Absolute Value

Linear Things: Absolute Value Solving

The absolute value represents the distance a number is from 0 on a number line. The notation \( |x|=5 \) represents all numbers that are a distance of 5 units from 0. On a number line, there are two numbers with a distance of 5 units from 0: when \( x=5 \) and when \( x=-5 \). Students often forget about the negative number, or moving left of 0 on the number line. This same principle applies to solving equations involving absolute value.

Solving Absolute Value Equations

Example

Solve \(|3x-4|=11\).

Step 1: Create two equations: Distance to right of 0:   \(3x-4=11\) Distance to left of 0:   \(3x-4=-11\)
Step 2: Solve using Algebra: $$3x=15$$ $$3x= -7$$
$$x=5$$ $$x=-\frac{7}{3}$$

Often the equation needs to be rewritten using algebra before solving.

Example

Solve \(2|5x+3|-9=7\).

Step 1: Algebra to isolate absolute value: \(2|5x+3|=16\) Add 9
\(|5x+3|=8\) Divide by 2
Step 2: Solve the rewritten Absolute Value Equation: \(5x+3=8\) \(5x+3=-8\)
\(5x=5\) \(5x=-11\)
\(x=1\) \(x=-\frac{11}{5}\)

Solving Absolute Value Inequalities

Now we combine the last two topics together: inequalities and absolute value. There are two types of problems:

  1. Less than/less than or equal:   \(\implies|6x+5|\leq 19\)

  2. Greater than/greater than or equal:   \(\implies|7-3x|>11\)

Recall that absolute value means distance from 0, so the first inequality is looking for a distance from 0 to be less than or equal to 19. There are two numbers that give a "boundary" around 0, at \(x=-19\) and \(x=19\). This gives all the values from -19 to 19 to be considered.

ABS Less Than

The second inequality shows a distance from 0 to be greater than 11, numbers "outside the boundary". So this starts with numbers left of -11 (less than) and right of 11 (greater than).

ABS Greater Than

Example

Solve \(|6x+5|\leq 19\).

Make two inequalities: \(\Longrightarrow -19\leq 6x+5\leq 19\)
Subtract 5 from all parts: \(-19 \color{red}{-5} \leq 6x+5 \color{red}{-5} \leq 19 \color{red}{-5}\)
Combine like terms: \(-24 \leq 6x \leq 14\)
Divide by 6 from all parts: \(-4 \leq x \leq \frac{7}{3}\)
Answer in Interval Notation: \(\left[-4,\frac{7}{3}\right]\)
Graph: [-4,7/3]

Example

Solve \(|7-3x|>11\).

Make two inequalities: Left of -11:   \(\Rightarrow 7-3x < -11\) Right of 11:   \(\Rightarrow 7-3x > 11 \)
Solve normal algebra: \(-3x < -18\) \(-3x > 4\)
Divide by -3 switches symbol: \(x > 6\) \(x < -\frac{4}{3}\)
Answer in Interval Notation: \(\left( 6,\infty\right)\) \(\left( -\infty,-\frac{4}{3}\right)\)
Graph: [-4,7/3]