12 — 2𝑣) / 3 Putting value of u in (4) 2u + 3v = 13 2 ((12 -2𝑣) / 3) + 3v = 13 Multiplying both sides by 3 3 × 2 ((12 — 2𝑣) / 3) + 3 × 3v = 3 × 13 2 (12 — 2v) + 9v = 39 24 — 4v + 9v = 39 — 4v + 9v = 39 — 24 5v = 15 v = 15/5 v = 3 Putting v = 3 in (3) 3u + 2v = 12 3u + 2 (3) = 12 3u + 6 = 12 3u = 12 — 6 3u = 6 u = 6/3 u = 2 Hence, v = 3, u = 2 But we have to find x & y We know that u = 𝟏 / 𝒙 2 = 1 / 𝑥 x = 𝟏 / 𝟐 v = 𝟏 / 𝒚 3 = 1 / 𝑦 y = 𝟏 / 𝟑 So, x = 𝟏 / 𝟐, y = 𝟏 / 𝟑 is the solution of the given equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (Ii) 2 / √𝑥 + 3 / √𝑦 = 2 4 / √𝑥 — 9 / √𝑦 = -1 2 / √𝑥 + 3 / √𝑦 = 2 4 / √𝑥 — 9 / √𝑦 = -1 Let 1 / √𝑥 = u & 1 / √𝑦 = v So, our equations become 2u + 3v = 2 4u — 9v = -1 Our equations 2u + 3v = 2 … (3) 4u — 9v = -1 … (4) From (3) 2u + 3v = 2 2u = 2 — 3v u = (2 — 3𝑣) / 2 Putting value of u in (4) 4u — 9v = — 1 4 ((2 — 3𝑣) / 2) — 9v = -1 2 (2 — 3v) — 9v = -1 4 — 6v — 9v = -1 — 6v — 9v = -1 — 4 -15v = — 5 v = (-5) / (- 15) v = 𝟏 / 𝟑 Putting v = 1/3 in (3) 2u + 3v = 2 2u + 3 (1/3) = 2 2u + 1 = 2 2u = 2 — 1 u = 𝟏 / 𝟐 Hence, u = 1/2 & v = 1/3 But, we need to find x & y u = 𝟏 / √𝒙 1/2 = 1 / √𝑥 √𝑥 = 2 Squaring both sides (√𝑥) 2 = (2) 2 x = 4 v = 𝟏 / √𝒚 1/3 = 1 / √𝑦 √𝑦 = 3 Squaring both sides (√𝑦) 2 = (3) 2 y = 9 Therefore, x = 4, y = 9 is the solution of the given equation 90003 Show More .90000 Solve linear and quadratic equations with Step-by-Step Math Problem Solver 90001 90002 90003 Enter an equation along with the variable you wish to solve it for and click the Solve button. 90004 90003 The solution of equations is the central theme of algebra. In this chapter we will study some techniques for solving equations having one variable. To accomplish this we will use the skills learned while manipulating the numbers and symbols of algebra as well as the operations on whole numbers, decimals, and fractions that you learned in arithmetics.90004 90007 CONDITIONAL AND EQUIVALENT EQUATIONS 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to: 90004 90013 90014 Classify an equation as conditional or an identity. 90015 90014 Solve simple equations mentally. 90015 90014 Determine if certain equations are equivalent. 90015 90020 90003 90022 An equation is a statement in symbols that two number expressions are equal. 90023 90004 90003 Equations can be classified in two main types: 90004 90003 1.An 90022 identity 90023 is true for all values of the literal and arithmetical numbers in it. 90004 90003 90022 Example 1 90023 5 x 4 = 20 is an identity. 90004 90003 90022 Example 2 90023 2 + 3 = 5 is an identity. 90004 90003 90022 Example 3 90023 2x + 3x = 5x is an identity since any value substituted for x will yield an equality. 90004 90003 2. A 90022 conditional equation 90023 is true for only certain values of the literal numbers in it. 90004 90003 90022 Example 4 90023 x + 3 = 9 is true only if the literal number x = 6.90004 90003 90022 Example 5 90023 3x — 4 = 11 is true only if x = 5. 90004 90003 The literal numbers in an equation are sometimes referred to as 90022 variables 90023. 90004 90003 Finding the values that make a conditional equation true is one of the main objectives of this text. 90004 90003 A 90022 solution 90023 or 90022 root 90023 of an equation is the value of the variable or variables that make the equation a true statement. 90004 90003 The solution or root is said to 90068 satisfy the equation 90069.90004 90003 90068 Solving an equation 90069 means finding the solution or root. 90004 90003 90076 90004 90003 Many equations can be solved mentally. Ability to solve an equation mentally will depend on the ability to manipulate the numbers of arithmetic. The better you know the facts of multiplication and addition, the more adept you will be at mentally solving equations. 90004 90003 90022 Example 6 90023 Solve for x: x + 3 = 7 90004 90003 Solution 90004 90003 To have a true statement we need a value for x that, when added to 3, will yield 7.Our knowledge of arithmetic indicates that 4 is the needed value. Therefore the solution to the equation is x = 4. 90004 90088 90089 90090 What number added to 3 equals 7? 