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Inequality Basics: 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 Which number is larger

Compare $-\frac74$ and $-1.8$ by placing $<$ or $>$ between them.
1. Hint 1
  Put both numbers in a form that makes their positions easy to compare.
2. Hint 2
  Among negative numbers, the one closer to zero is greater.
Answer

$-\frac74>-1.8$ .

Full solution

Convert the fraction exactly:
$-\frac74=-1.75$
Since $-1.75$ is closer to zero than $-1.8$ ,
$-\frac74>-1.8$
As a check, their difference is $-1.75-(-1.8)=0.05$ , which is positive.

Answer

$-\frac74>-1.8$ .

Key idea

Comparing equivalent decimal forms can clarify the order of negative fractions.
Problem 1: Mark complete
Problem 2 An interval from two clues

An interval extends without bound in one direction and has its only real endpoint at 5. It contains 9 and excludes 5. Write the interval.
1. Hint 1
  The member you are given tells you which side of the endpoint the interval lies on.
2. Hint 2
  Match each end of the interval to its fence: one end is a number that is not a member, the other is not a number at all.
Answer

$(5,\infty)$ .

Full solution

The member 9 is larger than the endpoint 5, so the interval extends right without bound.
Because 5 itself is excluded, the real endpoint takes a parenthesis, and an unbounded end always takes one too:
$(5,\infty)$
This contains 9 and excludes 5, as required.

Answer

$(5,\infty)$ .

Key idea

Endpoint membership and one interior value can identify the direction and fences of an unbounded interval.
Problem 2: Mark complete
Problem 3 The lift and the wind

A ski lift runs only when the wind speed $w$ , measured in miles per hour, is at most 35. A wind speed is never negative. Give the possible values of $w$ while the lift is running, as an inequality and as an interval.
1. Hint 1
  Decide whether a wind speed of exactly 35 miles per hour still lets the lift run.
2. Hint 2
  The condition carries a lower bound as well as an upper one, and each end takes the fence its own symbol calls for.
Answer

$0\le w\le35$ ; $[0,35]$ .

Full solution

Two conditions hold at once.
A wind speed is never negative, which gives $w\ge0$ , and the lift runs when the speed is at most 35, which allows exactly 35 and everything below it, giving $w\le35$ .

Written as one line with the smaller bound on the left, that is
$0\le w\le35$
Both bounds are allowed, so both ends take brackets, and the interval is $[0,35]$ .
A speed of 35 still runs the lift, while 36 does not.

Answer

$0\le w\le35$ ; $[0,35]$ .

Key idea

A phrase such as at most keeps its boundary, so that end of the interval takes a bracket.
Problem 3: Mark complete
Problem 4 Two acceptance bands

The figure shows the numbers accepted by devices A and B. Give the numbers $x$ accepted by both devices as a compound inequality and an interval, and shade them on the blank number line C.

The numbers accepted by device A and by device B, with line C left blank.

Text description of this figure

Three number lines are drawn one above another and labeled A, B, and C. Each runs from negative 6 to 5, with a tick and a label at every integer and an arrowhead at each end. On line A an open circle sits at negative 5 and a filled circle sits at 1, and the stretch between them is shaded. On line B a filled circle sits at negative 2 and an open circle sits at 4, and the stretch between them is shaded. Line C carries no circles and no shading.
1. Hint 1
  A number must lie in both shaded bands to be accepted by both devices.
2. Hint 2
  Use the stricter lower bound and the stricter upper bound, then check those endpoints in both bands.
Answer

$-2\le x\le1$ ; $[-2,1]$ ; filled circles at $-2$ and 1, shaded between.

Full solution

A accepts numbers above $-5$ through 1, while B accepts numbers from $-2$ up to but not including 4.
Their common part starts at $-2$ , which A accepts, and ends at 1, which B accepts.

