Question

What is the solution set of the equation, $\left| 3-2x \right|=5$ ?

Answer 1

To find the solution set of the given equation in our problem, we need to first find out how a modulus is opened. Here, if we consider the quantity inside the modulus is positive, then it opens with a positive sign and if we consider the quantity inside the modulus is negative, then it opens with a negative sign. We shall use this to answer our question.

Complete step by step answer:
We have been given the mathematical expression as: $\left| 3-2x \right|=5$.
In the first case let us assume the term inside the modulus is positive. Then, it will open up with a positive sign. Thus, the condition for modulus to be positive is:
$\begin{aligned} & \Rightarrow 3-2x>0 \\\ & \Rightarrow 3>2x \\\ & \therefore x<\dfrac{3}{2} \\\ \end{aligned}$
Thus, our equation becomes:
$\begin{aligned} & \Rightarrow 3-2x=5 \\\ & \Rightarrow 2x=-2 \\\ & \therefore x=-1 \\\ \end{aligned}$
Here, our solution lies in the range of values of ‘x’ which are acceptable. Thus, we get the first solution of our expression as -1.
Now, in the second case let us assume the term inside the modulus is negative. Then, it will open up with a negative sign. Thus, the condition for modulus to be negative is:
$\begin{aligned} & \Rightarrow 3-2x<0 \\\ & \Rightarrow 3<2x \\\ & \therefore x>\dfrac{3}{2} \\\ \end{aligned}$
Thus, our equation becomes:
$\begin{aligned} & \Rightarrow -\left( 3-2x \right)=5 \\\ & \Rightarrow -3+2x=5 \\\ & \Rightarrow 2x=8 \\\ & \therefore x=4 \\\ \end{aligned}$
Here, our solution lies in the range of values of ‘x’ which are acceptable. Thus, we get the second and final solution of our expression as 4.
Thus our solution set is equal to \left\\{ -1,4 \right\\}.
Hence, the solution set of the equation, $\left| 3-2x \right|=5$ comes out to be \left\\{ -1,4 \right\\}.

Note: A very common mistake a student concurs while solving problems of inequalities is ignoring the fact that our solution may not lie in the domain of our function. So, it is always important to find the set of acceptable values of our function, and then check if our solution lies in that set or not.