How do you solve \[\left| {5x} \right| + 5 = 45\]?

Hint:The above given equation is an example of absolute value equation.
Absolute value equations: Absolute value equations are the equations where the variable is within an absolute value operator, also known as a modulus operator. So while considering an absolute value equation involving variables we have to consider both the cases of positive and negative signs.

Complete step by step solution:
Given
\[\left| {5x} \right| + 5 = 45.....................................\left( i \right)\]
Now in order to solve the given equation we need to solve for $x$.
Such that we have to manipulate the given equation in terms of only $x$, which can be achieved by performing different arithmetic operations on both LHS and RHS equally.
So to isolate the $x$ term from equation (i) we can subtract $5$ from both the LHS and RHS, since subtracting the term $5$ to the LHS will isolate the term $x$ alone by canceling the term $5$.
Subtracting $5$ to both LHS and RHS of equation (i) we get:
\[\left| {5x} \right| + 5 - 5 = 45 - 5............................\left( {ii} \right)\]
On simplifying (ii) we can write:
\[\left| {5x} \right| = 40...................................\left( {iii} \right)\]
Now on observing (iii) we can say that in the LHS there is the involvement of absolute or the modulus sign. Such that in order to remove the modulus sign we have to take two cases of both positive and negative signs.
So we can write:
\[
\Rightarrow \left| {5x} \right| = 40 \\
\Rightarrow 5x = \pm 40 \\
\Rightarrow 5x = + 40\;\;\;{\text{and}}\;\;5x = - 40 \\
\Rightarrow x = + \dfrac{{40\;}}{5}\;\;{\text{and}}\;\;x = - \dfrac{{40}}{5} \\
\Rightarrow x = + 8\;\;{\text{and}}\;\;x = - 8 \\
\Rightarrow x = \pm 8..............................................\left( {iv} \right) \\
\]
Therefore on solving \[\left| {5x} \right| + 5 = 45\]we get $x = \pm 8$.

Note: The equation is said to be true only if we find the value of the variable which makes the equation true. We can also check if the value of the variable that we got is true or not by substituting the value of the variable back into the equation and checking whether it satisfies the given equation or not.