Thus the common condition is
$-2\le x\le1$
The interval is $[-2,1]$ .
Shade between filled circles at $-2$ and 1 on C; a number such as 0 checks in both original bands.

Answer

$-2\le x\le1$ ; $[-2,1]$ ; filled circles at $-2$ and 1, shaded between.

Key idea

Two acceptance bands overlap where both lower and upper conditions hold.
Problem 4: Mark complete
Problem 5 The second display

A dial reading $x$ can be any real number with $-4<x\le2$ . A second display reports $r=-3x$ . Give the possible values of $r$ as an interval, and draw them on the blank number line in the figure.

A blank number line from $-8$ to $14$ , ticked every 2 units.

Text description of this figure

A single number line with an arrowhead at each end. It carries a tick and a label every 2 units, at negative 8, negative 6, negative 4, negative 2, 0, 2, 4, 6, 8, 10, 12, and 14. The line carries no circles and no shading.
1. Hint 1
  A negative scale reverses the order of the endpoint readings.
2. Hint 2
  Track which original endpoint is included when computing the two new bounds.
Answer

$[-6,12)$ ; filled circle at $-6$ , open circle at 12, shaded between.

Full solution

The condition $-4<x\le2$ is two conditions at once.
From $x\le2$ , multiplying by $-3$ reverses the direction and gives $-3x\ge-6$ .
From $x>-4$ , the same multiplier gives $-3x<12$ .
Both hold together, so
$-6\le r<12.$
Dividing any $r$ in that range by $-3$ gives back an allowed $x$ , so every one of those values does occur.

The possible readings are $[-6,12)$ .
Draw a filled circle at $-6$ and an open circle at 12, shading between.
Input 0 is permitted and gives output 0 inside the result.

Answer

$[-6,12)$ ; filled circle at $-6$ , open circle at 12, shaded between.

Key idea

A negative scale reverses endpoint order while preserving which endpoint values are included.
Problem 5: Mark complete
Problem 6 A gauge with a cutoff

A gauge takes a real input $x$ and reports $r=x-5$ . A monitor rejects every reading with $r\ge2$ and accepts all the others. Describe the accepted values of $x$ in words, as an inequality, and as an interval, and draw them on the blank number line in the figure.

A blank number line from 1 to 10, ticked at every integer.

Text description of this figure

A single number line with an arrowhead at each end. It carries a tick and a label at every integer from 1 through 10. The line carries no circles, no shading, and no other marks.
1. Hint 1
  Find the input that puts the reading exactly at the cutoff value, then decide which side of that input survives.
2. Hint 2
  Now decide whether the boundary input itself is accepted, and let that choose its fence.
Answer

Numbers less than 7; $x<7$ ; $(-\infty,7)$ ; open circle at 7, shaded left.

Full solution

A reading is rejected exactly when $r\ge2$ , that is when
$x-5\ge2$
Adding 5 to both sides leaves the direction alone, so the rejected inputs are those with $x\ge7$ .

Every other input is accepted, so the accepted values are the numbers less than 7:
$x<7$
The boundary 7 is itself rejected, so it is excluded, and the accepted values run down without a lower bound, giving $(-\infty,7)$ .

Draw an open circle at 7 and shade left with an arrow.
As a check, $x=6$ gives $r=1$ , which is accepted, while $x=7$ gives $r=2$ , which is rejected.

Answer

Numbers less than 7; $x<7$ ; $(-\infty,7)$ ; open circle at 7, shaded left.

Key idea

The accepted values are everything the rejection rule leaves out, so an inclusive rejection boundary becomes an excluded acceptance boundary.
Problem 6: Mark complete
Problem 7 A shift and a scale

Two readings satisfy $a<b$ . A processor subtracts 8 from each reading and then divides each result by $-2$ . State the comparison between the two processed readings and say whether their order has changed from $a<b$ , justifying each step.
1. Hint 1
  Handle the two moves one after the other, and ask what each one does to the order.
2. Hint 2
  Compare the shifted readings first, then look at the sign of the divisor before writing the second comparison.
Answer

$\frac{a-8}{-2}>\frac{b-8}{-2}$ ; the order has reversed.

