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Unit notes and practice questions

Grade 9 Mathematics unit 4: Solving Inequalities

Everything here is for unit 4 of the Grade 9 Mathematics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

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5
Cards
10
Practice questions on unit 4
130–161
The textbook pages for unit 4

What unit 4 covers

  1. 1.Revision on Linear Inequalities in One Variable
  2. 2.Systems of Linear Inequalities in Two Variables
  3. 3.Inequalities Involving Absolute Value
  4. 4.Quadratic Inequalities
  5. 5.Applications

Short notes for unit 4

The cards filed under this chapter on the Grade 9 Mathematics page. Open one at a time.

∣k∣<c  ⟺  −c<k<cand∣k∣>c  ⟺  k>c or k<−c|k| < c \iff -c < k < c \quad \text{and} \quad |k| > c \iff k > c \text{ or } k < -c
kk
algebraic expressiondimensionlessdimensionless
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positive real number constantdimensionlessdimensionless

When you use it

Use when solving linear inequalities where an expression is enclosed in absolute value bars.

Watch out

Treating a greater-than absolute value inequality as a single compound inequality instead of two separate disjoint conditions.

Drafted from Grade 9 Mathematics, pages 130-161, then checked twice before it went up

When you use it

Use when finding the solution region of a linear inequality or system of linear inequalities on the coordinate plane.

Watch out

Selecting a test point that lies directly on the boundary line fails to determine which side of the region to shade.

Drafted from Grade 9 Mathematics, pages 130-161, then checked twice before it went up

IntervalSet InequalityBoundary Dots
(a,b)(a, b)a<x<ba < x < bOpen dots at aa and bb
[a,b][a, b]a≤x≤ba \le x \le bSolid dots at aa and bb
(a,b](a, b]a<x≤ba < x \le bOpen dot at aa, solid dot at bb
[a,b)[a, b)a≤x<ba \le x < bSolid dot at aa, open dot at bb

When you use it

Use when expressing the solution set of an inequality on a number line or in interval format.

Watch out

Using square brackets or solid dots for strict inequalities includes boundary points that are not part of the solution.

Drafted from Grade 9 Mathematics, pages 130-161, then checked twice before it went up

When you use it

Use when simplifying or solving an algebraic inequality by multiplying or dividing both sides by a negative value.

Watch out

Failing to reverse the inequality direction when multiplying or dividing by a negative number changes the solution set.

Drafted from Grade 9 Mathematics, pages 130-161, then checked twice before it went up

When you use it

Use when solving factored inequalities of the form m times n greater than 0 or m times n less than 0.

Watch out

Assuming both factors have the same sign when their product is negative. They must have opposite signs.

Drafted from Grade 9 Mathematics, pages 130-161, then checked twice before it went up

Practice questions on unit 4

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

According to the properties of inequalities, if a > b and m < 0, which of the following is true?

  1. A.ma > mb
  2. B.ma < mb
  3. C.a + m < b + m
  4. D.a - m < b - m
Show the answer and why

Answer: B ma < mb

Why

When both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality sign is reversed. Since m < 0, multiplying a > b by m gives ma < mb.

Question 2

Which interval notation represents the set of all real numbers x such that -3 ≤ x < 2?

  1. A.[-3, 2]
  2. B.(-3, 2]
  3. C.[-3, 2)
  4. D.(-3, 2)
Show the answer and why

Answer: C [-3, 2)

Why

A square bracket is used for included endpoints (≤ or ≥) and a round parenthesis is used for excluded endpoints (< or >). Therefore, -3 ≤ x < 2 is written as [-3, 2).

Question 3

What is the solution set of the absolute value inequality |6x + 8| > -2?

  1. A.No solution
  2. B.x > -1
  3. C.x < -1
  4. D.All real numbers
Show the answer and why

Answer: D All real numbers

Why

The absolute value of any real number is always non-negative (greater than or equal to 0). Since 0 is always greater than -2, the inequality holds true for all real numbers.

