Solving Linear and Quadratic Equations resources... · In the linear equation 3 2 we showed and 4 2 3 6 82 tested that 2 is the solution or root for this equation. 33 x x x x §·

Solving Linear and

Quadratic Equations

Grade 10 CAPS Mathematics

Video Series

Outcomes for this Video

In this DVD you will:

• Revise factorization.

LESSON 1.

• Revise simplification of algebraic fractions.

LESSON 2.

• Discuss when trinomials can be factorized.

LESSON 3.

2

In this Video we will solve:

Linear Equations (Lesson 1)

Quadratic Equations (Lesson 2)

Two Linear Equations Simultaneously (Lesson 3)

Solving Linear

Equations


Video Series

Lesson 1

Outcomes for Lesson 1

In this lesson we will focus on:

• Examples of linear equations

• General steps to solve linear equations

• How to verify the correctness of a solution

• How to verify the solution (root) graphically

Examples of Linear Equations

1 3 4 5

32 4

3 1

43 6 7 4

3

x

x

x

xx

Simplest equation to solve is a linear equation

A linear equation is an equation where the power

of the variable is one

Also known as an equation of the first degree

We will le

General remarks :

arn how to find what value of makes

both sides of the given linear equation true

There is at most or for a linear equation

x

one solution root

3 4 5 2

2 5

1 From solution to equation:

5 3 4 25 3 3 9 3 4 5 5 3 4 25

2 2

2 From equation to solution (Reversing process above):

5 3 4 25 5 3 4 25 3

2 2

xx x x x

xx

Consider the following :

4 34 5 3 9 3x x x

General steps to solve linear equations

1. If there are any fractions multiply all the terms

by the LCM of the denominators.

2. Remove brackets where possible.

3. Transfer terms containing the variable to the

LHS of the e

General steps :

quation (Transpose, change signs).

4. Transfer the other terms to the RHS of the equations

(Transpose, change sign).

5. Simplify both sides where possible.

6. Divide by the coefficient of the unknown.

7. Test the correctness of the solution.

Apply general steps to solve a linear equation

3 2 3 Solve for if 3 2

4 2 3 6

x x xx

M 3 3 18 4 24 2 2 3 ultiply each term by LCM 1 2x x x Step 1:

9 54 4 2 Remove brack4 4 6 e s tx x x Step 2 :

9 4 4 24 54 6 Transpose, change signs x x x Steps 3 & 4 :

Sim 9 24 plify both si desx Step 5 :

D24 8 2

2 9 3 3

ivide by coefficient of x x

Step 6 :

8 162 3 3

8 1 3 8 1 9 8 5 8 61 16 9 25 613 3LHS 3 2 and RHS 2 2 2 23 4 2 3 3 2 9 2 9 18 6 6 18 18 18

LHS RHS

Substitute solution in

Test correctne

to both sides

ss of sol

of original equatio

n

n

io

:

utStep 7 :

Checking the solution graphically

3 2 3In the linear equation 3 2 we showed and

4 2 3 6

8 2tested that 2 is the solution or root for this equation.

3 3

x x x

x

In the solution process the linear equation

is reduced to the form , .

A linear equation of the form ,

is a straight line to the axis.

is the root for this linear equation.

x k k

x k k

y

k

;0 is the point of intersection

between and 0 axis .

k

x k y x

Solve a more complex linear equation

2 2 2

1 1 1 Solve for if

6 15 9 4 6 19 15x

x x x x x

Factorize denominators:

1 1 1

3 5 2 3 3 2 3 2 3 5 2 3x x x x x x

Multiply by LCM 3 5 2 3 2 3 :

2 3 3 5 2 3 5 3

and 3 2

x x

x

x

x x xx

2 3 2 3 3 5x x x 3 11x

Use your calcul

11 is the

ator to te

solution to given eq

st the correctness of solution.

uation.3

x

Apply some or all of the general steps,

where applicable, to solve the following

linear equations. Test in each case whether

the solution is correct:

1 10 1 3 2 3

3 2 4 32 3 2

2 3 6

1 2 53 5

4 3 3 2

x x

x xx

yy

2 2 2

06

1 1 14

24 2 15 9 16 24 38 15

y

x x x x x

Tutorial 1: Solving Linear Equations

PAUSE VIDEO

• Do Tutorial 1

• Then View Solutions

1 Solve for and test solution : 10 1 3 2 3x x x

Tutorial 1: Problem 1: Suggested Solution

10 10 3 2 3

10 2 3 3 10

12 4

Remove brackets

Transpose and change signs

Simplify both sides

Divide by coefficient of unknown

1

3

x x

x x

x

x

1 10 1 3 2 3

1 2 20 2LHS 10 1 10 6 6,6

3 3 3 3

1 2 2 2RHS 3 2 3 3 3 3 3 6

3 3 3 3

LHS RHS

x x

TEST

3 2 4 3

2 Solve the equation and test the solution: 2 3 22 3 6

x xx


9 3 4 12 4 3

9 27 4 12 4 3

9 4 4 1

Multiply both sides by LCD 6

Remove brackets

Transpose and change si2 3 27

9 12

4

des

Simplify both sides

x x x

x x x

x x x

x

x

Divide by coefficient of unk3

nown

163

3 4 8 61 613LHS 3 RHS 2 LHS RHS2 3 9 18 6 18

TEST (Use Calculator)