90091 90092 90093 90003 90022 Example 7 90023 Solve for x: x — 5 = 3 90004 90003 Solution 90004 90003 What number do we subtract 5 from to obtain 3? Again our experience with arithmetic tells us that 8 — 5 = 3. Therefore the solution is x = 8. 90004 90003 90022 Example 8 90023 Solve for x: 3x = 15 90004 90003 Solution 90004 90003 What number must be multiplied by 3 to obtain 15? Our answer is x = 5.90004 90003 90111 90004 90003 Solution 90004 90003 What number do we divide 2 by to obtain 7? Our answer is 14. 90004 90003 90022 Example 10 90023 Solve for x: 2x — 1 = 5 90004 90003 Solution 90004 90003 We would subtract 1 from 6 to obtain 5. Thus 2x = 6. Then x = 3. 90004 90003 Regardless of how an equation is solved, the solution should always be checked for correctness. 90004 90003 90022 Example 11 90023 A student solved the equation 5x — 3 = 4x + 2 and found an answer of x = 6.Was this right or wrong? 90004 90003 Solution 90004 90003 Does x = 6 satisfy the equation 5x — 3 = 4x + 2? To check we substitute 6 for x in the equation to see if we obtain a true statement. 90004 90003 90136 90004 90003 This is not a true statement, so the answer x = 6 is wrong. 90004 90003 Another student solved the same equation and found x = 5. 90004 90003 90143 90004 90003 This is a true statement, so x = 5 is correct. 90004 90088 90089 90090 Many students think that when they have found the solution to an equation, the problem is finished.90068 Not so! 90069 The final step should always be to check the solution. 90091 90092 90093 90003 Not all equations can be solved mentally. We now wish to introduce an idea that is a step toward an orderly process for solving equations. 90004 90088 90089 90090 Is x = 3 a solution of x — 1 = 2? 90160 Is x = 3 a solution of 2x + I = 7? 90160 What can be said about the equations x — 1 = 2 and 2x + 1 = 7? 90091 90092 90093 90003 Two equations are 90022 equivalent 90023 if they have the same solution or solutions 90004 90003 90022 Example 12 90023 3x = 6 and 2x + 1 = 5 are equivalent because in both cases x = 2 is a solution.90004 90003 Techniques for solving equations will involve processes for changing an equation to an equivalent equation. If a complicated equation such as 2x — 4 + 3x = 7x + 2 — 4x can be changed to a simple equation x = 3, and the equation x = 3 is equivalent to the original equation, then we have solved the equation. 90004 90003 Two questions now become very important. 90004 90013 90014 Are two equations equivalent? 90015 90014 How can we change an equation to another equation that is equivalent to it? 90015 90020 90003 The answer to the first question is found by using the substitution principle.90004 90003 90022 Example 13 90023 Are 5x + 2 = 6x — 1 and x = 3 equivalent equations? 90004 90003 Solution 90004 90003 90192 90004 90003 The answer to the second question involves the techniques for solving equations that will be discussed in the next few sections. 90004 90088 90089 90090 To use the substitution principle correctly we must substitute the numeral 3 for x wherever x appears in the equation. 90091 90092 90093 90003 90203 90004 90007 THE DIVISION RULE 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to: 90004 90013 90014 Use the division rule to solve equations.90015 90014 Solve some basic applied problems whose solutions involve using the division rule. 90015 90020 90003 As mentioned earlier, we wish to present an orderly procedure for solving equations. This procedure will involve the four basic operations, the first of which is presented in this section. 90004 90003 90022 If each term of an equation is 90068 divided 90069 by the same nonzero number, the resulting equation is 90068 equivalent 90069 to the original equation. 90023 90004 90003 To prepare to use the division rule for solving equations we must make note of the following process: 90004 90003 90230 90004 90003 (We usually write 1x as x with the coefficient 1 understood.) 90004 90003 90022 Example 1 90023 Solve for x: 3x = 10 90004 90003 Solution 90004 90003 Our goal is to obtain x = some number. The division rule allows us to divide each term of 3x = 10 by the same number, and our goal of finding a value of x would indicate that we divide by 3. This would give us a coefficient of 1 for x. 90004 90003 90243 90004 90003 Check: 3x = 10 and x = 90246 these equivalent equations? 