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Linear 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 divided comparison

Solve $\frac{5-2x}{3}\le x+1$ for real $x$ .
1. Hint 1
  Clear the positive denominator without reversing the direction.
2. Hint 2
  Collect the variable terms, noting the sign of the final coefficient.
Answer

$x\ge\frac25$ .

Full solution

Multiply by the positive number $3$ .
$\begin{aligned}5-2x&\le3x+3\\2&\le5x\\\frac25&\le x\end{aligned}$

The endpoint gives equality in the original; $x=1$ satisfies it and $x=0$ does not.

Answer

$x\ge\frac25$ .

Key idea

Positive scaling keeps an inequality pointing in the same direction.
Problem 1: Mark complete
Problem 2 A bracketed condition

Solve $7-2(3x+1)>4(x-2)$ for real $x$ .
1. Hint 1
  First rewrite each side as an equal expression.
2. Hint 2
  Collecting the variable on the left gives a negative coefficient, which matters when dividing.
Answer

$x<\frac{13}{10}$ .

Full solution

Expand and collect.
$\begin{aligned}5-6x&>4x-8\\-10x&>-13\\x&<\frac{13}{10}\end{aligned}$

Dividing by $-10$ reverses the direction.
At the endpoint the sides are equal, so it is excluded; $x=0$ gives $5>-8$ .

Answer

$x<\frac{13}{10}$ .

Key idea

The sign of the actual divisor determines whether the inequality reverses.
Problem 2: Mark complete
Problem 3 Two boundary rules

Find all real $x$ satisfying $2x+1\le9$ and $5-x<7$ .
1. Hint 1
  Both conditions must hold at once.
2. Hint 2
  Solve each inequality separately, then retain their overlap.
Answer

$-2<x\le4$ .

Full solution

The first condition gives
$\begin{aligned}2x&\le8\\x&\le4\end{aligned}$

The second gives $-x<2$ , hence $x>-2$ after division by $-1$ .
Their intersection is $-2<x\le4$ .
The left endpoint fails the strict second condition, while the right endpoint satisfies both.

Answer

$-2<x\le4$ .

Key idea

An and condition keeps the intersection of the two solution sets.
Problem 3: Mark complete
Problem 4 A delivery allowance

A courier has a budget of 42 dollars for a trip. A fixed fee is 6 dollars and the distance fee is 1.5 dollars per kilometer. A coupon removes 3 dollars from the total. Find the allowed real distances $d\ge0$ in kilometers.
1. Hint 1
  The discounted total must stay at or below the budget.
2. Hint 2
  Write the total as fixed fee plus distance fee minus coupon.
Answer

$0\le d\le26$ kilometers.

Full solution

The budget condition is
$\begin{aligned}6+1.5d-3&\le42\\1.5d&\le39\\d&\le26\end{aligned}$

Combine this with $d\ge0$ .
At 26 kilometers the total is exactly 42 dollars.

Answer

$0\le d\le26$ kilometers.

Key idea

A practical inequality must be combined with the domain allowed by the situation.
Problem 4: Mark complete
Problem 5 Two entry routes

An entry rule accepts real $x$ when either $3(x-1)<3x-4$ or $8-2x\le0$ . Find the accepted set.
1. Hint 1
  An or rule accepts everything accepted by either piece.
2. Hint 2
  Check whether the first inequality can ever be true after its variable cancels.
Answer

$x\ge4$ .

Full solution

The first inequality reduces to $-3<-4$ , which is false, so it contributes no values.
The second gives
$\begin{aligned}-2x&\le-8\\x&\ge4\end{aligned}$

The union of the empty set and $[4,\infty)$ is $[4,\infty)$ .

Answer

$x\ge4$ .

Key idea

A false piece contributes nothing to an or condition.
Problem 5: Mark complete
Problem 6 A ratio threshold

Solve $\frac{x+4}{x-2}\le2$ for real $x$ , explaining how the sign of a multiplier affects the steps.
1. Hint 1
  The denominator is forbidden at zero and has different signs on its two sides.
2. Hint 2
  Compare the resulting linear inequality separately for $x>2$ and $x<2$ .
Answer

$x<2$ or $x\ge8$ ; $x=2$ is excluded.

Full solution

Exclude $x=2$ .
When $x>2$ , multiplying by $x-2$ keeps the direction.
$\begin{aligned}x+4&\le2x-4\\x&\ge8\end{aligned}$

When $x<2$ , multiplication reverses it: $x+4\ge2x-4$ , so $x\le8$ .
Every $x<2$ meets this condition.
The union is $x<2$ or $x\ge8$ .
Checking $x=0,4,8$ gives respectively true, false, and equality.

