Equations and Inequalities: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

Classify the equation $4(x+3) - x = 3x + 12$ .

Answer choices for question 1 A $x = 0$ B Every real number C No solution D $x = 3$
2

A student is about to clear a fraction by multiplying or dividing both sides of an equation. Which of the following is guaranteed to produce an equivalent equation, whatever equation it is applied to?

Answer choices for question 2 A Multiply both sides by $x-6$ B Multiply both sides by $0$ C Divide both sides by $-4$ D Divide both sides by $x+2$
3

Which step, applied to both sides of a true inequality, is guaranteed to keep both its solution set and the direction of its symbol?

Answer choices for question 3 A Add $-13$ to both sides B Multiply both sides by $-5$ C Multiply both sides by $0$ D Divide both sides by $x$
4

A rectangle's perimeter is $P = 2w + 2(kw + L)$ , where $w$ is the width and $k$ , $L$ are fixed numbers. Solving for $w$ requires collecting every $w$ term and dividing by a single quantity. Which quantity is it, and what condition does that place on $k$ ?

Answer choices for question 4 A $2k$ , requiring $k \neq 0$ B $2$ , with no condition on $k$ C $2-2k$ , requiring $k \neq 1$ D $2+2k$ , requiring $k \neq -1$
5

What is the solution set of $\lvert x+6 \rvert = 3x-2$ ?

Answer choices for question 5 A $\{4,-1\}$ B $\{4\}$ C $\{-1\}$ D $\varnothing$
6

What is the solution set of $\dfrac{2}{x} > 4$ ?

Answer choices for question 6 A $0<x<\frac12$ B $x>\frac12$ C $x<0$ or $x>\frac12$ D $x<\frac12$
7

Solve $\lvert x-9 \rvert < 4$ .

Answer choices for question 7 A $x<5$ or $x>13$ B $-4<x<4$ C $5<x<9$ D $5<x<13$
8

How many solutions does $\lvert x-3 \rvert + \lvert x+5 \rvert = 6$ have?

Answer choices for question 8 A Exactly two B Infinitely many C None D Exactly one
9

Solve $\dfrac{9}{x-2} = \dfrac{3}{x+4}$ .

Answer choices for question 9 A $x=2$ B $x=-7$ C $x=-4$ D $\varnothing$
10

Solve $ax > 14$ for $x$ , in the case $a<0$ .

Answer choices for question 10 A $x<\dfrac{14}{a}$ B $x>\dfrac{14}{a}$ C $x<-\dfrac{14}{a}$ D $x>-\dfrac{14}{a}$
11

For which value of $m$ does $mx+11=5x+m$ have no solution?

Answer choices for question 11 A $m=11$ B $m=-6$ C no such $m$ D $m=5$
12

Rearrange $px+q=rx+s$ to isolate $x$ . At $p=9$ , $r=9$ , $q=4$ , $s=22$ , what happens to that rearranged formula, even though neither $p$ nor $r$ is zero?

Answer choices for question 12 A It gives $x=1$ B It gives $x=18$ C It is undefined, since $p-r=0$ D It is well-defined, since $p \neq 0$ and $r \neq 0$
13

Start from the true statement $-8<-2$ . Multiply both sides by $-5$ , then add $3$ to both sides. What is the resulting true statement?

Answer choices for question 13 A $43>13$ B $43<13$ C $37>7$ D $13>43$
14

Find every $x$ satisfying $4x-9>11$ AND $2x+15<7$ , or state that none exist.

Answer choices for question 14 A $x>5$ B No such $x$ exists C $x<-4$ D $-4<x<5$
15

Solve $\lvert 2x-5 \rvert = \lvert 3x+1 \rvert$ .

Answer choices for question 15 A $\{-6\}$ B $\{\frac45\}$ C $\varnothing$ D $\{-6,\frac45\}$
16

For which of the following is the solution set all of $\mathbb{R}$ ?