and

1 2 5

3 Solve equation and test solution: 5 04 3 3 2 6

y yy


Often better to first remove brackets

Mult

5 2 50

4 4 9 3 6

9 45 8 60 6 0

9 8 6 45 60

7 105

iply by LCD 36

Transpose and chang

e sig

ns

y y y

y y y

y y y

y

15

Simplify both sides

Divide by coefficient of unk own ny

1 2 5 15

LHS 15 5 5 0 RHS LHS RHS4 3 2 6

TEST (Use Calculator)

2 2 2

1 1 14 Solve equation and test solution:

24 2 15 9 16 24 38 15x x x x x


1 1 1

4 3 6 5 4 3 4 3 4 3 6 5

4 3 6 5 4 3

Factorize denominators

Multiply by LCD 4 3 4 3 6 5

Transpose and change sign4 6 4 3 3 5

116 11

s

Simplify 6

x x x

x x x x x x

x x x

x x x

x x

and Divide by coefficient of unknown

2 2

2

1 1 1LHS

2611 11 112 2 15 9 16

6 6 6

1 1RHS LHS RHS

2611 1124 38 15

6 6

TEST (Use Calculator) :

Solving

Quadratic

Equations


Video Series

Lesson 2



• Examples of quadratic equations

• The standard form of a quadratic equation

• Zero products

• General steps to solve quadratic equations

• Different formats of quadratic equations

• Graphic interpretation of quadratic equations

Examples of Quadratic Equations

2

2

1 3 4 5

3 12 4 3 4 3 1 with

3 1 3

43 6 7 4

3

x x

xx x x x x

x

xx

Examples of quadratic equations :

A quadratic equation is an equation where the power

of the variable is at most two

Such an equation is also called an equation of the second degree

We will ony solve quadratic e

General remarks :

quations by factorization

Quadratic equations has at most or two solutions roots

Standard form of a Quadratic Equation

2

The of a quadratic equation is

0ax bx c

standard form

2

where:

is a variable

is the coefficient of

is the coefficient of

is independent of also known as the constant

x

a x

b x

c x

Zero Products

For all real numbers and it follows that:

If 0 then 0 or 0

a b

a b a b So we have that:

If 0 then may have any real value

If 0 and 0 then 0

If 0 and 0 then 0

a b

ab a b

ab b a

If the product of two or more factors is equal to zero,

at least one of the factors is zero.

In general :

Given 2 3 2 1 0

then because 2 0 either 3 0 or 2 1 0

x x

x x

Example :

Apply general steps to solve a quadratic equation

2 Solve for if 6 6 13 x x x 2 Write equati 6 1 on 3 6 0 in standard form x x Step 1:

Factorize the LHS of the equati 3 2 2 3 0 n ox x Step 2 :

2 3 3 2 0 or 2 3 0

3 2

2 3

3 2Determine the two possible solutions

x x x x

x x

Step 3 :

or

2

2

2 2 2 is a solution because

T

6 13 6 0 3 3 3

3 3 3 is a solution because 6 13 6 0

2 2 2

est correctness of solution(s): Use calculator where needed

x

x

Step 4 :

and

Quadratic Equations in Disguise

Sometimes an equation might not

look quadratic at first glance!!

NOTE :

2 2

Multiply both sides by and write in sta

0 where 0

ndard formc

ax bx

ax bx c ax bx c

x

x

Case 1 :

2

2 2

1

1

Cross Multiply

1 0 where 0 0 and

cax bx

bc ax bx ac x bc x x ax b x x

a

Case 2 :

2 2 2 2

Square both sid

0

es

where 0

ax b cx

ax b c x c x ax b ax b

Case 3 :

Example 1: Solving Quadratic Equations

2 Solve for if 2 5 12x x x Example 1:

2

2 5 12 0 2 3 4 0

32 3 0 or 4 0 or 4

2

x x x x

x x x x

Solution :

2

2

3 3 2 5 12 0

2 2

and 2 4 5 4 12 0

Check :

3 or 4 are both solutions

2x x

Example 1: Graphical Solution

2 3We showed algebraically that for 2 5 12, both and 4 are solutions.

2x x x x

coordinates of parabola and straight line

3 are equal when or 4.

2

y

x x

2

Consider the graphs of the parabola and line

defined by 2 5 and 12 respectively.y x x y

2

Then

; / 2 5 ; / 12

3;12 ; 4;12 the solution set

2

x y y x x x y y


Solve for if 14 5x x x Example 2 :

2 2

14 5 5 14 0 7 2 0

7 0 or 2 0 7 or 2

x x x x x x

x x x x

Solution :

14 5 7 49 7

and 14 5 2 4 2

Check :

2 is the only solutionx

Positive square root


We showed algebraically that 14 5

2 is a solution but 7 is not a solution.

x x

x x

Consider the graphs of one branch of a parabola defined

by 14 5 and the straight line defined by .