Answer

$x<2$ or $x\ge8$ ; $x=2$ is excluded.

Key idea

An unknown-sign denominator requires separate sign cases before clearing it in an inequality.
Problem 6: Mark complete
Problem 7 Two control settings

For real parameters $p$ and $q$ , solve $(p-1)x>q+2$ for real $x$ in all parameter cases.
1. Hint 1
  The sign of the whole coefficient decides the direction after division.
2. Hint 2
  If $p=1$ , test the remaining statement $0>q+2$ .
Answer

$x>\frac{q+2}{p-1}$ if $p>1$ ; $x<\frac{q+2}{p-1}$ if $p<1$ ; all real $x$ if $p=1,q<-2$ ; none if $p=1,q\ge-2$ .

Full solution

For $p>1$ , divide by the positive coefficient $p-1$ to obtain
$x>\frac{q+2}{p-1}.$
For $p<1$ , it is negative, so division gives
$x<\frac{q+2}{p-1}.$

At $p=1$ , no division is allowed.
$0>q+2$
This holds for every $x$ if $q<-2$ and for no $x$ if $q\ge-2$ , including the strict boundary $q=-2$ .

Answer

$x>\frac{q+2}{p-1}$ if $p>1$ ; $x<\frac{q+2}{p-1}$ if $p<1$ ; all real $x$ if $p=1,q<-2$ ; none if $p=1,q\ge-2$ .

Key idea

A parameter inequality has two sign cases and two possible outcomes when its coefficient vanishes.
Problem 7: Mark complete
Problem 8 A comparison of gaps

For real $a<b$ , a student claims $7-3a>7-3b$ . Is this correct? Justify the direction by comparing the difference of the two new quantities.
1. Hint 1
  A larger quantity minus a smaller one has positive difference.
2. Hint 2
  Compute $(7-3a)-(7-3b)$ and relate it to $b-a$ .
Answer

Yes; $7-3a>7-3b$ .

Full solution

The original order gives $b-a>0$ .
The new difference is
$\begin{aligned}(7-3a)-(7-3b)&=3(b-a)\end{aligned}$

This is positive, so the first new quantity is larger.
The negative scaling reverses the original order, and the added $7$ cancels from the comparison.

Answer

Yes; $7-3a>7-3b$ .

Key idea

Comparing the difference proves why negative scaling reverses order.
Problem 8: Mark complete
Problem 9 A proposed shortcut

A solver multiplies $x+1\le4$ by $0$ and replaces it with $0\le0$ . Does the replacement preserve the solution set? Explain with a specific input.
1. Hint 1
  Compare the inputs allowed by the original and the replacement.
2. Hint 2
  Find a number that is too large for the original but is still accepted by the replacement.
Answer

No; the original gives $x\le3$ , while the replacement accepts every real $x$ ; for example $x=4$ .

Full solution

Subtracting $1$ in the original gives
$x\le3.$

At $x=4$ , the original reads $5\le4$ , false, while $0\le0$ is true.
Thus multiplication by zero has admitted a value the original rejects.

Answer

No; the original gives $x\le3$ , while the replacement accepts every real $x$ ; for example $x=4$ .

Key idea

Multiplication by zero can erase the information that distinguishes accepted inputs from rejected ones.
Problem 9: Mark complete
Problem 10 A restricted comparison

On the domain $x>3$ , a solver replaces $x<5$ by $\frac{x}{x-3}<\frac5{x-3}$ . Are the two inequalities equivalent on this domain? Would the same replacement be equivalent on the larger domain $x\ne3$ ? Justify each verdict, using a counterexample if a verdict is no.
1. Hint 1
  Determine the sign of the divisor on each stated domain.
2. Hint 2
  For a counterexample, try an input below $3$ and compare the truth of the two inequalities.
Answer

Yes on $x>3$ ; no on $x\ne3$ , with $x=0$ as a counterexample.

Full solution

On $x>3$ , the divisor $x-3$ is positive.
Division and its inverse, multiplication, preserve the direction, so both inequalities have the same solution set there:
$3<x<5.$

On the larger domain the divisor can be negative.
At $x=0$ , the original statement $0<5$ is true, but the replacement becomes
$0<-\frac53,$
which is false.
Therefore the replacement is not equivalent on that domain.

Answer

Yes on $x>3$ ; no on $x\ne3$ , with $x=0$ as a counterexample.

Key idea

A domain restriction can establish the sign needed for a reversible inequality step.
Problem 10: Mark complete

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