Full solution

Subtracting 8 shifts both readings equally, leaving their difference unchanged:
$(b-8)-(a-8)=b-a$
This is positive, so $a-8<b-8$ and the first move keeps the direction.

Dividing by $-2$ scales by a negative number, which reverses the direction:
$\frac{a-8}{-2}>\frac{b-8}{-2}$
So the processed readings compare the opposite way round from $a<b$ .

Take $a=0$ and $b=2$ as a check.
Subtracting 8 gives $-8<-6$ , the same direction.
Dividing by $-2$ gives 4 and 3, and $4>3$ , so the order has reversed.

Answer

$\frac{a-8}{-2}>\frac{b-8}{-2}$ ; the order has reversed.

Key idea

An equal shift leaves the order alone, so a negative scale applied after it is what reverses the comparison.
Problem 7: Mark complete
Problem 8 A list of members

A student lists $-1$ , 0 and 1 and claims these are all the real numbers in $[-1,1]$ . Is the claim correct? If not, give two numbers of the interval that are missing from the list, and say what kind of number the list leaves out.
1. Hint 1
  Decide whether an interval of real numbers can be listed one member at a time.
2. Hint 2
  Look for a value between two of the listed numbers and test it against both bounds.
Answer

No; for example $\frac14$ and $\frac12$ are missing, and any list of integers leaves out the non-integer values between them.

Full solution

The claim is not correct.
Both $\frac14$ and $\frac12$ satisfy the bounds:
$-1\le\frac14\le1$ $-1\le\frac12\le1$
Neither appears in the list.
Between any two such values lie further real numbers, so a list of integers leaves out every non-integer value the interval holds.

Answer

No; for example $\frac14$ and $\frac12$ are missing, and any list of integers leaves out the non-integer values between them.

Key idea

An interval of real numbers holds every value between its bounds, not only the integers a student can list.
Problem 8: Mark complete
Problem 9 Swapped fences

A student says that $[-1,2)$ and $(-1,2]$ have exactly the same members except for the two endpoints. Is the claim correct? Describe precisely what changes.
1. Hint 1
  Compare the endpoint fences separately from the interior.
2. Hint 2
  Test each endpoint in both interval records.
Answer

Yes; $-1$ belongs only to $[-1,2)$ , and 2 belongs only to $(-1,2]$ .

Full solution

Both intervals contain every number strictly between $-1$ and 2, and both exclude every number outside those bounds.
The first interval includes $-1$ and excludes 2; the second makes the opposite endpoint choices.

Therefore membership changes at exactly those two numbers.
A shared interior value such as 0 remains a member of both.

Answer

Yes; $-1$ belongs only to $[-1,2)$ , and 2 belongs only to $(-1,2]$ .

Key idea

Changing endpoint fences changes endpoint membership without changing the interior of an interval.
Problem 9: Mark complete
Problem 10 A zero multiplier

A student starts with $a<b$ and multiplies both sides by zero, claiming the result is still a strict inequality in one direction or the other. Is this possible? State the comparison between the two products and justify it.
1. Hint 1
  The rules for keeping or reversing a direction were stated for positive and for negative multipliers.
2. Hint 2
  Work out each product separately, then ask which of $<$ , $>$ and $=$ can hold between them.
Answer

No; both products are 0, and the comparison is $0=0$ .

Full solution

Both products equal zero, so the resulting comparison is
$0=0$
Neither $0<0$ nor $0>0$ is true.
Multiplication by zero removes the distinction between the original numbers rather than preserving or reversing a strict order.

Answer

No; both products are 0, and the comparison is $0=0$ .

Key idea

Multiplication by zero collapses unequal real numbers to equal products.
Problem 10: Mark complete

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