90004 90003 We substitute 90246 for x in the first equation obtaining 90004 90251 90003 The equations are equivalent, so the solution is correct.90004 90003 90251 90004 90003 90258 90004 90003 90022 Example 2 90023 Solve for x: 5x = 20 90004 90003 Solution 90004 90003 90267 90004 90088 90089 90090 Notice that the division rule does not allow us to divide by zero. Since dividing by zero is not allowed in mathematics, expressions such as 90272 are meaningless. 90091 90092 90093 90003 90022 Example 3 90023 Solve for x: 8x = 4 90004 90003 Solution 90004 90003 90283 90004 90088 90089 90090 Errors are sometimes made in very simple situations.Do not glance at this problem and arrive at x = 2! 90160 Note that the division rule allows us to divide each term of an equation by any nonzero number and the resulting equation is equivalent to the original equation. 90160 Therefore we could divide each side of the equation by 5 and obtain 90290, which is equivalent to the original equation. 90160 Dividing by 5 does not help find the solution however. What number should we divide by to find the solution? 90091 90092 90093 90003 90022 Example 4 90023 Solve for x: 0.5x = 6 90004 90003 Solution 90004 90003 90302 90004 90003 90305 90004 90003 90022 Example 6 90023 The formula for finding the circumference (C) of a circle is C = 2πr, where π represents the radius of the circle and it is approximately 3.14. Find the radius of a circle if the circumference is measured to be 40.72 cm. Give the answer correct to two decimal places. 90004 90003 Solution 90004 90003 To solve a problem involving a formula we first use the substitution principle. 90004 90003 90316 90004 90088 90089 90090 Circumference means «distance around.»It is the perimeter of a circle. 90160 The radius is the distance from the center to the circle. 90091 90092 90093 90003 90326 90004 90007 THE SUBTRACTION RULE 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to use the subtraction rule to solve equations. 90004 90003 The second step toward an orderly procedure for solving equations will be discussed in this section. You will use your knowledge of like terms from chapter l as well as the techniques from section 90022 THE DIVISION RULE 90023.Notice how new ideas in algebra build on previous knowledge. 90004 90003 90022 If the same quantity is 90068 subtracted 90069 from both sides of an equation, the resulting equation will be 90068 equivalent 90069 to the original equation. 90023 90004 90003 90022 Example 1 90023 Solve for x if x + 7 = 12. 90004 90003 Solution 90004 90003 Even though this equation can easily be solved mentally, we wish to illustrate the subtraction rule. We should think in this manner: 90004 90003 «I wish to solve for x so I need x by itself on one side of the equation.But I have x + 7. So if I subtract 7 from x + 7, I will have x alone on the left side. «(Remember that a quantity subtracted from itself gives zero.) But if we subtract 7 from one side of the equation, the rule requires us to subtract 7 from the other side as well. So we proceed as follows: 90004 90003 90357 90004 90088 90089 90090 Note that x + 0 may be written simply as x since zero added to any quantity equals the quantity itself. 90091 90092 90093 90003 90022 Example 2 90023 Solve for x: 5x = 4x + 3 90004 90003 Solution 90004 90003 Here our thinking should proceed in this manner.»I wish to obtain all unknown quantities on one side of the equation and all numbers of arithmetic on the other so I have an equation of the form x = some number. I thus need to subtract Ax from both sides.» 90004 90003 90374 90004 90088 90089 90090 Our goal is to arrive at x = some number. 90160 Remember that checking your solution is an important step in solving equations. 90091 90092 90093 90003 90022 Example 3 90023 Solve for x: 3x + 6 = 2x + 11 90004 90003 Here we have a more involved task.First subtract 6 from both sides. 90004 90003 90374 90004 90003 Now we must eliminate 2x on the right side by subtracting 2x from both sides. 