Question

$\left| 2x-1 \right|=4x+5$ has how many numbers in its solution set?

Answer 1

Hint: Problems on solving equations of absolute values can be easily done by using the property of absolute values. We will take two cases into consideration for this problem. In the first case we consider a positive value of the term $2x-1$ and in the second case we consider the term $2x-1$ to be a negative one and solve both of the equations to get the solutions of the given equation.

Complete step by step solution:
The expression we have is
$\left| 2x-1 \right|=4x+5$
We apply the property of absolute value in the above equation to solve the equation.
First, we remove the absolute value term in the given equation. This creates two cases among which the first one is where we consider the positive value of the term $2x-1$ and in the second one is where we consider the term $2x-1$ to be negative one.
Therefore, the given equation becomes
$2x-1=4x+5......\left( \text{1} \right)$ , when $\left( 2x-1 \right)>0$
And $-\left( 2x-1 \right)=4x+5......\left( 2 \right)$ , when $\left( 2x-1 \right)<0$
Considering equation $\left( 1 \right)$ we get
$\Rightarrow 4x+5=2x-1$
Subtracting $2x$ from both the sides of the above equation we get
$\Rightarrow 4x+5-2x=2x-1-2x$
Further simplifying we get
$\Rightarrow 2x+5=-1$
Subtracting $5$ from both the sides of the above equation we get
$\Rightarrow 2x+5-5=-1-5$
Further simplifying we get
$\Rightarrow 2x=-6$
Dividing both sides of the above equation by $2$ we get
$\Rightarrow x=-\dfrac{6}{2}$
Further simplifying
$\Rightarrow x=-3$
Considering equation $\left( 2 \right)$ we get
$\Rightarrow 4x+5=-\left( 2x-1 \right)$
$\Rightarrow 4x+5=-2x+1$
Adding $2x$ to both the sides of the above equation we get
$\Rightarrow 4x+5+2x=-2x+1+2x$
Further simplifying we get
$\Rightarrow 6x+5=1$
Subtracting $5$ from both the sides of the above equation we get
$\Rightarrow 6x+5-5=1-5$
Further simplifying
$\Rightarrow 6x=-4$
Dividing both sides of the above equation by $6$ we get
$\Rightarrow x=\dfrac{-4}{6}$
 $\Rightarrow x=-\dfrac{2}{3}$
Also, we put both the solutions in the given equation. The solution $x=-\dfrac{2}{3}$ satisfies the given equation but the solution $x=-3$ does not satisfy the given equation as the RHS becomes $-7$ , and any absolute value cannot be negative. Hence, the solution $x=-3$ is an extraneous one.

Therefore, the number of solutions in the equation’s solution set is one.

Note: We have to keep in mind that while removing the absolute term we have to take both the cases $\left( 2x-1 \right)>0$ and $\left( 2x-1 \right)<0$ into account so that we get all the solutions. Also, we must properly simplify the obtained equations so that mistakes are avoided.