Solve absolute value equations

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Difficulty: Intermediate
Estimated time: 27 minutes
Domains: Advanced Math
Techniques: Absolute-value-equationsDistanceTwo-case-solvingSolution-countParameter-equations

What you’ll learn

Read absolute value as a distance on the number line.
Start by graphing both sides exactly as written when the other side has the variable too.
Get the absolute value by itself before you split it into two cases.
Tell whether an equation has zero, one, or two real solutions.
Solve by hand when the other side is a plain number or the question wants an exact constant.
Give the answer the question asks for: every solution, one chosen solution, or the count.

Prerequisites

Solve linear equations
Isolate and solve the linear equation that remains in each case.

Why this matters on the SAT

Turn one distance equation into two linear cases

An absolute value equation often hides two possible answers. The SAT might ask for one possible solution, the positive or the negative one, all of them, or how many there are. Whatever it asks, your first move is the same: get the absolute value by itself.

Solution to the example

Clear away the $4$ and the $2$ that sit outside the bars:

\begin{aligned} 2\lvert 3x-1\rvert+4&=18\\[1.4em] 2\lvert 3x-1\rvert&=14\\[1.4em] \lvert 3x-1\rvert&=7. \end{aligned}

Absolute value measures distance from $0$ . Both $7$ and $-7$ are $7$ units from $0$ , so the inside, $3x-1$ , could be either one. Solve both:

\begin{aligned} 3x-1=7 &\quad\Longrightarrow\quad x=\frac83,\\[1.4em] 3x-1=-7 &\quad\Longrightarrow\quad x=-2. \end{aligned}

Only $\frac83$ is positive, so the answer is D. Choice B, $-2$ , is a real solution too. It’s there to catch you if you solve only one case and stop.

SAT example

What is the positive solution to the equation

2\lvert 3x-1\rvert+4=18?

A
$-3$
B
$-2$
C
$2$
D
$\frac{8}{3}$

Let the other side choose the method

Look at the side of the equation without the bars. It tells you where to start.

If it’s a plain number, like the $7$ in $\lvert 3x-1\rvert=7$ , work by hand. Get the bars alone and solve two short cases.
If it has the variable too, like the $x+4$ in $\lvert 2x+1\rvert=x+4$ , graph the two original sides and click every intersection.
If the question wants an exact constant, like the $k$ that gives exactly one solution, work through the cases by hand. A graph can hint at that value, but it can’t pin it down exactly.

Here’s how that looks for $\lvert 2x+1\rvert=x+4$ . Enter each side on its own line:

y=|2x+1|
y=x+4

Click every intersection and read its $x$ -coordinate. The graphs meet at

x=-\frac53 \qquad\text{and}\qquad x=3.

Both are real solutions. You graphed the equation exactly as it was given, so wherever the two graphs meet, the two sides are equal. There’s nothing to double-check or throw out.

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Click both intersections before you pick one, such as the positive, greatest, or least.

Recognize distance from a value

Absolute value gives distance from $0$ . For example, $\lvert -5\rvert=5$ because $-5$ is $5$ units from $0$ .

The same idea works with any center. An expression like

\lvert x-3\rvert

gives the distance between $x$ and $3$ . Subtracting finds the gap between the two numbers, and the bars remove any minus sign. So

\lvert x-3\rvert=5

asks, “Which numbers are $5$ units away from $3$ ?”

Step

5

left from

3

to land on

3-5=-2

, or

5

right to land on

3+5=8

That’s why there are two cases. The gap $x-3$ is either $5$ or $-5$ :

x-3=5 \qquad\text{or}\qquad x-3=-5.

Solving gives $x=8$ or $x=-2$ , the same two points as on the number line.

Check your understanding:

Without any algebra, find the two numbers that are $6$ units from $-4$ . Then write an absolute value equation that describes them.

Isolate, split, and solve

For a linear expression $A$ and a positive number $d$ ,

\lvert A\rvert=d

means

A=d \qquad\text{or}\qquad A=-d.