Answer choices for question 16 A $\lvert 3x+1 \rvert < -4$ B $\lvert x-6 \rvert < 7$ C $\lvert 2x-5 \rvert > -9$ D $\lvert x+2 \rvert > 7$
17

Both $\dfrac{x^2-49}{x-7}=x+7$ and $\lvert x-7 \rvert = 2x-20$ ask a student to accept a candidate without a further check. Which statement is correct?

Answer choices for question 17 A The first equation's solution set is $\mathbb{R}\setminus\{7\}$ , and the second's is $\{13\}$ B Both equations have solution set $\{13\}$ C The first equation has no solution, since $x=7$ is excluded D The second equation's solution set is $\{13,9\}$
18

A solver multiplies both sides of $\dfrac{2}{x+3}<5$ by $x+3$ without checking its sign, reaching $x>-\dfrac{13}{5}$ . Testing $x=-10$ in the original gives $\dfrac{2}{-7}\approx -0.29<5$ , which is true. What does this reveal?

Answer choices for question 18 A The naive step correctly identified every solution B The naive step lost solutions, since $x=-10$ solves the original but fails $x>-\frac{13}{5}$ C $x=-10$ is extraneous and should be discarded D The multiplier $x+3$ must have been positive at $x=-10$
19

Let $d$ be the target in $\lvert x-4 \rvert + \lvert x+10 \rvert = d$ (anchors $4$ and $-10$ , gap $g=14$ ), and let $c$ be the bound in $\lvert x-4 \rvert \ge c$ . For which pair $(d,c)$ does the first equation have exactly two solutions while the second inequality's solution set is all of $\mathbb{R}$ ?

Answer choices for question 19 A $(d,c)=(10,-1)$ B $(d,c)=(15,3)$ C $(d,c)=(15,-1)$ D $(d,c)=(14,0)$
20

Consider the equation $kx=9k$ and the inequality $kx>9k$ , both at the specific value $k=0$ . Which statement correctly describes their solution sets there?

Answer choices for question 20 A Both are $\mathbb{R}$ B Both are $\varnothing$ C The equation is $\varnothing$ and the inequality is $\mathbb{R}$ D The equation is $\mathbb{R}$ and the inequality is $\varnothing$

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Where you stand

State which operations on an equation are guaranteed to produce an equivalent one (adding an expression defined for every $x$ , multiplying or dividing by a nonzero constant), and apply the excluded-value discipline when a step multiplies by an expression that can be zero: list the excluded values before clearing a denominator, then discard any candidate that lands on that list. Review: Solving Linear Equations
Reduce a linear equation, with numeric or with parameter coefficients, to the form $ax = b$ , and classify its solution set as one number, no solution, or every real number from the values of $a$ and $b$ , including finding for which parameter values each outcome occurs. Review: Solving Linear Equations , Literal Equations and Formulas
In a literal equation, treat the chosen unknown and the remaining parameters correctly, rearrange for the unknown, and state any new nonzero condition the rearrangement acquires as a single statement about the whole factor divided by, not the individual letters composing it. Review: Literal Equations and Formulas
Determine whether adding to, or multiplying or dividing, both sides of an inequality preserves its solution set and its direction, including why scaling by a negative constant forces the direction to reverse and why scaling by zero destroys the inequality. Review: Linear Inequalities
Recognize that multiplying an inequality by an expression of unknown sign is not a single legal step, and solve such an inequality by excluding any forbidden value and splitting into sign-based cases before combining the results. Review: Linear Inequalities
Solve a linear inequality $ax > b$ whose coefficient is a parameter by casing on the sign of $a$ , and in the boundary case $a = 0$ use the sign of $b$ to decide between every real number and no solution. Review: Linear Inequalities
Combine two inequalities joined by 'and' or 'or' into the intersection or union of their solution sets, including recognizing when a piece that loses its variable makes the compound all reals or empty. Review: Linear Inequalities
Solve an absolute-value equation by the split its right side requires: check each candidate of $\lvert X \rvert = Y$ against $Y \ge 0$ , but recognize that $\lvert X \rvert = \lvert Y \rvert$ needs no such check since both sides are already nonnegative. Review: Absolute Value Equations and Inequalities
Solve an absolute-value inequality $\lvert X \rvert < c$ or $\lvert X \rvert > c$ by reading it as a distance band or a pair of rays, correctly handling every real $c$ , including where $c \le 0$ . Review: Absolute Value Equations and Inequalities
Solve an equation with two absolute values added, $\lvert x-p \rvert + \lvert x-q \rvert = d$ , by locating the breakpoints, solving the resulting linear equation on each interval, keeping only roots inside their own interval, and predicting none, the whole segment, or two points by comparing $d$ with the gap $g$ . Review: Absolute Value Equations and Inequalities

