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Solving 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 Decimal coefficients on both sides

Solve $1.2x-3.5<0.7x+1$ over the real numbers.
1. Hint 1
  Decimal coefficients need no new rule: the same inverse moves isolate the variable.
2. Hint 2
  Gather the variable terms on the left, where the coefficient is larger, then divide.
Answer

$x<9$ , the interval $(-\infty,9)$ .

Full solution

Subtracting $0.7x$ from both sides keeps the direction:
$0.5x-3.5<1$
Adding 3.5 to both sides gives
$0.5x<4.5$
Dividing by the positive number 0.5 leaves the symbol pointing the same way:
$x<9$

Clearing the decimals first is a second route: multiplying both sides by 10 gives $12x-35<7x+10$ , which leads to $5x<45$ and the same bound.

At $x=9$ both original sides equal 7.3, so the strict symbol leaves 9 out.
At $x=0$ the comparison reads $-3.5<1$ , true, while at $x=10$ it reads $8.5<8$ , false.

Answer

$x<9$ , the interval $(-\infty,9)$ .

Key idea

Decimal coefficients follow the same inverse moves, and a positive divisor keeps the direction.
Problem 1: Mark complete
Problem 2 An expression in parentheses

Solve $8-3(2x-1)\ge29$ over the real numbers.
1. Hint 1
  Remove the parentheses first, taking care with the sign of each product.
2. Hint 2
  Isolate the variable term, then watch what dividing by its negative coefficient does to the symbol.
Answer

$x\le-3$ , the interval $(-\infty,-3]$ .

Full solution

Distributing $-3$ across the parentheses gives
$8-6x+3\ge29$
Combining the constants on the left gives
$11-6x\ge29$
Subtracting 11 from both sides keeps the direction:
$-6x\ge18$

Dividing both sides by the negative number $-6$ reverses the symbol:
$x\le-3$
The earlier subtraction reversed nothing; only the negative divisor does.

At $x=-3$ the left side is $8-3(-7)=29$ , so the endpoint is included.
At $x=-4$ it is 35, which is at least 29, while at $x=0$ it is 11, which is not.

Answer

$x\le-3$ , the interval $(-\infty,-3]$ .

Key idea

Distribute first, and reverse the symbol only at a step that multiplies or divides both sides by a negative number.
Problem 2: Mark complete
Problem 3 A fraction and a linear expression

Solve $\frac{x-2}{2}<2x+5$ over the real numbers.
1. Hint 1
  Clear the positive denominator first.
2. Hint 2
  The larger variable coefficient is on the right, so remove the left variable term.
Answer

$x>-4$ .

Full solution

Multiply by 2, preserving the direction:
$x-2<4x+10$
Subtract $x$ from both sides and then subtract 10:
$-12<3x$
Dividing by positive 3 gives $-4<x$ , or
$x>-4$

At $x=0$ , the original comparison is $-1<5$ , true.
At $x=-5$ , it is $-7/2<-5$ , false.
No step scales by a negative number.

Answer

$x>-4$ .

Key idea

Collecting on the side with the larger coefficient can leave a positive final divisor.
Problem 3: Mark complete
Problem 4 Comparing two rental offers

For a rental lasting $h$ hours, offer A costs $8+3h$ dollars and offer B costs $14+2h$ dollars. Any real duration $h\ge0$ is allowed. For which durations does A cost no more than B? Give an interval.
1. Hint 1
  Compare the complete costs, including the starting charges.
2. Hint 2
  Solve the cost comparison and combine it with the nonnegative duration restriction.
Answer

$[0,6]$ hours.

Full solution

Offer A costs no more when
$8+3h\le14+2h$
Subtracting $2h$ and 8 gives
$h\le6$
Together with $h\ge0$ , this gives the interval $[0,6]$ hours.

At 6 hours both offers cost 26 dollars.
At 0 hours A costs 8 dollars and B costs 14 dollars; at 7 hours A costs 29 dollars and B costs 28 dollars, confirming the bound.

Answer

$[0,6]$ hours.

Key idea

A cost comparison must be combined with the allowed domain of the duration.
Problem 4: Mark complete
Problem 5 Sensor settings

A sensor with a real setting $x\ge0$ reports $r=\frac{5-2x}{3}$ . It accepts readings from $-1$ inclusive to 3 exclusive. Find all allowed settings that produce accepted readings, as an interval.
1. Hint 1
  Put the reported expression between the two acceptance bounds.
2. Hint 2
  Operate on all three parts, reversing both symbols when dividing by a negative, then apply $x\ge0$ .
Answer

$[0,4]$ .

Full solution

The reading condition is
$-1\le\frac{5-2x}{3}<3$
Multiplying by 3 and subtracting 5 throughout gives
$-8\le-2x<4$
Dividing all parts by $-2$ reverses both symbols, giving
$4\ge x>-2$

Rewriting in increasing order gives $-2<x\le4$ .
With the setting restriction $x\ge0$ , the answer is $[0,4]$ .
Setting 4 gives reading $-1$ , included; setting 0 gives $5/3$ , also accepted.