90004 90003 90395 90004 90003 We now look at a solution that requires the use of both the subtraction rule and the division rule. 90004 90088 90089 90090 Note that instead of first subtracting 6 we could just as well first subtract 2x from both sides obtaining 90160 3x — 2x + 6 = 2x — 2x + 11 90160 x + 6 = 11. 90160 Then subtracting 6 from both sides we have 90160 x + 6 — 6 = 11 — 6 90160 x = 5.90003 Keep in mind that our goal is x = some number. 90004 90091 90092 90093 90003 90022 Example 4 90023 Solve for x: 3x + 2 = 17 90004 90003 Solution 90004 90003 We first use the subtraction rule to subtract 2 from both sides obtaining 90004 90003 90421 90004 90003 Then we use the division rule to obtain 90004 90003 90426 90004 90003 90022 Example 5 90023 Solve for x: 7x + 1 = 5x + 9 90004 90003 Solution 90004 90003 We first use the subtraction rule. 90004 90003 90437 90004 90003 Then the division rule gives us 90004 90003 90442 90004 90003 90022 Example 6 90023 The perimeter (P) of a rectangle is found by using the formula P = 2l + 2w, where l stands for the length and w stands for the width.If the perimeter of a rectangle is 54 cm and the length is 15 cm, what is the width? 90004 90003 Solution 90004 90003 90451 90004 90003 90454 90004 90088 90089 90090 Perimeter is the distance around. Do you see why the formula is P = 2l + 2w? 90091 90092 90093 90007 THE ADDITION RULE 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to use the addition rule to solve equations. 90004 90003 We now proceed to the next operation in our goal of developing an orderly procedure for solving equations.Once again, we will rely on previous knowledge. 90004 90003 90022 If the same quantity is 90068 added 90069 to both sides of an equation, the resulting equation will be 90068 equivalent 90069 to the original equation. 90023 90004 90003 90022 Example 1 90023 Solve for x if x — 7 = 2. 90004 90003 Solution 90004 90003 As always, in solving an equation we wish to arrive at the form of «x = some number.» We observe that 7 has been subtracted from x, so to obtain x alone on the left side of the equation, we add 7 to both sides.90004 90003 90487 90004 90088 90089 90090 Remember to always check your solution. 90091 90092 90093 90003 90022 Example 2 90023 Solve for x: 2x — 3 = 6 90004 90003 Solution 90004 90003 Keeping in mind our goal of obtaining x alone, we observe that since 3 has been subtracted from 2x, we add 3 to both sides of the equation. 90004 90003 90504 90004 90003 Now we must use the division rule. 90004 90003 90509 90004 90088 90089 90090 Why do we add 3 to both sides? 90160 Note that in the example just using the addition rule does not solve the problem.90091 90092 90093 90003 90022 Example 3 90023 Solve for x: 3x — 4 = 11 90004 90003 Solution 90004 90003 We first use the addition rule. 90004 90003 90527 90004 90003 Then using the division rule, we obtain 90004 90003 90532 90004 90088 90089 90090 Here again, we needed to use both the addition rule and the division rule to solve the equation. 90091 90092 90093 90003 90022 Example 4 90023 Solve for x: 5x = 14 — 2x 90004 90003 Solution 90004 90003 Here our goal of obtaining x alone on one side would suggest we eliminate the 2x on the right, so we add 2x to both sides of the equation.90004 90003 90549 90004 90003 We next apply the division rule. 90004 90003 90554 90004 90088 90089 90090 Here again, we needed to use both the addition rule and the division rule to solve the equation. 90160 Note that we check by always substituting the solution in the original equation. 90091 90092 90093 90003 90022 Example 5 90023 Solve for x: 3x — 2 = 8 — 2x 90004 90003 Solution 90004 90003 Here our task is more involved. We must think of eliminating the number 2 from the left side of the equation and also the lx from the right side to obtain x alone on one side.We may do either of these first. If we choose to first add 2x to both sides, we obtain 90004 90003 90572 90004 90003 We now add 2 to both sides. 90004 90003 90577 90004 90003 Finally the division rule gives 90004 90003 90582 90004 90088 90089 90090 Could we first add 2 to both sides? Try it! 90091 90092 90093 90007 THE MULTIPLICATION RULE 90008 90009 OBJECTIVES 90010 90003 Upon completing this section you should be able to: 90004 90013 90014 Use the multiplication rule to solve equations.90015 90014 Solve proportions. 