Core practice

10 problems from across the chapter. 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 divided record

Solve $\frac{x+1}{x-2}=1+\frac6{x+3}$ over the real numbers, listing excluded values first and classifying the equation.
1. Hint 1
  List the values that make either original denominator zero.
2. Hint 2
  After subtracting $1$ , combine the left side into one fraction before clearing both denominators.
Answer

Excluded: $x=2,-3$ ; solution: $x=7$ ; conditional.

Full solution

Exclude $x=2$ and $x=-3$ .
Subtracting $1$ gives
$\frac3{x-2}=\frac6{x+3}.$
On the permitted domain, multiply by the nonzero product $(x-2)(x+3)$ .
$\begin{aligned}3(x+3)&=6(x-2)\\3x+9&=6x-12\\3x&=21\\x&=7\end{aligned}$

The candidate is allowed and is unique.
In the original, the left side is $\frac85$ and the right side is
$1+\frac6{10}=\frac85.$
The equation is conditional because some, but not all, allowed inputs satisfy it.

Answer

Excluded: $x=2,-3$ ; solution: $x=7$ ; conditional.

Key idea

Collecting a fractional equation into linear form preserves the original denominator exclusions.
Problem 1: Mark complete
Problem 2 A scale setting

The real quantities satisfy $W=a(t+u)+b(t-u)$ . Find $t$ whenever it is uniquely determined, stating the condition as one nonzero factor.
1. Hint 1
  Keep the requested unknown separate from the fixed parameters.
2. Hint 2
  Expand, collect the terms containing $t$ , and identify their combined coefficient.
Answer

$t=\frac{W-(a-b)u}{a+b}$ , requiring $a+b\ne0$ .

Full solution

Here $t$ is the unknown and $W,a,b,u$ are parameters.
$\begin{aligned}W&=(a+b)t+(a-b)u\\(a+b)t&=W-(a-b)u\\t&=\frac{W-(a-b)u}{a+b}\end{aligned}$

Division requires the whole factor $a+b$ to be nonzero.
Substituting the formula replaces the first product by $W-(a-b)u$ , recovering $W$ .

Answer

$t=\frac{W-(a-b)u}{a+b}$ , requiring $a+b\ne0$ .

Key idea

Collecting all occurrences of an unknown identifies the factor that controls uniqueness.
Problem 2: Mark complete
Problem 3 Two signed readings

Two readings are $2x-5$ and $x+1$ , where $x$ is real. Find every setting at which the readings are unequal but their absolute values agree.
1. Hint 1
  Equal magnitudes can correspond to equal readings or opposite readings.
2. Hint 2
  Use the requirement that the readings themselves are unequal to decide which candidate to retain.
Answer

$x=\frac43$ .

Full solution

Equal absolute values give two possible equations.
Equal readings give $2x-5=x+1$ , hence $x=6$ , but this violates the requirement that the readings be unequal.
Opposite readings give $\begin{aligned}2x-5&=-x-1\\3x&=4\\x&=\frac43\end{aligned}$

At $x=\frac43$ , the readings are $-\frac73$ and $\frac73$ .
They are unequal and have the same absolute value, so this candidate works.
Both equality cases have been covered.

Answer

$x=\frac43$ .

