MATH 96 | Solving Absolute Value Equations

Step	Operation	Perform	Result
Given			$\|x\| = 7$
	Apply absolute value property	$\|x\| = 7$	$x = 7$ or $x = - 7$

Solution: $x = 7$ or $x = - 7$

Step	Operation	Perform	Result
Given			$3 \|x\| = 12$
	Divide both sides by $3$	$\frac{3 \|x\|}{3} = \frac{12}{3}$	$\|x\| = 4$
	Apply absolute value property	$\|x\| = 4$	$x = 4$ or $x = - 4$

Solution: $x = 4$ or $x = - 4$

Step	Operation	Perform	Result
Given			$2 \|x\| - 7 = 15$
	Add $7$ to both sides	$2 \|x\| - 7 + 7 = 15 + 7$	$2 \|x\| = 22$
	Divide both sides by $2$	$\frac{2 \|x\|}{2} = \frac{22}{2}$	$\|x\| = 11$
	Apply absolute value property	$\|x\| = 11$	$x = 11$ or $x = - 11$

Solution: $x = 11$ or $x = - 11$

Step	Operation	Perform	Result
Given			$\|x + 5\| = 7$
	Apply absolute value property	$\|x + 5\| = 7$	$x + 5 = 7$ or $x + 5 = - 7$
	Subtract $5$ from both sides for both equations	$x + 5 - 5 = 7 - 5$ or $x + 5 - 5 = - 7 - 5$	$x = 2$ or $x = - 12$

Solution: $x = 2$ or $x = - 12$

Step	Operation	Perform	Result
Given			$2 \|x - 3\| = 15$
	Divide both sides by $2$	$\frac{2 \|x - 3\|}{2} = \frac{15}{2}$	$\|x - 3\| = \frac{15}{2} = 7 \frac{1}{2}$
	Apply absolute value property	$\|x - 3\| = \frac{15}{2}$	$x - 3 = \frac{15}{2}$ or $x - 3 = - \frac{15}{2}$
	Add $3$ to both sides for both equations	$x - 3 + 3 = \frac{15}{2} + 3$ or $x - 3 + 3 = - \frac{15}{2} + 3$	$x = \frac{15}{2} + 3$ or $x = - \frac{15}{2} + 3$
	Simplify and subsitute $\frac{6}{2}$ for $3$	$x = \frac{15}{2} + \frac{6}{2}$ or $x = - \frac{15}{2} + \frac{6}{2}$	$x = \frac{21}{2}$ or $x = - \frac{9}{2}$

Solution: $x = \frac{21}{2} = 10 \frac{1}{2}$ or $x = - \frac{9}{2} = - 4 \frac{1}{2}$

Step	Operation	Perform	Result
Given			$\|2 x + 7\| = - 5$
	Not applicable	Nothing	$2 x + 7 \neq - 5$

Solution: No solution. Why? Because the absolute value of a quantity can never be negative.

Step	Operation	Perform	Result
Given			$\|x + 5\| = \|3 x - 1\|$
	Apply absolute value property	$\|x + 5\| = \|3 x - 1\|$	$x + 5 = 3 x - 1$ or $x + 5 = - (3 x - 1)$
	Subtract $5$ from both sides of both equations	$x + 5 - 5 = 3 x - 1 - 5$ or $x + 5 - 5 = - (3 x - 1) - 5$	$x = 3 x - 6$ or $x = - 3 x - 4$
	Subtract $3 x$ from both sides in the first equation	$x - 3 x = 3 x - 6 - 3 x$ or $x = - 3 x - 4$	$- 2 x = - 6$ or $x = - 3 x - 4$
	Add $3 x$ to both sides in the second equation	$- 2 x = - 6$ or $x + 3 x = - 3 x - 4 + 3 x$	$- 2 x = - 6$ or $4 x = - 4$
	Divide first equation by $- 2$ and second equation by $4$	$\frac{- 2 x}{- 2} = \frac{- 6}{- 2}$ or $\frac{4 x}{4} = \frac{- 4}{4}$	$x = 3$ or $x = - 1$

Solution: $x = 3$ or $x = - 1$