That split only works once the absolute value is by itself. Here’s the whole routine:

\boxed{\text{isolate the bars}\ \longrightarrow\ \text{check the other side}\ \longrightarrow\ \text{write two cases}\ \longrightarrow\ \text{solve}.}

Take

5\lvert 2x+3\rvert-10=30.

The $5$ and the $-10$ are outside the bars, so clear them first. Add $10$ , then divide by $5$ :

\begin{aligned} 5\lvert 2x+3\rvert&=40\\[1.4em] \lvert 2x+3\rvert&=8. \end{aligned}

Now split into the two cases:

\begin{aligned} 2x+3=8 &\quad\Longrightarrow\quad x=\frac52,\\[1.4em] 2x+3=-8 &\quad\Longrightarrow\quad x=-\frac{11}{2}. \end{aligned}

Both values work. Plug either one in and the inside of the bars becomes $8$ or $-8$ , and both have an absolute value of $8$ .

Common mistake:

It’s tempting to drop the bars right away and write $5(2x+3)-10=30$ and $5(2x+3)-10=-30$ . But the plus-or-minus choice belongs to the absolute value alone, not to the $5$ and the $-10$ around it. That second equation gives $x=-\frac72$ , and $5\lvert 2(-\frac72)+3\rvert-10=10$ , not $30$ . Get to $\lvert 2x+3\rvert=8$ first, then use $8$ and $-8$ .

Count zero, one, or two solutions

Once the bars are alone, you have

\lvert A\rvert=d.

This is the “check the other side” step. When $A$ is a linear expression with the variable in it, the sign of $d$ tells you how many solutions there are:

If $d>0$ , there are two real solutions, because $A=d$ and $A=-d$ are different equations.
If $d=0$ , there’s one real solution, because both cases become $A=0$ .
If $d<0$ , there’s no real solution, because a distance can’t be negative.

Positive, zero, negative means two, one, none. For example:

\begin{array}{c|c|c} \text{Equation} & \text{Sign of the right side} & \text{Number of solutions}\\[0.9em] \hline \lvert 4x-1\rvert=6 & 6>0 & 2\\[0.9em] \lvert 4x-1\rvert=0 & 0=0 & 1\\[0.9em] \lvert 4x-1\rvert=-6 & -6<0 & 0 \end{array}

Do this check before you split into cases. Take

3\lvert x+2\rvert+7=1.

Getting the bars alone gives

\lvert x+2\rvert=-2.

Stop there. A distance can’t be $-2$ , so there’s no real solution. If you split anyway, $x+2=-2$ and $x+2=2$ give $x=-4$ and $x=0$ , and neither one works: each makes the left side $13$ , not $1$ .

Check your understanding:

What value of the constant $c$ gives $2\lvert 5x-4\rvert+c=11$ exactly one real solution? Explain why.

Example: Isolate and select the negative solution

Often the SAT wants only one of the two solutions. The trap is choosing too early.

Worked example

What is the negative solution to the equation

4\lvert 2x-3\rvert+1=29?

A
$-5$
B
$-2$
C
$2$
D
$5$

Step 1

Get the absolute value alone

Subtract $1$ from both sides, then divide by $4$ :

\begin{aligned} 4\lvert 2x-3\rvert&=28\\[1.4em] \lvert 2x-3\rvert&=7. \end{aligned}

The right side, $7$ , is positive, so there are two cases.

Step 2

Write both cases

The inside of the bars can equal $7$ or $-7$ :

2x-3=7 \qquad\text{or}\qquad 2x-3=-7.

Hold off on the word negative for now. It describes the $x$ you report at the end, not which case to solve.

Step 3

Solve both cases

Solve each linear equation:

\begin{aligned} 2x-3=7 &\quad\Longrightarrow\quad 2x=10 \quad\Longrightarrow\quad x=5,\\[1.4em] 2x-3=-7 &\quad\Longrightarrow\quad 2x=-4 \quad\Longrightarrow\quad x=-2. \end{aligned}

Try it yourself:

Which of $5$ and $-2$ does the question want? Decide, then plug it into the original equation to check before you move on.