Key idea

Equal absolute values permit either equal or opposite inputs; an additional condition can select between these cases.
Problem 3: Mark complete
Problem 4 An adjustable acceptance rule

For a real threshold $c$ , a setting $x$ is accepted when $\lvert x-1\rvert<c$ . Find all values of $c$ for which no real setting is accepted.
1. Hint 1
  Compare the threshold with the smallest possible value of the left side.
2. Hint 2
  The setting at the center has distance zero; distinguish a strict inequality from a non-strict one.
Answer

$c\le0$ .

Full solution

Every distance $\lvert x-1\rvert$ is nonnegative.
If $c\le0$ , a nonnegative number cannot be strictly less than $c$ , so no setting works.

If $c>0$ , the setting $x=1$ gives $\lvert1-1\rvert=0<c$ , so at least one setting is accepted.
Therefore exactly $c\le0$ gives no accepted setting.

Answer

$c\le0$ .

Key idea

A strict distance bound has no solutions when its threshold is zero or negative.
Problem 4: Mark complete
Problem 5 A two-part filter

A filter accepts real $x$ if $\frac{2x+1}{x-1}<3$ and $2(x+2)\ge2x+3$ . Find its full accepted set, explaining the restrictions needed in the calculation.
1. Hint 1
  Each part must hold, so determine what each contributes.
2. Hint 2
  Exclude the denominator zero and compare separately on its positive and negative sides.
Answer

$x<1$ or $x>4$ ; $x=1$ is excluded.

Full solution

The second part simplifies to $4\ge3$ , true for every real $x$ .
The first excludes $x=1$ .
For $x>1$ , multiplication by $x-1$ gives
$\begin{aligned}2x+1&<3x-3\\x&>4\end{aligned}$

For $x<1$ , the multiplier is negative, so the inequality reverses to $2x+1>3x-3$ , giving $x<4$ .
Combined with the case assumption, every $x<1$ works.
Intersecting this union with the all-real second part leaves $x<1$ or $x>4$ .

Answer

$x<1$ or $x>4$ ; $x=1$ is excluded.

Key idea

A sign-case solution still has to be combined with the other part of an and condition.
Problem 5: Mark complete
Problem 6 An adjustable balance

For fixed real parameters $a,b,c$ , determine every real $u$ satisfying $a(u-2)+b(u+2)=c$ . State exactly when there is one solution, no solution, or every real number as a solution.
1. Hint 1
  The target appears in two products, so identify its full coefficient.
2. Hint 2
  When that coefficient is zero, compare the remaining constant with $c$ .
Answer

If $a+b\ne0$ , $u=\frac{c+2a-2b}{a+b}$ ; if $a+b=0,c=2b-2a$ , all real $u$ ; otherwise no solution.

Full solution

Expand and collect.
$\begin{aligned}(a+b)u-2a+2b&=c\\(a+b)u&=c+2a-2b\end{aligned}$

For $a+b\ne0$ , divide by that whole factor to obtain the stated unique value.
For $a+b=0$ , the left side is zero.
Every $u$ works exactly when $c+2a-2b=0$ ; otherwise no $u$ works.

Answer

If $a+b\ne0$ , $u=\frac{c+2a-2b}{a+b}$ ; if $a+b=0,c=2b-2a$ , all real $u$ ; otherwise no solution.

Key idea

The same collected coefficient determines the formula condition and the exceptional solution cases.
Problem 6: Mark complete
Problem 7 A parameter comparison

For fixed real $r$ and $s$ , solve $r(x-2)>s$ in every parameter case. Explain why the direction changes in one case and why multiplying by zero cannot settle the boundary case.
1. Hint 1
  The sign of $r$ must be settled before it is used as a divisor.
2. Hint 2
  When $r=0$ , evaluate the original statement rather than altering it.
Answer

If $r>0$ , $x>2+\frac sr$ ; if $r<0$ , $x<2+\frac sr$ ; if $r=0,s<0$ , all real $x$ ; if $r=0,s\ge0$ , none.

Full solution

If $r>0$ , division gives
$x-2>\frac sr.$
If $r<0$ , division reverses the order, giving
$x-2<\frac sr.$
Adding $2$ yields the stated rays.

