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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 Four numbers to test

Which of the numbers $-\frac{4}{5}$ , $-\frac{3}{4}$ , $-0.7$ and $0$ make $t\le-0.75$ true?
1. Hint 1
  The symbol compares $t$ with the number on its right, and the line beneath it allows equality.
2. Hint 2
  Write each fraction as a decimal so that all four numbers can be compared with $-0.75$ .
3. Hint 3
  On the number line, a number to the left of $-0.75$ is less than $-0.75$ .
Answer

$-\frac{4}{5}$ and $-\frac{3}{4}$ .

Full solution

The symbol $\le$ means less than or equal to.
Therefore the inequality describes every number below $-0.75$ together with $-0.75$ itself.

As decimals, $-\frac{4}{5}$ is $-0.8$ and $-\frac{3}{4}$ is $-0.75$ .

The number $-0.8$ lies to the left of $-0.75$ on the number line, so it is less than $-0.75$ and $-\frac{4}{5}$ makes the inequality true.
The number $-\frac{3}{4}$ equals $-0.75$ , and equality is allowed, so it makes the inequality true as well.

The numbers $-0.7$ and $0$ lie to the right of $-0.75$ , so they are greater than $-0.75$ and make the inequality false.

Answer

$-\frac{4}{5}$ and $-\frac{3}{4}$ .

Key idea

A bar beneath an inequality symbol includes the boundary value.
Problem 1: Mark complete
Problem 2 A number line note

Describe the number-line graph of $-\frac{1}{3}\ge h$ in words, stating its boundary circle and shading direction.
1. Hint 1
  Read the comparison as a statement about $h$ , even though $h$ is on the right.
2. Hint 2
  Decide whether equality is permitted and which side contains the smaller numbers.
Answer

Closed circle at $-\frac{1}{3}$ ; shade to the left.

Full solution

The comparison says that $-\frac{1}{3}$ is greater than or equal to $h$ , so $h$ is less than or equal to $-\frac{1}{3}$ .
Equivalently,
$h\le-\frac{1}{3}$

The bar under the symbol includes the boundary, so the circle is closed.
Smaller numbers lie to the left, so the shading extends leftward from that circle.

Answer

Closed circle at $-\frac{1}{3}$ ; shade to the left.

Key idea

Read what an inequality says about the variable before choosing a graph direction.
Problem 2: Mark complete
Problem 3 A fraction in front

Solve $-\frac{3}{4}x+2<-4$ , then test one value inside your range and one outside it in the original inequality.
1. Hint 1
  Use the same balance moves as for an equation, clearing the constant before the coefficient.
2. Hint 2
  Subtracting the same number from both sides leaves the symbol as it is.
3. Hint 3
  To undo the coefficient, multiply by its reciprocal, and check the sign of that reciprocal first.
Answer

$x>8$ ; for example, $12$ makes the original true and $0$ makes it false.

Full solution

Subtract $2$ from both sides.
Subtraction does not change the symbol:
$-\frac{3}{4}x<-6$

Multiply both sides by $-\frac{4}{3}$ , the reciprocal of the coefficient.
It is a negative number, so reverse the symbol:
$x>8$

For an inside value, $x=12$ gives a left side of $-9+2$ , or $-7$ , and $-7<-4$ is true.
For an outside value, $x=0$ gives a left side of $2$ , and $2<-4$ is false.
Other correctly chosen inside and outside values are acceptable.

Answer

$x>8$ ; for example, $12$ makes the original true and $0$ makes it false.

Key idea

Multiplying by the reciprocal of a negative coefficient is multiplying by a negative number, so the symbol reverses.
Problem 3: Mark complete
Problem 4 A controller alert

A controller computes $1.2t-0.6$ from a temperature of $t$ degrees Celsius and issues an alert when the computed number is less than $-3$ . Find the temperature range that triggers the alert, describe its number-line graph, and test $t=-3$ and $t=0$ .
1. Hint 1
  The alert condition compares the complete computed expression with its limit.
2. Hint 2
  Clear the constant, then use the sign of the coefficient to decide whether the symbol changes.
3. Hint 3
  Test each temperature in the original expression.
Answer

$t<-2$ , temperatures below $-2$ degrees Celsius; open circle at $-2$ , shade left; $t=-3$ triggers an alert and $t=0$ does not.

Full solution

The alert condition is
$1.2t-0.6<-3$
Add $0.6$ to both sides, then divide by positive $1.2$ , keeping the symbol:
$\begin{aligned}1.2t&<-2.4\\t&<-2\end{aligned}$

The temperatures are below $-2$ degrees Celsius.
The boundary is excluded, so use an open circle at $-2$ and shade left.

