10-Week Mathematics Catch-Up Programme (CAPS)

Week 2 – Day 3

This lesson marks another important step towards Grade 10 and Grade 11 mathematics. Two topics that many learners find difficult—inequalities and changing the subject of a formula—are introduced in Grade 9 and then used extensively in later grades.

The goal today is not mastery, but to become comfortable with the ideas.

Inequalities and Formula Rearrangement

Duration: 1 Hour

Grade Level: Grade 9 Foundation

Learning Outcomes

By the end of this lesson you should be able to:


Part A – Study Guide

1. What is an Inequality?

An inequality compares two quantities that are not necessarily equal.

Symbols

SymbolMeaningExample
=Equal to(x = 5)
<Less than(x < 5)
>Greater than(x > 5)
≤Less than or equal to(x \leq 5)
≥Greater than or equal to(x \geq 5)

2. Solving Inequalities

The process is almost the same as solving equations.

Example 1

Solve

x+4>10x+4>10

Subtract 4 from both sides.

x>6x>6


Example 2

Solve

x−7≤9x-7\le9

Add 7.

x≤16x\le16


Example 3

Solve

3x<213x<21

Divide by 3.

x<7x<7


Important Rule

When multiplying or dividing by a negative number, the inequality sign reverses direction.

Example

Solve

−2x>8-2x>8

Divide by -2.

Remember to reverse the sign.

x<−4x<-4


3. Number Line Representation

Example

x>4x>4

Draw an open circle at 4 and shade to the right.

Example

x≤4x\le4

Draw a closed circle at 4 and shade to the left.

(Sketch these in your workbook.)


Formula Rearrangement

Formulae are used throughout Grade 10 and 11.

Sometimes you need to make a different variable the subject.


Example 1

Given

y=x+4y=x+4

Make xx the subject.

Subtract 4.

x=y−4x=y-4


Example 2

Given

A=l×wA=l\times w

Make ll the subject.

Divide both sides by (ww).

l=Awl=\frac{A}{w}


Example 3

Given

P=2l+2wP=2l+2w

Make (ll) the subject.

Subtract (2w2w):

P−2w=2lP-2w=2l

Divide by 2:

l=P−2w2l=\frac{P-2w}{2}


Common Mistakes

Mistake 1

Forgetting to reverse the inequality sign when dividing by a negative.

Example

−5x<20-5x<20

Correct answer:

x>−4x>-4


Mistake 2

Only moving one term in a formula.

Remember:

Treat formula rearrangement exactly like solving an equation.


Part B – Worked Examples

Example 1

Solve

x+12>18x+12>18

Subtract 12.

x>6x>6


Example 2

Solve

4x≤204x\le20

Divide by 4.

x≤5x\le5


Example 3

Solve

−3x<15-3x<15

Divide by -3.

Reverse sign.

x>−5x>-5


Example 4

Make (aa) the subject.

b=a+8b=a+8

Subtract 8.

a=b−8a=b-8


Example 5

Make (ww) the subject.

A=lwA=lw

Divide by (ll).

w=Alw=\frac{A}{l}


Example 6

Make (rr) the subject.

C=2πrC=2\pi r

Divide by (2π2\pi).

r=C2πr=\frac{C}{2\pi}


Part C – Practice Questions

Section A – Solve the Inequalities

x+5>12x+5>12
x−9<4x-9<4
2x>182x>18
5x≤355x\le35
3x+6>183x+6>18
4x−8≤204x-8\le20
7+x≥187+x\ge18
9−x>39-x>3
−2x<10-2x<10
−5x≥25-5x\ge25

Section B – Draw on a Number Line

For each inequality, draw a number line and show the solution.

x>2x>2
x≤7x\le7
x<−1x<-1
x≥5x\ge5
x≤0x\le0

Section C – Rearranging Formulae

Make the variable shown in bold the subject.

y=x+9y=x+9

Make xx the subject.


A=lwA=lw

Make ll the subject.


V=lwhV=lwh

Make hh the subject.


C=2πrC=2\pi r

Make rr the subject.


P=2l+2wP=2l+2w

Make ww the subject.


F=maF=ma

Make mm the subject.


I=VRI=\frac{V}{R}

Make VV the subject.


d=vtd=vt

Make tt the subject.


A=12bhA=\frac12 bh

Make hh the subject.


M=x+y2M=\frac{x+y}{2}

Make xx the subject.


Part D – Mixed Questions

Solve

5x−10>205x-10>20

Solve

4x+12≤364x+12\le36

Make (bb) the subject.

A=12bhA=\frac12 bh

A cinema allows entry only to learners aged 13 years or older.

Write an inequality for the learner's age (aa).


A learner must score at least 50% to pass.

If the learner's percentage is (pp), write an inequality.


Challenge Questions

Solve

3x−5>163x-5>16

Solve

−4x+8≤24-4x+8\le24

The perimeter of a rectangle is

P=2l+2wP=2l+2w

Make (ww) the subject.


The formula for simple interest is

I=PrtI=Prt

Make (rr) the subject.


A plumber charges a call-out fee of R250 plus R180 per hour.

Write a formula for the total cost (CC).

If a customer only has R970, what inequality can you write to find the maximum number of hours (hh) the plumber can work?


Answers

Section A

  1. (x>7)(x>7)

  2. (x<13)(x<13)

  3. (x>9)(x>9)

4.(x≤7) (x\le7)

  1. (x>4)(x>4)

  2. (x≤7)(x\le7)

  3. (x≥11)(x\ge11)

  4. (x<6)(x<6)

  5. (x>−5)(x>-5)

  6. (x≤−5)(x\le-5)


Section B

Learner should draw:

  1. Open circle at 2, shade right.

  2. Closed circle at 7, shade left.

  3. Open circle at -1, shade left.

  4. Closed circle at 5, shade right.

  5. Closed circle at 0, shade left.


Section C

  1. (x=y−9)(x=y-9)

  2. (l=Aw)(l=\frac{A}{w})

  3. (h=Vlw)(h=\frac{V}{lw})

  4. (r=C2π)(r=\frac{C}{2\pi})

  5. (w=P−2l2)(w=\frac{P-2l}{2})

  6. (m=Fa)(m=\frac{F}{a})

  7. (V=IR)(V=IR)

  8. (t=dv)(t=\frac{d}{v})

  9. (h=2Ab)(h=\frac{2A}{b})

  10. (x=2M−y)(x=2M-y)


Section D

5x>305x>30 x>6x>6
4x≤244x\le24 x≤6x\le6
b=2Ahb=\frac{2A}{h}
a≥13a\ge13
p≥50p\ge50

Challenge

3x>213x>21 x>7x>7
−4x≤16-4x\le16

Divide by -4 and reverse the sign:

x≥−4x\ge-4
w=P−2l2w=\frac{P-2l}{2}
r=IPtr=\frac{I}{Pt}

Formula:

C=250+180hC=250+180h

With a budget of R970:

250+180h≤970250+180h\le970

Subtract 250:

180h≤720180h\le720

Divide by 180:

h≤4h\le4

The plumber can work for 4 hours or less.


Parent's Notes

Today's lesson introduces skills that will be used throughout Grades 10 and 11, especially in:

Encourage your student to write every step clearly. Most mistakes happen when learners try to do too much in their heads.

Looking Ahead – Week 2, Day 4

Tomorrow we'll cover Ratio, Proportion and Rates, including: