How To Write An Absolute-Value Equation That Has Given Solutions

You can denote absolute value by a pair of vertical lines bracketing the number in question. When you take the absolute value of a number, the result is always positive, even if the number itself is negative. For a random number ​x​, both the following equations are true: |-​x​| = ​x​ and |​x​| = ​x​. This means that any equation that has an absolute value in it has two possible solutions. If you already know the solution, you can tell immediately whether the number inside the absolute value brackets is positive or negative, and you can drop the absolute value brackets.

Consider the equality

\(|x + y| = 4x – 3y\)

To solve this, you have to set up two equalities and solve each separately.

1. Set Up Two Equations

Set up two separate (and unrelated) equations for ​x​ in terms of ​y​, being careful not to treat them as two equations in two variables:

\(1: \quad (x + y) = 4x – 3y\)
\(2: \quad (x + y) = -(4x – 3y)\)

2. Solve One Equation for the Positive Value

\(x + y = 4x -3y\)
\(4y = 3x\)
\(x = \frac{4}{3}y\)

This is solution for equation 1.

3. Solve the Other Equation for the Negative Value

\(x + y = -4x +3y\)
\(5x = 2y\)
\(x = \frac{2}{5}y\)

This is the solution for equation 2.

Because the original equation contained an absolute value, you're left with two relationships between ​x​ and ​y​ that are equally true. If you plot the above two equations on a graph, they will both be straight lines that intersect the origin. One has a slope of 4/3 while the other has a slope of 2/5.

If you have values for x and y for the above example, you can determine which of the two possible relationships between x and y is true, and this tells you whether the expression in the absolute value brackets is positive or negative.

Suppose you know the point ​x​ = 4, ​y​ = 10 is on the line. Plug these values into both equations.

\(1: \quad 4 = \frac{4}{3}×10 = 40/3 = 14.33 \rarr \text{ False!}\)
\(2: \quad 4 = \frac{2}{5}×10 = 20/5 = 4 \rarr \text{ True!}\)

Equation 2 is the correct one. You can now drop the absolute value brackets from the original equation and write instead:

\((x + y) = -(4x – 3y)\)