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Absolute Value Equations and Inequalities: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A root expression

For real $t<1$ , simplify $\sqrt{(2t-2)^2}+t$ to an expression without a root or absolute-value bars.
1. Hint 1
  The square root of a square gives the magnitude of the original quantity.
2. Hint 2
  Use $t<1$ to decide the sign of $2t-2$ before removing the bars.
Answer

$2-t$ .

Full solution

Since $2t-2<0$ , its magnitude is its opposite.
$\begin{aligned}\sqrt{(2t-2)^2}+t&=-(2t-2)+t\\&=2-t\end{aligned}$

At $t=0$ , both the original and simplified expressions equal $2$ , consistent with the sign choice.

Answer

$2-t$ .

Key idea

Removing the root from a square requires the sign of the quantity being squared.
Problem 1: Mark complete
Problem 2 A shifted reading

Solve $2\lvert x+1\rvert-3=7$ for real $x$ .
1. Hint 1
  Isolate the absolute value before interpreting its distance.
2. Hint 2
  The inside must be either the positive isolated value or its opposite.
Answer

$x=-6$ or $x=4$ .

Full solution

Add $3$ and divide by $2$ .
$\lvert x+1\rvert=5$

Thus $x+1=5$ or $x+1=-5$ , giving $x=4$ or $x=-6$ .
Each makes the original left side $2(5)-3=7$ .

Answer

$x=-6$ or $x=4$ .

Key idea

Isolating the absolute value makes its two sign cases easier to solve.
Problem 2: Mark complete
Problem 3 A restricted reading

Solve $5-\lvert 2x+3\rvert\ge1$ for real $x$ .
1. Hint 1
  Rearrange the inequality so the absolute value stands alone.
2. Hint 2
  The resulting upper bound puts the inside between two opposite numbers.
Answer

$-\frac72\le x\le\frac12$ .

Full solution

Rearranging gives
$\lvert2x+3\rvert\le4.$
This means
$-4\le2x+3\le4.$
Subtract $3$ and divide by positive $2$ to get
$-\frac72\le x\le\frac12.$

Both endpoints make the original left side $1$ , so they are included.

Answer

$-\frac72\le x\le\frac12$ .

Key idea

A nonnegative upper bound on magnitude places the expression inside the bars in a band.
Problem 3: Mark complete
Problem 4 A calibration setting

A technician chooses a real constant $c$ so that the readings $\lvert2x-1\rvert$ and $\lvert x+c\rvert$ agree at $x=2$ . Find every possible $c$ and, for each, all other real settings where the readings agree.
1. Hint 1
  Substitute the recorded setting to find the possible calibration constants.
2. Hint 2
  For each constant, equal magnitudes allow the inside expressions to be equal or opposite.
Answer

$c=1$ , with other setting $x=0$ ; or $c=-5$ , with other setting $x=-4$ .

Full solution

At $x=2$ , agreement requires
$\lvert2+c\rvert=3.$
Thus $2+c=3$ or $2+c=-3$ , giving $c=1$ or $c=-5$ .

For $c=1$ , the equal-inside case gives $2x-1=x+1$ , hence $x=2$ .
The opposite case gives
$\begin{aligned}2x-1&=-x-1\\3x&=0\\x&=0\end{aligned}$
At $x=0$ , both readings are $1$ .

For $c=-5$ , the equal-inside case gives $2x-1=x-5$ , hence $x=-4$ .
The opposite case gives
$\begin{aligned}2x-1&=-x+5\\3x&=6\\x&=2\end{aligned}$
At $x=-4$ , both readings are $9$ .
These two sign cases exhaust the possibilities; neither inside expression needs to be nonnegative.

Answer

$c=1$ , with other setting $x=0$ ; or $c=-5$ , with other setting $x=-4$ .

Key idea

An observed agreement can determine several calibration constants, each with its own set of agreement settings.
Problem 4: Mark complete
Problem 5 A calculation record

A record for $\lvert2x-3\rvert=x+c$ lists $x=-4$ from the branch $2x-3=x+c$ and $x=\frac{10}{3}$ from the branch $-(2x-3)=x+c$ . Recover the real constant $c$ , and decide which recorded candidates solve the original equation.
1. Hint 1
  The branch equations determine the constant, but satisfying a branch alone does not settle the original equation.
2. Hint 2
  Substitute a recorded candidate into its branch to find $c$ , then check the original right side at both candidates.
Answer

$c=-7$ ; neither candidate solves the original equation, which has no real solution.

Full solution

The first record gives
$\begin{aligned}2(-4)-3&=-4+c\\-11&=-4+c\\c&=-7\end{aligned}$
The other record agrees: at $x=\frac{10}{3}$ , both sides of its branch equal $-\frac{11}{3}$ .

In the original equation the right side is $x-7$ .
It is $-11$ at $x=-4$ and $-\frac{11}{3}$ at
$x=\frac{10}{3}.$
Neither can equal a nonnegative absolute value.
Both branches have been solved, and neither candidate works, so there is no real solution.

Answer

$c=-7$ ; neither candidate solves the original equation, which has no real solution.

Key idea

Candidates from equal-or-opposite equations still need a nonnegative right side in the original absolute-value equation.
Problem 5: Mark complete
Problem 6 A separation rule

A track marker has real coordinate $x$ in meters. It must be at least 5 meters from the marker at $-2$ , and it must lie between coordinates $-10$ and $8$ , including the endpoints. Find all allowed coordinates.
1. Hint 1
  A minimum distance leaves two possible sides of the fixed marker.
2. Hint 2
  First solve $\lvert x+2\rvert\ge5$ , then keep the portion inside the track.
Answer

$[-10,-7]\cup[3,8]$ meters.

