How do you solve $\left| 2x-3 \right|=7$ ?

Hint: We recall the definition of absolute value function and the break point of absolute value function. We recall that the break point of $\left| ax+b \right|$ is the solution of $ax+b=0$ where the definition of absolute value function changes. We find the break point of $\left| 2x-3 \right|$ and solve accordingly.

Complete step by step solution:
We know that of absolute value function of modulus function denoted as $\left| x \right|$ returns a non-negative value without regard for the input of the functions. It is defined in piecewise manner as;
\[\left| x \right|=\left\{ \begin{matrix}
   -x & \text{if }x<0\text{ } \\
   x & \text{if }x\ge 0 \\
\end{matrix} \right.\]
We see that the function has a breakpoint at $x=0$ which means the absolute value function has different definitions for $x\in \left( -\infty ,0 \right)$ and for $x\in \left[ 0,\infty \right)$. If we define an absolute value function $\left| ax+b \right|$ for any real numbers $a,b$ then the breakpoint will be at the solution of $ax+b=0$ which is $x=\dfrac{-b}{a}$ because $\left| ax+b \right|$ has to return non-negative outputs and for that we have to define the function as
\[\left| x-a \right|=\left\{ \begin{matrix}
   -\left( ax+b \right) & \text{if }x<\dfrac{-b}{a}\text{ } \\
   ax+b & \text{if }x\ge \dfrac{b}{a} \\
\end{matrix} \right.\]
We are given the following equation in the question
\[\begin{align}
  & \left| 2x-3 \right|=7 \\
 & \Rightarrow \left| 2x+\left( -3 \right) \right|=7 \\
\end{align}\]
 We see that in the left hand side of the above equation the absolute value function is in the form $\left| ax+b \right|$ with $a=2,b=-3$. So the break point is
\[x=\dfrac{-b}{a}=\dfrac{-\left( -3 \right)}{2}=\dfrac{3}{2}\]
So the definition of the given absolute value function is
\[\left| 2x-3 \right|=\left\{ \begin{matrix}
   -\left( 2x-3 \right) & \text{if }x<\dfrac{3}{2}\text{ } \\
   2x-3 & \text{if }x\ge \dfrac{3}{2} \\
\end{matrix} \right.\]
So we get linear equations to solve
\[\begin{align}
  & -\left( 2x-3 \right)=7\text{ if }x<\dfrac{3}{2}............\left( 1 \right) \\
 & 2x-3=7\text{ if }x\ge \dfrac{3}{2}.......\left( 2 \right) \\
\end{align}\]
We solve equation (1) to have
\[\begin{align}
  & -\left( 2x-3 \right)=7 \\
 & \Rightarrow -2x+3=7 \\
 & \Rightarrow -2x=4 \\
 & \Rightarrow x=-2 \\
\end{align}\]
We see that above solution $x=-2$ satisfies the condition $x<\dfrac{3}{2}$.We solve equation (2) to have ;
\[\begin{align}
  & 2x-3=7 \\
 & \Rightarrow 2x=10 \\
 & \Rightarrow x=5 \\
\end{align}\]
We see that the above solution satisfies $x=5$ the condition $x\ge \dfrac{3}{2}$. So we got a solution to have $x=-2,5$.

Note: We can alternatively by using the property $\left| a \right|=b\Rightarrow a=\pm b$ to have $2x-3=\pm 7$ from where we can proceed to solve linear equations. The break point of $\left| x-a \right|$ is $x=a$. We should remember that a piecewise defined function has different definitions in different intervals. In absolute value function if $x=a$ is the break point it has two different definitions in $\left( -\infty ,a \right),\left( a,\infty \right)$.