90015 90014 Solve some basic applied problems using the multiplication rule. 90015 90020 90003 We now come to the last of the four basic operations in developing our procedure for solving equations. We will also introduce ratio and proportion and use the multiplication rule to solve proportions. 90004 90003 90022 If each term of an equation is 90068 multiplied 90069 by the same nonzero number, the resulting equation is 90068 equivalent 90069 to the original equation.90023 90004 90003 In elementary arithmetic some of the most difficult operations are those involving fractions. The multiplication rule allows us to avoid these operations when solving an equation involving fractions by finding an equivalent equation that contains only whole numbers. 90004 90003 Remember that when we multiply a whole number by a fraction, we use the rule 90617 90004 90003 90620 90004 90003 We are now ready to solve an equation involving fractions. 90004 90088 90089 90090 Note that in each case only the numerator of the fraction is multiplied by the whole number.90091 90092 90093 90003 90022 Example 4 90023 90633 90004 90003 Solution 90004 90003 Keep in mind that we wish to obtain x alone on one side of the equation. We also would like to obtain an equation in whole numbers that is equivalent to the given equation. To eliminate the fraction in the equation we need to multiply by a number that is divisible by the denominator 3. We thus use the multiplication rule and multiply each term of the equation by 3. 90004 90003 90640 90004 90003 We now have an equivalent equation that contains only whole numbers.Using the division rule, we obtain 90004 90003 90645 90004 90088 90089 90090 To eliminate the fraction we need to multiply by a number that is divisible by the denominator. 90160 In the example we need to multiply by a number that is divided by 3. 90160 We could have multiplied both sides by 6, 9, 12, and so on, but the equation is simpler and easier to work with if wc use the smallest multiple. 90091 90092 90093 90003 90022 Example 5 90023 90658 90004 90003 Solution 90004 90003 90663 90004 90088 90089 90090 See if you obtain the same solution by multiplying each side of the original equation by 16.90160 Always check in the original equation. 90091 90092 90093 90003 90022 Example 6 90023 90675 90004 90003 Solution 90004 90003 Here our task is the same but a little more complex. We have two fractions to eliminate. We must multiply each term of the equation by a number that is divisible by both 3 and 5. It is best to use the least of such numbers, which you will recall is the 90022 least common multiple 90023. We will therefore multiply by 15. 90004 90003 90684 90004 90088 90089 90090 In arithmetic you may have referred to the least common multiple as the «lowest common denominator.»90091 90092 90093 90003 90022 Example 7 90023 90695 90004 90003 Solution 90004 90003 The least common multiple for 8 and 2 is 8, so we multiply each term of the equation by 8. 90004 90003 90702 90004 90003 We now use the subtraction rule. 90004 90003 90707 90004 90003 Finally the division rule gives us 90004 90003 90712 90004 90088 90089 90090 Before multiplying, change any mixed numbers to improper fractions. In this example change 90717. 90160 Remember that 90068 each 90069 term must be multiplied by 8.90160 Note that in this example we used three rules to find the solution. 90091 90092 90093 90003 Solving simple equations by multiplying both sides by the same number occurs frequently in the study of ratio and proportion. 90004 90003 A 90022 ratio 90023 is the quotient of two numbers. 90004 90003 The ratio of a number x to a number y can be written as x: y or 90732. In general, the fractional form is more meaningful and useful. Thus, we will write the ratio of 3 to 4 as 90733. 90004 90003 A 90022 proportion 90023 is a statement that two ratios are equal.90004 90003 90022 Example 8 90023 90742 90004 90003 Solution 90004 90003 We need to find a value of x such that the ratio of x to 15 is equal to the ratio of 2 to 5. 90004 90003 Multiplying each side of the equation by 15, we obtain 90004 90003 90751 90004 .