Question

How do you solve $\left| {3x - 4} \right| = 6$ ?

Answer 1

Hint: Absolute value of any value is written by using two straight bars and the value within them. The absolute value bracket makes any negative value positive and any positive value positive itself. So, here we consider two cases where we take the negative and positive value of the expression and then solve. We get two values of $x$ in our solution.

Complete step by step solution:
The given expression is $\left| {3x - 4} \right| = 6$
Since there is a modulus symbol and an expression inside it, we consider two cases here.
The first case will be the one where we take the positive value of the expression inside the modulus and solve
The second case will be the one where the negative value of the expression in the modulus is considered and then solve it.
Let’s first take positive value.
$ \Rightarrow \left| {3x - 4} \right| = 6$
$ \Rightarrow ( + )(3x - 4) = 6$
Now evaluate further,
$ \Rightarrow 3x - 4 = 6$
Bring the constants $ - 4,6$ to one side because they both are of the same degree which is $0$
And $3x\;$ will remain there itself since it is a term of degree $1$
Now add $4$ on both sides of the equation to get all the same degree terms to one side.
$ \Rightarrow 3x - 4 + 4 = 6 + 4$
$ \Rightarrow 3x = 10$
Now divide the entire equation with $3$
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{10}}{3}$
$ \Rightarrow x = \dfrac{{10}}{3}$
This is one value of $x$
The other value will be found out by taking the negative value of the expression inside the modulus.
$ \Rightarrow \left| {3x - 4} \right| = 6$
$ \Rightarrow ( - )(3x - 4) = 6$
Now evaluate further,
$ \Rightarrow - 3x + 4 = 6$
Bring the constants $4,6\;$ to one side because they both are of the same degree which is $0$
And $3x\;$ will remain there itself since it is a term of degree $1$
Now subtract $4$ on both sides of the equation to get all the same degree terms to one side.
$ \Rightarrow - 3x + 4 - 4 = 6 - 4$
$ \Rightarrow - 3x = 2$
Now divide the entire equation with $3$ and then negate it
$ \Rightarrow \dfrac{{3x}}{3} = - \dfrac{2}{3}$
$ \Rightarrow x = - \dfrac{2}{3}$
This is the other value of $x$

Hence the values of $x$ on solving the expression are $\dfrac{{10}}{3}, - \dfrac{2}{3}$

Note: The absolute value forces the value inside it to be positive and does not affect whatever is outside. Whenever you convert a negative value into a non-negative value inside the absolute value brackets, do not forget to write the non-negative value back into the absolute value brackets.
Substitute the values back into the expression to check the answers.
Let’s check for $x = \dfrac{{10}}{3}$
$ \Rightarrow \left| {3\left( {\dfrac{{10}}{3}} \right) - 4} \right| = 6$
$ \Rightarrow \left| {10 - 4} \right| = 6$
$ \Rightarrow \left| 6 \right| = 6$
Any positive value will remain positive inside the modulus.
$\Rightarrow 6 = 6$
LHS=RHS.
Let’s check for $x = - \dfrac{2}{3}$
$ \Rightarrow \left| {3\left( { - \dfrac{2}{3}} \right) - 4} \right| = 6$
$ \Rightarrow \left| { - 2 - 4} \right| = 6$
$ \Rightarrow \left| { - 6} \right| = 6$
Any negative value will become positive inside the modulus.
$ \Rightarrow 6 = 6$
LHS=RHS.
Hence both the solutions are correct.