6.03 Graphical approach to solving equations

It is crucial that we are able to connect our algebraic understanding to equivalent graphical representations. Graphical methods of solution often provide a more efficient path to an answer or can act as a sanity check on our solution.

Graphical solution approach

Let's take a look at the graph below of $y=\left|\frac{1}{2}x+1\right|$y=|12​x+1|. We are going to use it to help us solve the following equations: $\left|\frac{1}{2}x+1\right|=1$|12​x+1|=1, $\left|\frac{1}{2}x+1\right|=0$|12​x+1|=0 and $\left|\frac{1}{2}x+1\right|=-1.5$|12​x+1|=−1.5.

 

Let's start with $\left|\frac{1}{2}x+1\right|=1$|12​x+1|=1. The graph of the absolute value function $y=\left|\frac{1}{2}x+1\right|$y=|12​x+1|, the left hand side of this equation, is the solid line. We then want to sketch a graph of the right hand side, by thinking about it as a separate function. In this case we can represent it as the straight line $y=1$y=1, which is the top dashed line in the diagram above. The solutions to the equation are given by the $x$x coordinates of the points of intersection of the two functions.

And so, we can see that there are two solutions $x=-4$x=−4 and $x=0$x=0 to the equation $\left|\frac{1}{2}x+1\right|=1$|12​x+1|=1.

Two other lines $y=0$y=0 and $y=-1.5$y=−1.5 have also been sketched. Using the same thought process, we can see that $\left|\frac{1}{2}x+1\right|=0$|12​x+1|=0 has one solution at $x=-2$x=−2, while $\left|\frac{1}{2}x+1\right|=-1.5$|12​x+1|=−1.5 has no solutions.

Solving absolute value equations

Although this chapter is focusing on graphs, it is a good time to show you the algebraic method of solving absolute value equations. This is useful if you are not able to sketch the functions in the equation you are trying to solve. You may remember that we can sketch absolute value functions by considering their piecewise definition, with:

So, we will solve the equation from our first example $\left|\frac{1}{2}x+1\right|=1$|12​x+1|=1 by creating two related equations: $-\left(\frac{1}{2}x+1\right)=1$−(12​x+1)=1 and $\frac{1}{2}x+1=1$12​x+1=1.

Solving the first, we get:

$\frac{1}{2}x+1$12​x+1	$=$=	$-1$−1
$\frac{1}{2}x$12​x	$=$=	$-2$−2
$x$x	$=$=	$-4$−4

And from the second, we get;

$\frac{1}{2}x+1$12​x+1	$=$=	$1$1
$\frac{1}{2}x$12​x	$=$=	$0$0
$x$x	$=$=	$0$0

Not every absolute value equation will have solutions. Again, thinking about the previous examples using graphs, we saw that $\left|\frac{1}{2}x+1\right|=-1.5$|12​x+1|=−1.5 had no solutions. This is because of the range of the absolute value function. So even if you are solving equations algebraically it is very helpful to have your graphical understanding at the front of your thoughts so that you can be confident in the accuracy of the answers you find.

We can use the graphical approach with all kinds of graphs, not just absolute value functions. In practice question 3 below, it will be helpful to remember that when we set $y=0$y=0, we are solving for the $x$x intercepts of the function.

 

Practice questions

Question 1

Question 2

Question 3

Question 4

Question 5