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Absolute Value Equations and Graphs: 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 Bars alone first

Solve $4|x-7|-5=15$ , and check each value in the original equation.
1. Hint 1
  The bars are not alone yet, so the two-case rule does not apply to the equation as written.
2. Hint 2
  Undo the subtraction and then the multiplication, so that the absolute value stands by itself.
3. Hint 3
  Once the right side is a positive number, the inside equals that number or its opposite.
Answer

$x=2$ or $x=12$ .

Full solution

Add $5$ to both sides and divide by $4$ , which leaves the bars alone:
$|x-7|=5$
The right side is positive, so the inside is $5$ or $-5$ .

One case is
$\begin{aligned}x-7&=5\\x&=12\end{aligned}$
The other case is
$\begin{aligned}x-7&=-5\\x&=2\end{aligned}$

Substituting $x=12$ gives $4|12-7|-5=15$ , and substituting $x=2$ gives $4|2-7|-5=15$ , so both values satisfy the original equation.

Answer

$x=2$ or $x=12$ .

Key idea

Isolate the absolute value before splitting into cases, since splitting first solves a different equation.
Problem 1: Mark complete
Problem 2 The moving labels

Two labels lie at $2a+1$ and $1-2a$ on a number line, where $a$ is real. Write their distance as a constant multiple of $|a|$ .
1. Hint 1
  Distance is the absolute value of the difference of the coordinates.
2. Hint 2
  Simplify that difference before taking its nonnegative size.
Answer

$4|a|$ .

Full solution

The difference of the coordinates is
$(2a+1)-(1-2a)=4a$
Its absolute value is $|4a|$ .

Multiplication by the positive number $4$ scales distance by $4$ , so
$|4a|=4|a|$
This also gives zero when the labels coincide at $a=0$ .

Answer

$4|a|$ .

Key idea

The distance between two moving coordinates is found by subtracting their rules before taking the absolute value.
Problem 2: Mark complete
Problem 3 The paired records

The equation $|x-h|=3$ has solutions $x=1$ and $x=7$ . Find $h$ .
1. Hint 1
  The two solutions are equally far from the center $h$ .
2. Hint 2
  The center lies midway between them.
Answer

$h=4$ .

Full solution

The center is the average of the two locations:
$h=\frac{1+7}{2}$
Thus $h=4$ .

Checking, both $|1-4|$ and $|7-4|$ equal $3$ , as required.

Answer

$h=4$ .

Key idea

The two solutions of a positive-distance equation lie equally far on opposite sides of its center.
Problem 3: Mark complete
Problem 4 The two requirements

Find all real numbers satisfying both $|x-2|<4$ and $|x+1|>2$ .
1. Hint 1
  Each requirement gives an allowed region, and the input must belong to both.
2. Hint 2
  The first gives a band; the second gives two outside rays.
Answer

$1<x<6$ .

Full solution

The first inequality gives
$-2<x<6$
The second gives $x<-3$ or $x>1$ .

The ray $x<-3$ does not meet the band.
Intersecting the other ray with the band leaves
$1<x<6$
Both ends are excluded because one original inequality becomes equality there.

Answer

$1<x<6$ .

Key idea

Simultaneous absolute-value inequalities require the overlap of their bands and rays.
Problem 4: Mark complete
Problem 5 The marked roof

The figure shows $f(x)=a|x-h|+k$ with $a\ne0$ . Find its rule and all inputs where its output is $-7$ .

The graph of $f$ on a coordinate grid, with its corner and one other point marked.

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x axis runs from negative 2 to 6 and the vertical y axis from negative 3 to 6, with gridlines, tick marks and number labels at every whole number and the origin labeled 0. A graph labeled f is drawn as two straight branches that meet at a sharp corner and open downward, forming an upside down V. A filled dot marks the corner and is labeled (2, 5), and a second filled dot on the vertical axis is labeled (0, 1). The left branch rises from the lower left corner of the grid to the sharp corner, and the right branch falls from there to the lower right corner of the grid, each ending in an arrowhead to show that it continues. No other point, height, crossing or solution is marked.
1. Hint 1
  The vertex identifies $h$ and $k$ .
2. Hint 2
  Use the marked nonvertex point to find $a$ , then isolate the absolute value for the requested output.
Answer

$f(x)=-2|x-2|+5$ ; $x=-4$ or $x=8$ .

Full solution

The vertex $(2,5)$ gives $h=2$ , $k=5$ .
The point $(0,1)$ gives
$1=2a+5$
so $a=-2$ .

For output $-7$ , the rule gives $-2|x-2|+5=-7$ , hence
$|x-2|=6$
The two cases give $x=-4$ and $x=8$ .
Both are six units from the vertex input and return $-7$ .

Answer

$f(x)=-2|x-2|+5$ ; $x=-4$ or $x=8$ .

Key idea

A vertex and one further point determine an absolute-value graph, after which a target height gives a distance equation.
Problem 5: Mark complete
Problem 6 The allowed positions

A marker has coordinate $x$ on a number line. It must stay at least $2$ units and at most $4$ units from the location $1$ . On the axes in the figure, sketch the graph whose height is the distance from the marker to $1$ , then find all allowed coordinates.

Blank axes on which to graph the distance from $1$ .