Step 4

Pick the one the question asks for

The negative solution is $-2$ , so the answer is B.

Check it in the original equation:

4\lvert 2(-2)-3\rvert+1 =4\lvert -7\rvert+1 =29.

Common mistake:

It’s tempting to solve only the $-7$ case because the question says negative. It worked here, but a negative case doesn’t guarantee a negative $x$ . In $\lvert 4-x\rvert=6$ , the case $4-x=-6$ gives $x=10$ . Find them all, then pick: solve both cases first, and only then apply words like positive, negative, greater, or least.

Graph the original sides when both sides have the variable

You’ve already graphed one of these, $\lvert 2x+1\rvert=x+4$ . For any equation like

\lvert A\rvert=B,

where $B$ has the variable in it, the graph is still your first move:

y=|A|
y=B

It shows you every solution in view at once. Sometimes, though, the question needs an exact value the graph doesn’t give you. Then solve the two cases by hand afterward:

A=B \qquad\text{or}\qquad A=-B.

Here’s the catch. These cases can hand you an answer that doesn’t work in the original equation. An absolute value is never negative, so at a real solution the right side $B$ can’t be negative either. Check every answer from the cases in the original equation.

Take

\lvert x-2\rvert=2x-1.

Graph $y=\lvert x-2\rvert$ and $y=2x-1$ , and you’ll see only one intersection, at $x=1$ .

Now solve the two cases by hand and watch what happens:

\begin{aligned} x-2=2x-1 &\quad\Longrightarrow\quad x=-1,\\[1.4em] x-2=-(2x-1) &\quad\Longrightarrow\quad x=1. \end{aligned}

Algebra gave you two answers, so check both in the original equation:

\begin{aligned} x=-1:&\quad \lvert -3\rvert=3,\quad 2(-1)-1=-3 &&\text{not equal},\\[1.4em] x=1:&\quad \lvert -1\rvert=1,\quad 2(1)-1=1 &&\text{equal}. \end{aligned}

Only $x=1$ works. At $x=-1$ the right side is $-3$ , and no absolute value can equal a negative number. The graph never offers $x=-1$ at all, because the two graphs don’t meet there.

Check your understanding:

Before you solve the two cases for $\lvert 2x+3\rvert=x$ , what must be true of every real solution? Why?

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Click the one intersection, at

x=1

. The graphs don’t meet anywhere else.

Practice problems

Before each problem, look at the side without the bars to choose your method. Then read the last sentence to see exactly what to report.

Graph the original sides

Practice problem

What is the positive solution to the equation

\lvert 3x-2\rvert=x+6?

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The variable is on both sides, so graph both sides as written.

Use the greater solution in an expression

Practice problem

The greater solution to

\lvert 5-2t\rvert=17

is $t$ . What is the value of $t-3$ ?

Your answer

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The other side is a plain number, so two short cases by hand are quickest.

Create exactly one real solution

Practice problem

For what value of $k$ does the equation

\lvert 3x-4\rvert=2x+k

have exactly one real solution for $x$ ?

Your answer

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Find the exact

k

by hand. Afterward, a graph can confirm that the line meets the V-shaped graph once.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Absolute value is distance, so a positive distance usually gives two answers.
Get the absolute value by itself before you split into cases.
For $\lvert A\rvert=d$ , where $A$ is linear and has the variable in it: positive, zero, negative $d$ means two, one, none.
Find them all, then pick. Solve both cases before you apply words like positive, negative, greater, or least.
When the other side has the variable, graph both original sides and click every intersection.
Solve $\lvert A\rvert=\text{a number}$ by hand, and find an exact constant like $k$ by hand too. A graph can only estimate that boundary.

Next lesson

Solve higher-degree equations from structure

Use structure to shrink a higher-degree equation once two-case thinking feels solid.

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153 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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