At $t=-3$ , the computation gives
$1.2(-3)-0.6=-4.2$
This is less than $-3$ , so the alert occurs.
At $t=0$ , the computation gives $-0.6$ , which is not less than $-3$ , so no alert occurs.

Answer

$t<-2$ , temperatures below $-2$ degrees Celsius; open circle at $-2$ , shade left; $t=-3$ triggers an alert and $t=0$ does not.

Key idea

A strict condition excludes its boundary, and test values check the solved direction.
Problem 4: Mark complete
Problem 5 A stretched strip

A strip is $l$ cm long, where $l$ is greater than $3$ . First $3$ cm is trimmed off, then the remaining piece is stretched to $\frac{3}{2}$ of its length. The stretched piece must be at least $15$ cm long. Find the permitted original lengths and describe their number-line graph.
1. Hint 1
  Write the final length after both changes, keeping the length of the remaining piece in parentheses.
2. Hint 2
  Undo the stretch before undoing the trim.
Answer

$l\ge13$ ; original lengths of at least $13$ cm; closed circle at $13$ , shade right.

Full solution

After the trim the piece is $l-3$ cm long, and after the stretch it is $\frac{3}{2}(l-3)$ cm, so the condition is
$\frac{3}{2}(l-3)\ge15$

Multiply both sides by $\frac{2}{3}$ , a positive number, so the symbol stays.
Then add $3$ to both sides:
$\begin{aligned}l-3&\ge10\\l&\ge13\end{aligned}$

The boundary $13$ is included, so the graph has a closed circle at $13$ with shading to the right.
At $l=15$ , the remaining piece is $12$ cm and the stretched piece is $18$ cm, which works.
At $l=13$ , the stretched piece is exactly $15$ cm, which is allowed.
At $l=9$ , the remaining piece is $6$ cm and the stretched piece is only $9$ cm, which is too short.

Answer

$l\ge13$ ; original lengths of at least $13$ cm; closed circle at $13$ , shade right.

Key idea

A positive scale factor preserves the direction of an inequality.
Problem 5: Mark complete
Problem 6 A written condition

Find every number satisfying $4-(2-3x)+x\le14$ . Describe the number-line graph and check $x=2$ and $x=4$ .
1. Hint 1
  The group must be handled before the variable term is isolated.
2. Hint 2
  The outside minus changes each sign in the group; then collect the matching terms.
3. Hint 3
  Check the sign of the coefficient before you divide; negatives inside the group do not decide it.
Answer

$x\le3$ ; closed circle at $3$ , shade left; $2$ is a solution and $4$ is not.

Full solution

The outside minus changes each sign in the group, so $-(2-3x)$ is $-2+3x$ .
The left side becomes $4-2+3x+x$ , or $4x+2$ .
Subtract $2$ from both sides:
$4x\le12$

Divide both sides by $4$ .
It is a positive number, so the symbol stays, even though the original inequality contains minus signs:
$x\le3$
The graph has a closed circle at $3$ and shading to the left.

At $x=2$ , the group is $2-6$ , or $-4$ , so the original left side is $4-(-4)+2=10$ , and $10\le14$ is true.
At $x=4$ , the group is $2-12$ , or $-10$ , so the left side is $4-(-10)+4=18$ , and $18\le14$ is false.

Answer

$x\le3$ ; closed circle at $3$ , shade left; $2$ is a solution and $4$ is not.

Key idea

Clean up a grouped expression first; then the sign of the number you divide by, not the signs inside the group, decides whether the symbol flips.
Problem 6: Mark complete
Problem 7 A transit card

A prepaid transit card holds $18$ dollars, and each ride costs $4.50$ dollars. The card is allowed a negative balance, but the balance may never be less than $-9$ dollars. If $n$ is the whole number of rides taken, give every possible value of $n$ .
1. Hint 1
  The balance is the starting amount minus the total cost of the rides.
2. Hint 2
  After solving the balance condition, restrict the answer to whole numbers.
3. Hint 3
  A whole-number count can be zero.
Answer

$n=0,1,2,3,4,5,6$ .

Full solution

After $n$ rides, the balance in dollars is $18-4.5n$ , and it must be at least $-9$ :
$18-4.5n\ge-9$
Subtract $18$ from both sides and divide by $-4.5$ , reversing the symbol:
$\begin{aligned}-4.5n&\ge-27\\n&\le6\end{aligned}$

A number of rides is zero or a positive whole number, so the possible values are $0$ through $6$ .