Full solution

The distance rule gives $x+2\le-5$ or $x+2\ge5$ , hence $x\le-7$ or $x\ge3$ .
Intersect these rays with the track interval.
$x\in[-10,-7]\cup[3,8]$

The inner endpoints are exactly 5 meters from $-2$ and all endpoints are permitted.

Answer

$[-10,-7]\cup[3,8]$ meters.

Key idea

Combine distance rays with any bounds imposed by the physical setting.
Problem 6: Mark complete
Problem 7 An incomplete station record

One station on a straight path is at coordinate $-1$ meter and another is at an unknown real coordinate $c$ meters. At each checkpoint $0$ and $4$ , the sum of the distances to the stations is 8 meters. Find $c$ , then find every checkpoint coordinate $x$ with the same distance total. Justify the latter set by splitting at the station coordinates.
1. Hint 1
  Use each recorded checkpoint to restrict the unknown station coordinate.
2. Hint 2
  After finding the station, split the path into the intervals before, between, and after the two stations.
Answer

$c=7$ meters; every $x$ with $-1\le x\le7$ meters.

Full solution

At checkpoint $0$ , the known distance is $1$ , so
$\lvert c\rvert=7.$
This gives $c=-7$ or $c=7$ .
At checkpoint $4$ , the known distance is $5$ , so $\lvert4-c\rvert=3$ gives $c=1$ or $c=7$ .
Only $c=7$ fits both records.

The required total is
$\lvert x+1\rvert+\lvert x-7\rvert=8.$
For $x<-1$ , its left side is $6-2x$ , which equals $8$ only at $x=-1$ , outside that open interval.
For $-1\le x\le7$ , the sum is $(x+1)+(7-x)=8$ , so every point works.
For $x>7$ , the sum is $2x-6$ , which equals $8$ only at $x=7$ , outside that open interval.
Thus exactly $-1\le x\le7$ works.

Answer

$c=7$ meters; every $x$ with $-1\le x\le7$ meters.

Key idea

Distance records can locate an unknown anchor, after which the anchor coordinates determine the sign intervals.
Problem 7: Mark complete
Problem 8 A proposed bound

For a real parameter $c$ , a student claims that $2-\lvert x\rvert<c$ is true for every real $x$ exactly when $c>2$ . Decide whether the claim is correct and justify the boundary.
1. Hint 1
  Find the largest value the left side actually reaches.
2. Hint 2
  Check the input where the absolute value is zero.
Answer

Correct; exactly $c>2$ .

Full solution

Since $\lvert x\rvert\ge0$ ,
$2-\lvert x\rvert\le2.$
The upper value is attained at $x=0$ .

If $c>2$ , every left-side value is less than $c$ .
If $c\le2$ , the input $x=0$ fails because it would require $2<c$ .
Thus the strict boundary is necessary.

Answer

Correct; exactly $c>2$ .

Key idea

A strict inequality holding everywhere must exceed an attained maximum strictly.
Problem 8: Mark complete
Problem 9 A solver record

A solver replaces $4-\lvert2x+1\rvert=x$ by $4-(2x+1)=x$ and reports $x=1$ as the sole real solution. Are any solutions missing? Give the full solution set and explain.
1. Hint 1
  Removing the bars in one way assumes a sign for their inside.
2. Hint 2
  Isolate the absolute value and consider both signs, checking that its other side is nonnegative.
Answer

Yes; $x=-5$ is missing. The full solution set is $\{-5,1\}$ .

Full solution

Isolating the bars gives $\lvert2x+1\rvert=4-x$ , which requires $x\le4$ .
The equal-inside branch gives $2x+1=4-x$ , so $x=1$ .
The opposite branch gives
$\begin{aligned}2x+1&=x-4\\x&=-5\end{aligned}$

Both candidates meet $x\le4$ .
Direct substitution gives $4-3=1$ at $x=1$ and $4-9=-5$ at $x=-5$ .
The original step used only the branch with $2x+1\ge0$ , missing the negative-inside solution.

Answer

Yes; $x=-5$ is missing. The full solution set is $\{-5,1\}$ .

Key idea

Removing absolute-value bars without checking the sign can discard an entire branch of solutions.
Problem 9: Mark complete
Problem 10 A total distance claim

A student says $\lvert x+3\rvert+\lvert x-5\rvert=d$ has exactly one real solution for some real $d$ . Is that possible? Explain all ranges of $d$ .
1. Hint 1
  Compare the target total with the distance between the two anchors.
2. Hint 2
  Between the anchors the total is constant; outside them it increases in two matching directions.
Answer

No; none for $d<8$ , every $x\in[-3,5]$ for $d=8$ , and two solutions for $d>8$ .

Full solution

The anchors are $-3$ and $5$ , so their gap is $8$ .
On $[-3,5]$ , the total is
$(x+3)+(5-x)=8.$
Outside this interval the total exceeds $8$ .

For $d>8$ , the left interval gives $2-2x=d$ , hence
$x=1-\frac d2<-3.$
The right interval gives $2x-2=d$ , hence
$x=1+\frac d2>5.$
Thus there are two points, while $d=8$ gives the whole interval and $d<8$ gives none.

Answer

No; none for $d<8$ , every $x\in[-3,5]$ for $d=8$ , and two solutions for $d>8$ .

Key idea

Two distinct anchors yield zero solutions, a whole segment, or two solutions for a specified distance total.
Problem 10: Mark complete

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