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How do you solve $\left| {2x - 3} \right| = 5$ ?

Answer
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462.9k+ views
Hint: Here we are given an equation $\left| {2x - 3} \right| = 5$ and we are asked to find the value of $x$ that satisfies $\left| {2x - 3} \right| = 5$. That is we need to calculate the solution of the given equation. To find the desired solutions, we need to deal with two cases because the given equation contains absolute (modulus) values.

Complete answer:
The given equation is $\left| {2x - 3} \right| = 5$ . Here we are asked to find the solution of the given equation.
Let us deal with cases to remove the absolute bars.
Case a: $2x - 3 > 0$
The given equation is $\left| {2x - 3} \right| = 5$.
Let us assume that $x$ is greater than zero. That is $x > 0$.
We know that the absolute value $x$ is equal to $x$ when $x$ is greater than zero.
That means $\left| x \right| = x$ when $x > 0$….. $\left( 1 \right)$
Let us substitute $\left( 1 \right)$ in the given equation.
$2x - 3 = 5$
$ \Rightarrow 2x = 5 + 3$
$ \Rightarrow 2x = 8$
$ \Rightarrow x = \dfrac{8}{2}$
Hence we get $x = 4$.
 Therefore $x = 4$ is the required solution.
Case b: $2x - 3 < 0$
The given equation is $\left| {2x - 3} \right| = 5$.
Let us assume that $x$ is less than zero. That is $x < 0$.
We know that the absolute value $x$ is equal to $ - x$when $x$ is less than zero.
That means $\left| x \right| = - x$ when $x < 0$….. $\left( 2 \right)$
Let us substitute $\left( 2 \right)$ in the given equation.
$ - \left( {2x - 3} \right) = 5$
$ \Rightarrow - 2x + 3 = 5$
$ \Rightarrow - 2x = 5 - 3$
$ \Rightarrow - 2x = 2$
$ \Rightarrow x = \dfrac{2}{{ - 2}}$
Hence, we get $x = - 1$
Therefore $x = - 1$ is the required solution.
Thus, $x = 4$ and $x = - 1$ are the desired solutions of the given equation $\left| {2x - 3} \right| = 5$

Note:
It is to be noted that the absolute value of $x$ is equal to $x$ when $x$ is greater than zero, the absolute value of $x$ is equal to $ - x$ when $x$ is less than zero, and the absolute value of $x$ is equal to zero when $x$ is equal to zero. That means $\left| x \right| = x$ when $x > 0$, $\left| x \right| = - x$ when $x < 0$, and $\left| x \right| = 0$ when $x = 0$.