Absolute Value Equations

The absolute value of a number, also called its modulus, is its distance from zero on the number line. The absolute value of a positive number or zero is the number itself, and the absolute value of a negative number is its opposite:

${|a|=a}$ if ${a\ge 0}$,   and   ${|a|=-a}$ if ${a<0}$

So the absolute value of every number except zero is positive, and opposite numbers have the same absolute value: ${|5|=|{-5}|=5}$. This gives the rule for the basic equation ${|X|=c}$, where $X$ is an expression with the unknown:

Right side	The equation ${|X|=c}$ means
${c>0}$	${X=c}$ or ${X=-c}$
${c=0}$	${X=0}$
${c<0}$	no solution, because an absolute value is never negative

Below are problems with full solutions. For equations whose right side contains a parameter, see linear equations with a parameter.

Problem 1

Solve the equation:

A) ${|x|=5}$   B) ${|3x+4|=7}$   C) ${\left|\frac{1}{3}x+4\right|=0}$
D) ${|2-5x|=-3}$   E) ${-|3x-1|=-11}$   F) ${|3x-3(x-1)|=3}$

Solution:

A) ${x=5}$ or ${x=-5}$: both 5 and $-5$ have absolute value 5, and no other number does.

B) ${3x+4=7}$ or ${3x+4=-7}$. The first gives ${3x=3}$ and ${x=1}$, the second gives ${3x=-11}$ and ${x=-\frac{11}{3}}$.

C) An absolute value is 0 only for the number 0, so ${\frac{1}{3}x+4=0}$, ${\frac{1}{3}x=-4}$ and ${x=-12}$.

D) No solution: the right side is negative, and an absolute value is never negative.

E) Multiply by $-1$: ${|3x-1|=11}$. So ${3x-1=11}$ or ${3x-1=-11}$, which give ${x=4}$ or ${x=-\frac{10}{3}}$.

F) Inside the absolute value, ${3x-3x+3=3}$, so the equation is ${|3|=3}$, which is true for every $x$. Every number is a solution.

Problem 2

Solve the equation:

A) ${3|5x|+4|5x|=35}$   B) ${{\frac{|2x|}{3}+\frac{3|2x|}{2}}={\frac{1}{2}}}$
C) ${3.7|x|-2.2|x|=22.5}$   D) ${\left|\frac{x+1}{3}\right|=5}$

Solution:

A) Collect the like terms: ${7|5x|=35}$, so ${|5x|=5}$. Then ${5x=5}$ or ${5x=-5}$, and ${x=1}$ or ${x=-1}$.

B) Multiply by 6: ${2|2x|+9|2x|=3}$, so ${11|2x|=3}$ and ${|2x|=\frac{3}{11}}$. Then ${2x=\frac{3}{11}}$ or ${2x=-\frac{3}{11}}$, and ${x=\frac{3}{22}}$ or ${x=-\frac{3}{22}}$.

C) ${1.5|x|=22.5}$, so ${|x|=15}$, and ${x=15}$ or ${x=-15}$.

D) ${\frac{x+1}{3}=5}$ or ${\frac{x+1}{3}=-5}$, so ${x+1=15}$ or ${x+1=-15}$, and ${x=14}$ or ${x=-16}$.

Problem 3

Prove that the equation has no solution:

A) ${-\left|\frac{2x+3}{14}\right|=5}$   B) ${|8x-4(2x+3)|=15}$

Solution:

A) Multiply by $-1$: ${\left|\frac{2x+3}{14}\right|=-5}$. No number has a negative absolute value, so there is no solution.

B) Inside the absolute value, ${8x-8x-12=-12}$, so the equation is ${|{-12}|=15}$, i.e. ${12=15}$, which is false whatever $x$ is. There is no solution.