Text description of this figure

Blank coordinate axes with equal unit lengths on both axes and a light square grid. The horizontal axis is labeled x and runs from negative 5 to 7. The vertical axis is labeled distance from 1, in units, and runs from negative 1 to 7. Both axes carry tick marks and number labels at every whole number, and the origin is labeled 0. Nothing is plotted: there is no graph, no marked point, no guide line, no highlighted tick and no shaded region.
1. Hint 1
  Express the distance from location $1$ as a nonnegative function of $x$ .
2. Hint 2
  Draw its two branches from the point where that distance is zero, then keep the inputs whose graph heights lie between $2$ and $4$ , including both heights.
Answer

Graph $y=|x-1|$ , a V with vertex $(1,0)$ ; allowed coordinates $[-3,-1]\cup[3,5]$ .

Full solution

Distance from $1$ is
$y=|x-1|$
Plot the vertex $(1,0)$ and draw two straight branches rising away from it, with slope $-1$ on the left and $1$ on the right.
Points $(0,1)$ and $(2,1)$ check the branches.

The upper distance bound gives
$-3\le x\le5.$
The lower bound gives $x\le-1$ or $x\ge3$ .
Their overlap is
$[-3,-1]\cup[3,5]$

These are the inputs where the V lies between heights $2$ and $4$ .
Each of the four interval endpoints gives one of those allowed heights, so all endpoints are included.

Answer

Graph $y=|x-1|$ , a V with vertex $(1,0)$ ; allowed coordinates $[-3,-1]\cup[3,5]$ .

Key idea

Graphing distance from a fixed location gives a V whose heights select the allowed positions.
Problem 6: Mark complete
Problem 7 The stored condition

A real input satisfies $|4x+6|=10$ . Find every possible value of $|x+9|$ .
1. Hint 1
  First recover every input allowed by the stored condition.
2. Hint 2
  Carry both signed cases into the new expression.
Answer

$5$ or $10$ .

Full solution

The inside of the given absolute value is either $10$ or $-10$ .
One case is
$\begin{aligned}4x+6&=10\\x&=1\end{aligned}$
The other case is
$\begin{aligned}4x+6&=-10\\x&=-4\end{aligned}$

The new expression gives $|1+9|=10$ or $|-4+9|=5$ .
Both inputs satisfy the original condition, so both outputs are possible.

Answer

$5$ or $10$ .

Key idea

A condition with two signed cases can lead to two different values of a later expression.
Problem 7: Mark complete
Problem 8 The opposite readings

Two nonzero numbers $u$ and $v$ have opposite signs. Kai claims that $|u+v|$ equals the distance between $|u|$ and $|v|$ . Is Kai correct? Explain.
1. Hint 1
  Adding numbers of opposite signs subtracts their sizes.
2. Hint 2
  Name the positive sizes $A$ and $B$ , then work out what the sum is in each of the two sign arrangements.
Answer

Yes; $|u+v|=\big||u|-|v|\big|$ .

Full solution

Write the positive sizes as $A=|u|$ and $B=|v|$ .
Since the signs are opposite, their sum is either $A-B$ or $B-A$ .

These two differences are opposites and have the same absolute value, so
$|u+v|=|A-B|$
The right side is exactly the distance between the two sizes.
Kai is correct.

Answer

Yes; $|u+v|=\big||u|-|v|\big|$ .

Key idea

For opposite signs, the size of the sum is the distance between the two input sizes.
Problem 8: Mark complete
Problem 9 Nora's comparison

Nora says $|x-2|$ and $|x|-2$ agree only at $x=2$ . Decide whether she is right and find every input where they agree.
1. Hint 1
  The signs of $x$ and $x-2$ split the line into three regions.
2. Hint 2
  On each region, rewrite both expressions without bars, then set the two results equal and keep only solutions lying in that region.
Answer

Nora is wrong; they agree for every $x\ge2$ .

Full solution

For $x<0$ , the expressions are $2-x$ and $-x-2$ , which differ by $4$ .
For $0\le x<2$ , equality would require
$2-x=x-2$
giving $x=2$ , outside that region.

For $x\ge2$ , both expressions are exactly $x-2$ .
Thus they agree throughout that ray, including the boundary $2$ .

Answer

Nora is wrong; they agree for every $x\ge2$ .

Key idea

Comparing shifted absolute values requires accounting for each interval where the inside signs are fixed.
Problem 9: Mark complete
Problem 10 Elena's comparison

For a real constant $c$ , the equation $3|x-1|+c=0$ has two distinct real solutions. Elena claims that replacing $c$ by $2c$ doubles the distance between the solutions and leaves their midpoint unchanged. Decide whether each claim is correct and justify your decisions.
1. Hint 1
  Two distinct solutions force the isolated absolute value to be positive.
2. Hint 2
  Name the original distance from $1$ and compare it with the distance after the constant changes.
3. Hint 3
  Write each pair as its center plus or minus its distance, then compare the two separations.
Answer

Both claims are correct; the distance doubles and the midpoint remains $1$ .

Full solution

Two distinct solutions require $c<0$ .
Set $d=-c/3$ , so $d>0$ .
The original solutions are $1-d$ and $1+d$ , with distance
$(1+d)-(1-d)=2d$
Their midpoint is $1$ .

Replacing $c$ by $2c$ gives
$|x-1|=2d$
The new solutions are $1-2d$ and $1+2d$ .
Their distance is $4d$ , twice the original distance, and their midpoint is still $1$ .

Both claims are therefore correct, and the new solutions remain distinct since $d>0$ .

Answer

Both claims are correct; the distance doubles and the midpoint remains $1$ .

Key idea

Scaling the positive distance in an absolute-value equation changes the separation of its solutions while preserving their center.
Problem 10: Mark complete

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