Answer

$[0,4]$ .

Key idea

A compound output requirement and an input restriction must both hold.
Problem 5: Mark complete
Problem 6 Two entrance rules

A real score $s$ is accepted when $s-4\ge2$ or $3s+1>10$ . Find the set of accepted scores, describe it as simply as you can, and justify that description.
1. Hint 1
  An or condition keeps a score that satisfies either rule.
2. Hint 2
  Solve each rule, then notice whether one accepted ray is contained in the other.
Answer

$(3,\infty)$ , that is $s>3$ .

Full solution

The first rule gives $s\ge6$ .
The second gives
$3s>9$ $s>3$
Every score at least 6 is already greater than 3, so the union is just $(3,\infty)$ .

Score 4 passes only the second rule and is accepted.
Score 6 passes both.
Score 3 passes neither, so the endpoint stays open.

Answer

$(3,\infty)$ , that is $s>3$ .

Key idea

The union of nested rays is the larger ray.
Problem 6: Mark complete
Problem 7 A parameter limit

A positive real input $x$ must satisfy $2(x+a)\le2x+6$ , where $a$ is fixed. Determine which values of $a$ permit an input, and give the permitted input set in each case.
1. Hint 1
  Expand and compare the variable terms before attempting to isolate the input.
2. Hint 2
  After the input cancels, the remaining comparison decides whether the positive-input condition is allowed or impossible.
Answer

If $a\le3$ , inputs are $(0,\infty)$ ; if $a>3$ , there are no inputs.

Full solution

Expand and subtract $2x$ :
$2a\le6$ $a\le3$
This condition is independent of $x$ .
If it holds, every positive input satisfies the original comparison.
If it fails, no input can change that failure.

For example, at $a=3$ both original sides agree for every $x$ , while $a=4$ makes the left side 2 larger for every $x$ .
The required positivity then gives precisely $(0,\infty)$ in the first case.

Answer

If $a\le3$ , inputs are $(0,\infty)$ ; if $a>3$ , there are no inputs.

Key idea

When the variable cancels, a parameter condition decides whether the remaining domain survives.
Problem 7: Mark complete
Problem 8 A requested solution set

Write one inequality of the form $ax+b<c$ , where $a$ is negative, whose solution set is $x>-2$ . Show that your inequality has exactly this solution set.
1. Hint 1
  Build backward from the desired solution using a negative scale.
2. Hint 2
  After multiplying by a negative number, add a constant to both sides.
Answer

For example, $-3x+1<7$ .

Full solution

Start with $x>-2$ .
Multiplying by $-3$ reverses the direction:
$-3x<6$
Adding 1 gives
$-3x+1<7$

Conversely, subtracting 1 and dividing by $-3$ in the constructed inequality returns $x>-2$ .
Thus no solutions were added or lost, and the coefficient of $x$ is negative as required.

Answer

For example, $-3x+1<7$ .

Key idea

Reversible inequality operations can construct a problem with a chosen solution set.
Problem 8: Mark complete
Problem 9 Scaling by two negatives

A student multiplies both sides of an inequality by $-2$ and then divides both sides by $-5$ . The student says the final comparison symbol points the same way as the original. Is this correct? Explain.
1. Hint 1
  Track the direction at each negative scaling step.
2. Hint 2
  Compare the two-step process with multiplication by their combined factor.
Answer

Yes; the direction reverses twice and returns to its original orientation.

Full solution

Multiplication by $-2$ reverses the comparison.
Division by $-5$ reverses it again.
The combined scale factor is
$\frac{-2}{-5}=\frac25$
This is positive, agreeing with preservation of the original direction.

For example, $1<3$ becomes $-2>-6$ and then $2/5<6/5$ .
Inclusive comparisons behave the same way, with equality still included.

Answer

Yes; the direction reverses twice and returns to its original orientation.

Key idea

Two negative scale changes combine to a positive scale change.
Problem 9: Mark complete
Problem 10 And versus or

A student claims that replacing "and" by "or" between two different-looking linear inequalities must change the solution set. Is the claim correct? Justify your answer with a specific pair of inequalities.
1. Hint 1
  Ask what each joining word demands of a number before deciding the two demands must differ.
2. Hint 2
  Choose one inequality and produce another by a reversible operation that preserves its solutions.
Answer

No; for example, $2x+6>0$ and $x>-3$ give $(-3,\infty)$ with either joining word.

Full solution

Take $2x+6>0$ and $x>-3$ .
The first simplifies as follows:
$2x>-6$ $x>-3$
Both inequalities therefore describe the same ray.

Requiring both of two identical solution conditions, or requiring either one, gives exactly $(-3,\infty)$ .
Thus different printed expressions do not ensure different solution sets, so the claim is not correct.

Answer

No; for example, $2x+6>0$ and $x>-3$ give $(-3,\infty)$ with either joining word.

Key idea

And and or give the same result when the two conditions have the same solution set.
Problem 10: Mark complete

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