Exercise 21 Page 247

When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. Although we can make $4$ statements about this equation, there are actually only two possible cases to consider.

Statement	Result
Both absolute values are positive.	$ax+b=cx+d$
Both absolute values are negative.	$\text{-}(ax+b)=\text{-}(cx+d)$
Only the left-hand side is negative.	$\text{-}(ax+b)=cx+d$
Only the right-hand side is negative.	$ax+b=\text{-}(cx+d)$

Because of the Properties of Equality, when the absolute values of two expressions are equal, either $\text{the}$ $\text{expressions}$ $\text{are}$ $\text{equal}$ or $\text{the}$ $\text{opposites}$ $\text{of}$ $\text{the}$ $\text{expressions}$ $\text{are}$ $\text{equal}.$ Now let's consider these two cases for the given equation. We will solve each of these equations by graphing separately.

First Equation

To graph the first equation, we create two functions out of the left- and right-hand sides of the equation. The $x\text{-}$coordinate of the point of intersection of the graphs of these functions is the solution to our equation. We can see that the graphs intersect at $(\text{-}2,\text{-}6)$ which means the solution to the equation is $x=\text{-}2.$ Let's check whether it is correct by substituting it into the original equation. The equation is true, so $x=\text{-}2$ is a solution.

Second Equation

In order to graph the second equation, we again create functions out of the left- and right-hand sides of the equation. Once more, the $x\text{-}$coordinate where the graphs of these functions intersect is the solution to our equation. From the graph, we can see that the lines intersect at $(1,\text{-}3),$ which means the solution to this equation is $x=1.$ Checking our solution, we confirmed that $x=1$ is a solution.