A negative scale reverses the sign of the difference between two unequal numbers, so their order reverses.
At $r=0$ , the original is $0>s$ , true exactly when $s<0$ .
Multiplying by zero would erase $s$ and cannot preserve that information.

Answer

If $r>0$ , $x>2+\frac sr$ ; if $r<0$ , $x<2+\frac sr$ ; if $r=0,s<0$ , all real $x$ ; if $r=0,s\ge0$ , none.

Key idea

A parameter inequality requires order-aware division and a direct check at a zero coefficient.
Problem 7: Mark complete
Problem 8 A comparison with a fraction

Find every real $x$ satisfying $\frac{\lvert x-1\rvert}{x-1}=\frac{3-x}{x-1}$ . State exclusions and explain why any algebraic candidate is kept or discarded.
1. Hint 1
  The denominator decides which inputs are available; the numerator equality has its own sign condition.
2. Hint 2
  Clear the nonzero denominator, then require $3-x\ge0$ in the resulting equation.
Answer

Excluded: $x=1$ ; solution: $x=2$ .

Full solution

Exclude $x=1$ .
Clearing the denominator gives $\lvert x-1\rvert=3-x$ , which requires $x\le3$ .
The equal-inside branch is
$\begin{aligned}x-1&=3-x\\2x&=4\\x&=2\end{aligned}$

The opposite branch gives $x-1=x-3$ , a contradiction.
The candidate $2$ is neither excluded nor disallowed by $x\le3$ .
In the original, both sides equal $1$ .

Answer

Excluded: $x=1$ ; solution: $x=2$ .

Key idea

Denominator restrictions and absolute-value sign restrictions must both survive a sequence of algebraic steps.
Problem 8: Mark complete
Problem 9 A changed inequality

A solver claims $\frac{3-x}{x+2}>1$ is equivalent to $3-x>x+2$ for every real $x\ne-2$ . Assess the claim and find the full solution set of the original.
1. Hint 1
  The proposed multiplication has a sign that depends on the input.
2. Hint 2
  Separate $x>-2$ from $x<-2$ , and keep the assumption attached to each result.
Answer

The claim is false; the original solution set is $-2<x<\frac12$ .

Full solution

Exclude $x=-2$ .
For $x>-2$ , multiplication keeps the direction.
$\begin{aligned}3-x&>x+2\\1&>2x\\x&<\frac12\end{aligned}$
Combined with this case,
$-2<x<\frac12.$

For $x<-2$ , the direction must reverse: $3-x<x+2$ gives $x>\frac12$ , incompatible with the case.
The claimed single inequality includes, for example, $x=-3$ , but the original there reads $-6>1$ , false.

Answer

The claim is false; the original solution set is $-2<x<\frac12$ .

Key idea

A positive-denominator calculation cannot be applied across a negative-denominator region.
Problem 9: Mark complete
Problem 10 A checkpoint restriction

A real coordinate $x$ must satisfy both $\lvert x+1\rvert+\lvert x-6\rvert=11$ and $\lvert2x-5\rvert<10$ . Find all such coordinates.
1. Hint 1
  Find the possible coordinates from the distance total, then apply the second condition.
2. Hint 2
  The total changes formula at $-1$ and $6$ ; the other inequality gives a band.
Answer

No coordinates satisfy both; solution set $\varnothing$ .

Full solution

For $x<-1$ , the distance sum is $5-2x$ , so $5-2x=11$ gives $x=-3$ , which belongs to that interval.
For $-1\le x\le6$ , the sum is $7$ , so it cannot equal $11$ .
For $x>6$ , $2x-5=11$ gives $x=8$ , also in its interval.

The second condition gives $-10<2x-5<10$ , hence
$-\frac52<x<\frac{15}{2}.$
Neither $-3$ nor $8$ lies in this band.
Each candidate in fact has $\lvert2x-5\rvert=11$ , so both fail the strict bound.

Answer

No coordinates satisfy both; solution set $\varnothing$ .

Key idea

Solutions to a distance-total equation must still be checked against an additional distance band.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra II. You can skip it. → Continue to Chapter 2: Functions and Their Graphs →

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