Solutions to 2-variable Linear Absolute Value Functions (|ax+b| only)

An equation of the form $|ax+b|=c$|ax+b|=c, where $a,b$a,b and $c$c are constants, can have zero, one or two solutions depending on the constants values.

 

Example 1

 

The graph of the linear absolute value function, depicted in green in the diagram, has the formula $y=|\frac{1}{2}x+1|$y=|12​x+1|. 

By letting the function value $y=c$y=c be fixed successively at $c=-1.5$c=−1.5, $c=0$c=0 and $c=1$c=1, we form three equations

$|\frac{1}{2}x+1|=-1.5$|12​x+1|=−1.5
$|\frac{1}{2}x+1|=0$|12​x+1|=0
$|\frac{1}{2}x+1|=1$|12​x+1|=1

From the graph we see that the first of these has no solution, the second has exacly one solution ($x=-2$x=−2) and the third has two solutions ($x=-4$x=−4 and $x=0$x=0).

 

It must always be true that an equation $|ax+b|=c$|ax+b|=c has no solutions if $c$c is negative, exactly one solution if $c=0$c=0 and two solutions if $c$c is a positive number.

 

When $c=0$c=0, the solution is $x=-\frac{b}{a}$x=−ba​. You should check that this is the value that makes the expression equal to zero.

Example 2

Solve $\left|\frac{2}{3}x+\frac{1}{3}\right|=0$|23​x+13​|=0.

We need to solve either $\frac{2}{3}x+\frac{1}{3}=0$23​x+13​=0 or $-\left(\frac{2}{3}x+\frac{1}{3}\right)=0$−(23​x+13​)=0. (The two equations are the same.)

Thus, $x=-\frac{1}{2}$x=−12​. 

Note that this is indeed $-\frac{b}{a}=-\frac{\frac{1}{3}}{\frac{2}{3}}$−ba​=−13​23​​.

 

Example 3

Solve $\left|\frac{2}{3}x+\frac{1}{3}\right|=12$|23​x+13​|=12.

To obtain both solutions, we consider the two parts of the function definition and hence write down the two equations

$\frac{2}{3}x+\frac{1}{3}=12$23​x+13​=12
$-\left(\frac{2}{3}x+\frac{1}{3}\right)=12$−(23​x+13​)=12

The first of these gives $x=\frac{3}{2}\left(12-\frac{1}{3}\right)=17\frac{1}{2}$x=32​(12−13​)=1712​.

The second equation gives $x=\frac{3}{2}\left(-12-\frac{1}{3}\right)=-18\frac{1}{2}$x=32​(−12−13​)=−1812​.

 

Worked Examples

Question 1

Question 2

Question 3