Question 4

Solve the linear inequality 3x - 9 > 7x - 1.

  1. A.x > -2
  2. B.x < -2
  3. C.x < 2
  4. D.x > 2
Show the answer and why

Answer: B x < -2

Why

Subtract 7x from both sides: -4x - 9 > -1. Add 9 to both sides: -4x > 8. Dividing both sides by -4 reverses the inequality symbol, giving x < -2.

Question 5

What is the solution of the absolute value inequality |x + 2| < 4?

  1. A.-6 < x < 2
  2. B.x < -6 or x > 2
  3. C.-2 < x < 6
  4. D.x > -6
Show the answer and why

Answer: A -6 < x < 2

Why

Using the theorem for |k| < c, we write -4 < x + 2 < 4. Subtracting 2 from all parts gives -4 - 2 < x < 4 - 2, which simplifies to -6 < x < 2.

Question 6

What is the solution in interval notation for the quadratic inequality (x - 3)(x - 1) < 0?

  1. A.(-infinity, 1) U (3, infinity)
  2. B.(1, 3)
  3. C.[-1, 3]
  4. D.(-3, -1)
Show the answer and why

Answer: B (1, 3)

Why

For the product of two factors to be negative, one factor must be positive and the other negative. This occurs when x - 1 > 0 and x - 3 < 0, which gives 1 < x < 3, or the interval (1, 3).

Question 7

What is the solution to the absolute value inequality |-5x - 7| ≤ 0?

  1. A.No solution
  2. B.All real numbers
  3. C.x = 7/5
  4. D.x = -7/5
Show the answer and why

Answer: D x = -7/5

Why

An absolute value cannot be strictly less than 0, so the inequality only holds when |-5x - 7| = 0. Solving -5x - 7 = 0 gives -5x = 7, so x = -7/5.

Question 8

What is the solution set of the quadratic inequality x2 - x ≥ 12?

  1. A.[-3, 4]
  2. B.(-infinity, -4] U [3, infinity)
  3. C.(-infinity, -3] U [4, infinity)
  4. D.[-4, 3]
Show the answer and why

Answer: C (-infinity, -3] U [4, infinity)

Why

Rewrite as x2 - x - 12 ≥ 0 and factor to get (x - 4)(x + 3) ≥ 0. The product is non-negative when x ≤ -3 or x ≥ 4, which is (-∞, -3] ∪ [4, ∞).

Question 9

Kebede has saved 6,000 birr for a tractor. He plans to save 1,000 birr each month until his total savings are more than 80,000 birr. If m is the number of months, which inequality and solution represent his saving plan?

  1. A.6,000 + 1,000m > 80,000; m > 74
  2. B.6,000 + 1,000m <= 80,000; m <= 74
  3. C.1,000m - 6,000 > 80,000; m > 86
  4. D.6,000 + 1,000m >= 80,000; m >= 80
Show the answer and why

Answer: A 6,000 + 1,000m > 80,000; m > 74

Why

Kebede starts with 6,000 birr and adds 1,000m birr. To have more than 80,000 birr: 6,000 + 1,000m > 80,000. Subtracting 6,000 gives 1,000m > 74,000, so m > 74.

Question 10

Chaltu has 50 birr in her phone account. She is charged 0.20 birr for each text message. If her total bill can be at most 80 birr, what is the maximum number of text messages she can send?

  1. A.300
  2. B.150
  3. C.400
  4. D.120
Show the answer and why

Answer: B 150

Why

At most means less than or equal to: 0.2m + 50 ≤ 80. Subtracting 50 gives 0.2m ≤ 30. Dividing by 0.2 gives m ≤ 150, so she can send a maximum of 150 messages.

Lesson video for unit 4

A lesson on this chapter from YouTube. It loads only when you press play.

Compound inequalities: Linear inequalities (part 2) · Watch on YouTube

The textbook pages for unit 4

Unit 4 runs from page 130 to page 161 of the Ministry of Education edition.

Ministry of Education edition, English, 337 pages

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