Problem 4

Solve the equation:

A) ${{2|x-1|+3}={9-|x-1|}}$
B) ${3|x|-(x+1)^2}={4|x|-(x^2-1)-2(x-5)}$
C) ${|{-3-5x}|=3}$
D) ${{2|x-1|}={9-|x-1|}}$
E) ${{|x|-\frac{3-x}{4}}={\frac{2x-1}{8}}}$

Solution:

A) Collect the absolute values on one side:

${{2|x-1|+|x-1|}={9-3}}$, so ${3|x-1|=6}$ and ${|x-1|=2}$

Then ${x-1=2}$ or ${x-1=-2}$, and ${x=3}$ or ${x=-1}$.

B) Move ${3|x|}$ to the right side and everything else to the left:

${x^2-1+2(x-5)-(x+1)^2}={4|x|-3|x|}$

${x^2-1+2x-10-x^2-2x-1}={|x|}$

${-12=|x|}$, which has no solution.

C) ${-3-5x=3}$ or ${-3-5x=-3}$. The first gives ${-6=5x}$ and ${x=-\frac{6}{5}}$, the second gives ${0=5x}$ and ${x=0}$.

D) ${{2|x-1|+|x-1|}={9}}$, so ${3|x-1|=9}$ and ${|x-1|=3}$. Then ${x-1=3}$ or ${x-1=-3}$, and ${x=4}$ or ${x=-2}$.

E) Move the fraction to the right side and add:

${{|x|}={\frac{2x-1}{8}+\frac{3-x}{4}}}$

${{|x|}={\frac{2x-1+2(3-x)}{8}}}$, so ${|x|=\frac{5}{8}}$

and ${x=\frac{5}{8}}$ or ${x=-\frac{5}{8}}$.

Problem 5

Solve the equation:

A) ${\left|4-|x|\right|=2}$   B) ${\left|9+|x|\right|=5}$

Solution:

A) ${4-|x|=2}$ or ${4-|x|=-2}$, so ${|x|=2}$ or ${|x|=6}$. The equation has four solutions: ${x=2}$, ${x=-2}$, ${x=6}$ and ${x=-6}$.

B) ${9+|x|=5}$ or ${9+|x|=-5}$, so ${|x|=-4}$ or ${|x|=-14}$. Neither of these has a solution, so the equation has no solution. (Faster: ${9+|x|}$ is at least 9, so its absolute value can't be 5.)

Problem 6

Solve the equation

${\left|(2x+1)^2-4x^2-2\right|-3|4x-1|}={-6}$

Solution: Simplify inside the first absolute value:

${(2x+1)^2-4x^2-2}={4x^2+4x+1-4x^2-2}={4x-1}$

The equation becomes

${{|4x-1|-3|4x-1|}={-6}}$, so ${-2|4x-1|=-6}$ and ${|4x-1|=3}$

Then ${4x-1=3}$ or ${4x-1=-3}$, and ${x=1}$ or ${x=-\frac{1}{2}}$.

Problem 7

Solve the equation:

A) ${|2x-(3x+2)|=1}$   B) ${{\frac{|x|}{3}-\frac{2|x|}{2}}={-1}}$   C) ${{|3x-1|}={2|3x-1|-2}}$

Solution:

A) ${|2x-3x-2|=1}$, i.e. ${|{-x-2}|=1}$. So ${-x-2=1}$ or ${-x-2=-1}$, and ${x=-3}$ or ${x=-1}$.

B) Multiply by 6: ${2|x|-6|x|=-6}$, so ${-4|x|=-6}$ and ${|x|=\frac{3}{2}}$. Then ${x=\frac{3}{2}}$ or ${x=-\frac{3}{2}}$.

C) ${{2}={2|3x-1|-|3x-1|}}$, so ${|3x-1|=2}$. Then ${3x-1=2}$ or ${3x-1=-2}$, and ${x=1}$ or ${x=-\frac{1}{3}}$.

More on equations: Solving Linear Equations · Equation Word Problems · Linear Equation Problems · Linear Equations with a Parameter