# Absolute Value Equations 3

#### harpazo

##### Pure Mathematics
Banned
Let's go back to high school together as we explore the world of absolute value equations. Ready to have fun with math?

#### TheJason

Staff member
Moderator
$$\displaystyle | x + 20| = \dfrac{3}{4}$$

$$\displaystyle | x + 20| = 0.75$$

$$\displaystyle (x + 20) =0.75$$

$$\displaystyle x + 20 = 0.75$$

$$\displaystyle x + 20 - 20 = 0.75 - 20$$

$$\displaystyle x = -19.25$$

Check

$$\displaystyle |(-19.25) + 20| = 0.75$$

$$\displaystyle |0.75| = 0.75$$

Negative side:

$$\displaystyle | x + 20| = 0.75$$

$$\displaystyle -(x + 20) = 0.75$$

$$\displaystyle -x - 20 = 0.75$$

$$\displaystyle -x - 20 + 20 = 0.75 + 20$$

$$\displaystyle -x = 20.75$$

$$\displaystyle \dfrac{x}{-1} = \dfrac{20.75}{-1}$$

$$\displaystyle x = -20.75$$

Check

$$\displaystyle |(-20.75) + 20| = 0.75$$

$$\displaystyle |-0.75| = 0.75$$

Two solutions of $$\displaystyle -19.25, -20.75$$

harpazo

#### harpazo

##### Pure Mathematics
Banned
$$\displaystyle | x + 20| = \dfrac{3}{4}$$

$$\displaystyle | x + 20| = 0.75$$

$$\displaystyle (x + 20) =0.75$$

$$\displaystyle x + 20 = 0.75$$

$$\displaystyle x + 20 - 20 = 0.75 - 20$$

$$\displaystyle x = -19.25$$

Check

$$\displaystyle |(-19.25) + 20| = 0.75$$

$$\displaystyle |0.75| = 0.75$$

Negative side:

$$\displaystyle | x + 20| = 0.75$$

$$\displaystyle -(x + 20) = 0.75$$

$$\displaystyle -x - 20 = 0.75$$

$$\displaystyle -x - 20 + 20 = 0.75 + 20$$

$$\displaystyle -x = 20.75$$

$$\displaystyle \dfrac{x}{-1} = \dfrac{20.75}{-1}$$

$$\displaystyle x = -20.75$$

Check

$$\displaystyle |(-20.75) + 20| = 0.75$$

$$\displaystyle |-0.75| = 0.75$$

Two solutions of $$\displaystyle -19.25, -20.75$$
Do we always get two solutions?

#### TheJason

Staff member
Moderator
Do we always get two solutions?
Not sure. I would say so because of the structure of it.

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
If we are taking the absolute value of a linear function, and we set this absolute value equal to a positive number, then we will get two solutions:

$$\displaystyle |ax+b|=c$$ where $$0<c$$ and $$a\ne0$$

$$\displaystyle ax+b=\pm c$$

$$\displaystyle x=\frac{-b\pm c}{a}$$

If $$c=0$$, then we get one solution, and if $$c<0$$ we get no solution.

harpazo and anemone

#### harpazo

##### Pure Mathematics
Banned
If we are taking the absolute value of a linear function, and we set this absolute value equal to a positive number, then we will get two solutions:

$$\displaystyle |ax+b|=c$$ where $$0<c$$ and $$a\ne0$$

$$\displaystyle ax+b=\pm c$$

$$\displaystyle x=\frac{-b\pm c}{a}$$

If $$c=0$$, then we get one solution, and if $$c<0$$ we get no solution.
Can you share an example when c = 0 and when c < 0?