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This is one of our many quizzes on Solving absolute value equations. Click on this to access more such competitive quizzes on Linear Inequalities in One Variable.

Solving absolute value equations Quiz 2

mathematics

 Solve $\left| {4x - 5} \right| = \left| {3x - 5} \right|$and choose the correct option.

 Apply the property $\left| a \right| = \left| b \right|$ which has the following four solutions $a = b$, $a =  - b$ , $ - a =  - b$ , $ - a = b$.

 So on applying the property $\left| a \right| = \left| b \right|$ we have the following four solutions and they are: $a = b$, $a =  - b$ , $ - a =  - b$ , $ - a = b$. On solving the equation given $4x - 5 = 3x - 5,4x - 5 =  - (3x - 5), - (4x - 5) =  - (3x - 5), - (4x - 5) = 3x - 5$. Here we get two equations only as equation one and equation three are the same and equation two and equation four are the same. So let us take equation one and equation two. $ \\  4x - 5 = 3x - 5 \\   \Rightarrow 4x - 3x - 5 + 5 = 0 \\   \Rightarrow x = 0 \\  $ On solving equation two we get, $ \\ 4x - 5 =  - (3x - 5) \\   \Rightarrow 4x - 5 =  - 3x + 5 \\   \Rightarrow 4x - 5 + 3x - 5 = 0 \\   \Rightarrow 7x = 10 \\   \Rightarrow x = \frac{{10}}{7} \\  $

 State the correct answer for the question $\left| {\dfrac{{2x - 1}}{3}} \right| = \frac{5}{6}$.

 Apply the absolute modulus property of $\left| { \pm a} \right| = a$.

 So firstly applying the property of modulus that is $\left| { \pm a} \right| = a$.  On applying this property we will get two equations that is :  $ \dfrac{{2x - 1}}{3} =  - \dfrac{5}{6}$ and $\dfrac{{2x - 1}}{3} = \dfrac{5}{6} $on solving equation one we get, $ \\  \dfrac{{2x - 1}}{3} =  - \dfrac{5}{6} \\   \Rightarrow 12x - 6 =  - 15 \\   \Rightarrow 12x =  - 15 + 6 \\   \Rightarrow 12x =  - 9 \\   \Rightarrow x = \dfrac{{ - 9}}{{12}} \\   \Rightarrow x =  - \dfrac{3}{4} \\  $ On solving equation two we get, $ \\ \dfrac{{2x - 1}}{3} = \dfrac{5}{6} \\   \Rightarrow 12x - 6 = 15 \\   \Rightarrow 12x = 15 + 6 \\   \Rightarrow 12x = 21 \\   \Rightarrow x = \dfrac{{21}}{{12}} \\   \Rightarrow x = \dfrac{7}{4} \\  $

 Solve the question $2\left| {2x - 6} \right| = 6$ and tick the correct answer.

 First simplify the modulus part such that you have only the modulus part on the left hand side. Do not multiply $2$ inside the modulus.

 We will simplify the question first. On simplifying we get,  $ \\  \left| {2x - 6} \right| = \dfrac{6}{2} \\   \Rightarrow \left| {2x - 6} \right| = 3 \\  $ The applying the property$\left| { \pm a} \right| = a$. We get two equations: $ \\ 2x - 6 = 3$and $2x - 6 =  - 3 \\ $ On solving equation one we get, $ \\  2x - 6 = 3 \\   \Rightarrow 2x = 3 + 6 \\   \Rightarrow 2x = 9 \\   \Rightarrow x = \dfrac{9}{2} \\  $ On solving equation two we get, $ \\ 2x - 6 =  - 3 \\   \Rightarrow 2x =  - 3 + 6 \\   \Rightarrow 2x = 3 \\   \Rightarrow x = \dfrac{3}{2} \\  $

 Mark the correct option for $4\left| {2(2x - 3)} \right| = 12$

 Initially simplify the given question such that we have the modulus part only at one side completely. Do not multiply $4$ inside the modulus. 

 Initially simplify the question we get,  $ \\ 4\left| {2(2x - 3)} \right| = 12 \\   \Rightarrow \left| {2(2x - 3)} \right| = 3 \\   \Rightarrow \left| {4x - 6} \right| = 3 \\  $ On applying the property $\left| { \pm a} \right| = a$ we get two equations as follows: $4x - 6 = 3$ and $4x - 6 =  - 3$. On solving equation one we get,  $ \\ 4x - 6 = 3 \\   \Rightarrow 4x = 3 + 6 \\   \Rightarrow 4x = 9 \\   \Rightarrow x = \dfrac{9}{4} \\  $ On solving equation two we get, $ \\  4x - 6 =  - 3 \\   \Rightarrow 4x =  - 3 + 6 \\   \Rightarrow 4x = 3 \\   \Rightarrow x = \dfrac{3}{4} \\  $

 Tick the correct answer for the question$10\left| {2(3x + 2)} \right| + 9 = 20$

 Firstly simplify the given question such that we have the modulus part only at one side completely. 

 Initially we will simplify the question and we get,  $ \\  10\left| {2(3x + 2)} \right| + 9 = 20 \\  \Rightarrow 10\left| {2(3x + 2)} \right| = 11 \\  \Rightarrow \left| {6x + 4} \right| = \dfrac{{11}}{{10}} \\  $ On applying the property$\left| { \pm a} \right| = a$ we get two equations as follows: $ \\ 6x + 4 = \dfrac{{11}}{{10}}$ and $6x + 4 =  - \dfrac{{11}}{{10}} \\ $ On solving equation one we get,  $ \\  6x + 4 = \dfrac{{11}}{{10}} \\   \Rightarrow 60x + 40 = 11 \\  \Rightarrow 60x =  - 29 \\   \Rightarrow x =  - \dfrac{{29}}{{60}} \\  $ On solving equation two we get, $ \\ 6x + 4 =  - \dfrac{{11}}{{10}} \\   \Rightarrow 60x + 40 =  - 11 \\ \Rightarrow 60x =  - 51 \\   \Rightarrow x =  - \dfrac{{51}}{{60}} \\  $