Six rides leave a balance of $18-27=-9$ dollars, which is allowed.
Seven leave $18-31.5=-13.5$ dollars, which is below the limit.

Answer

$n=0,1,2,3,4,5,6$ .

Key idea

A real-world count restricts a solved range to the whole numbers that make sense.
Problem 7: Mark complete
Problem 8 A proposed graph

A student describes the solutions of $18-3x<0$ as a closed circle at $6$ with shading to the left. Is the description correct? Explain, testing $x=0$ and $x=7$ in the original inequality, and describe the graph of the solutions yourself.
1. Hint 1
  Find the solution range before judging the proposed picture.
2. Hint 2
  The sign of the coefficient controls the direction after division, while strictness controls the circle.
Answer

No; $x>6$ , open circle at $6$ , shade right; $0$ is not a solution and $7$ is.

Full solution

Subtract $18$ from both sides, then divide by $-3$ , a negative number, and reverse the symbol:
$\begin{aligned}-3x&<-18\\x&>6\end{aligned}$

The inequality is strict, so $6$ is excluded and the circle is open.
The solutions are the numbers greater than $6$ , which lie to the right, so the shading runs right from the circle.
Both parts of the proposed graph need correction.

At $x=0$ , the original left side is $18$ , which is not less than zero.
At $x=7$ , it is $18-21=-3$ , which is less than zero.
So $0$ , which the proposed left shading covers, is not a solution, and $7$ , which it leaves out, is: the proposed shading points the wrong way.

Answer

No; $x>6$ , open circle at $6$ , shade right; $0$ is not a solution and $7$ is.

Key idea

The boundary circle and shading direction answer different questions about a solution range.
Problem 8: Mark complete
Problem 9 Tomas's symbol

Tomas says that placing $\ge$ in the box makes $-0.25x\;\square\;1.5$ true for exactly the numbers with $x\le-6$ . Is he right? Explain, and check one value that satisfies $x\le-6$ and one that does not.
1. Hint 1
  The completed comparison must be true for every number with $x\le-6$ and false for every other number.
2. Hint 2
  Find the value of $-0.25x$ at the boundary $x=-6$ to decide whether equality belongs.
3. Hint 3
  Then think about what multiplying both sides of $x\le-6$ by the negative number $-0.25$ does to the symbol.
Answer

Yes, $\ge$ is correct; for example, $-8$ makes $-0.25x\ge1.5$ true and $0$ makes it false.

Full solution

At the boundary $x=-6$ , the left side is
$-0.25(-6)=1.5$
The boundary is included, so the symbol must allow equality.

Multiplying both sides of $x\le-6$ by $-0.25$ , a negative number, reverses the order:
$-0.25x\ge1.5$

Dividing both sides by $-0.25$ undoes that step and reverses the symbol back, so $-0.25x\ge1.5$ holds for exactly the numbers with $x\le-6$ .
Tomas is right.

For a check, $x=-8$ satisfies $x\le-6$ , and $-0.25(-8)$ is $2$ , so $2\ge1.5$ is true.
The number $x=0$ does not satisfy $x\le-6$ , and it gives $0\ge1.5$ , which is false.
Other correctly chosen values are acceptable.

Answer

Yes, $\ge$ is correct; for example, $-8$ makes $-0.25x\ge1.5$ true and $0$ makes it false.

Key idea

Multiplying both sides of an inequality by a negative number reverses the symbol, and an included boundary stays included.
Problem 9: Mark complete
Problem 10 Dario's reading

A number-line graph has an open circle at $-7$ and shading to the right. Dario says the smallest number the graph includes is $-6$ , because $-7$ is left out. Is he right? Explain.
1. Hint 1
  The shaded part covers every number greater than the boundary, not only the integers.
2. Hint 2
  Look for an included number between $-7$ and $-6$ , and compare it with $-6$ .
Answer

No; for example, $-6.5$ is included and is less than $-6$ .

Full solution

The open circle at $-7$ and the shading to the right describe every number greater than $-7$ :
$x>-7$

The shading covers every number to the right of $-7$ , not only the integers.
For example, $-6.5$ is greater than $-7$ , so the graph includes it.

Since $-6.5$ is less than $-6$ , the number $-6$ is not the smallest number the graph includes, so Dario is wrong.
Any number between $-7$ and $-6$ is also a correct example.

Answer

No; for example, $-6.5$ is included and is less than $-6$ .

Key idea

An open circle leaves out the boundary, but not the numbers just beyond it.
Problem 10: Mark complete

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