The branch defined by 14 5 is excluded.

y x y x

y x

Note :

From the graphs it should

be clear why 2 is a

solution but 7

is not a solution.

x

x


28 Solve for if 3x x

x Example 3 :

2

3 28 3 28 0 7 4 0

7 0 or 4 0 7 or 4

x x x x x x

x x x x

Solution :

28 If 7 then 7 3 4

7

28and if 4 then 4 3 7

4

x

x

Check :

7 or 4 are both solutionsx x


28Algebraically we showed that 3 7 or 4x x x

x

Consider the graphs of the hyperbola defined by

28 and the straight line defined by 3y y x

x

From the graphs it should be clear why

7 and 4 are both solutions.x x

Tutorial 2: Solving Quadratic Equations

2

Solve the following quadratic equations

and test the correctness of the solutions:

1 14 17 6

2 2 2

153

2

x x

x x

xx

PAUSE VIDEO

• Do Tutorial 2



21 Solve 14 17 6 and test the correctness of the solutions. x x

2 214 17 6 14 17 6 0 7 2 2 3 0

2 37 2 0 or 2 3 0 or

7 2

x x x x x x

x x x x

2

2

2 2 14 17 6 and

7 7

3 3 14 17 6

2 2

Check (Use Calculator) :


2 Solve 2 2 and test the correctness of the solutions.x x

2 22 2 2 2 2 4 4x x x x x x x

If 2 then 2 0 and 2 0 2 is a solution

If 3 then 2 1 and 2 1 3 is not a solution

x x x x

x x x x

Check :

2 is the only solutionx

2 5 6 0 2 3 0 2 or 3x x x x x x


15

3 Solve and test the correctness of the solutions.2

xx

2152 15 2 15 0

2

5 3 0 5 0 or 3 0 5 or 3

x x x x xx

x x x x x x

15 15 5 5 5 is a solution

2 3

15 15 3 3 3 is a solution

2 5

x x xx

x x xx

Check :

5 and 3 are both solutionsx x

Solving

Simultaneous Linear

Equations


Video Series

Lesson 3



• The meaning of solving a system of linear

equations in two unknowns.

• Solving such a system:

• graphically,

• by means of the substitution method,

• by means of the elimination method.

System of Linear Equations in two unknowns

2 unknowns 3 unknowns 4 unknowns

2 3 5 2 3 4 7 2 2 3 4 12

3 5 7 3 5 8 12 3 4 2 7 11

5 6 10 15 5 3 2 4 19

6 4 7 8 16

x y x y z a b c d

x y x y z a b c d

x y z a b c d

a b c d

Examples of sytems of simultaneous linear equations :

Thus far we only solved equations in one unknown variable.

When two unknown variables need to be solved for, two equations

are required and these equations are known as simultane

General remarks :

ous equations.

The solutions to the system of simultaneous equations are the values

of the unknown variables which satisfy the system of equations

simultaneously, that means all equations at the

same time.

In general: If there are unknown variables, then equations are

required to (possibly) obtain a solution for each of the variables.

n n

n

Two variables

Two linear equations

Solving this 2 2 system of

linear equations simultaneously

Focus in this lesson will be on :

Method 1: Graphic Solution of

a 2x2 system of linear equations

1

2

Solve the following system

of linear equations graphically:

5

3

1

23

Assume:

; / 5 and

; / 3 3

x y

x y

x y x y

x y x y

Step 1: Sketch the two lines

Step 2: Find the point of intersection

Step 3: Write down the solution for the system

Step 4: Test the correctness of the solution

Method :

1 2

1 2

2;3

or 2 and 3 is a solution for the system

or 2;3 is a point on both lines and

Check: 2 3 5 and 3 3 2 3 3

x y

x y x y

Method 2: Solving a 2x2 system of linear equations by

means of substitution

Solve the following system of linear equations by means

of the substitution method: 2 7 and 3 2 1 27 x y x y

3 Use 1 and make or the subject of the formula

Substitute into 2 to obtain a linear equation in

Sol

7 2

3 2 7 2 7

3 14 4 7 7 21 3 v l e

y x

y x

y x

x x

x x x x

Method :

Step 1 :

Step 2 :

Step 3 :

inear equation in

Back-substitute 3 into 3 to determine corresponding

Write dow

7 2 3 1

3 and 1 is the soluti n the soluon for the system ti

2 3 1 7

on

y

x

x

x y

y

Step 4 :

Step 5 :

Step 6 :

solution satisfies 1

3 3 2 1 7 soluti Check con or sat rectisfies 2 ness of so u n l tio

Method 3: Solving a 2x2 system of linear equations by

means of elimination

Solve by means of the elimination method the following system

of linear equations: 12 3 5 and 3 2 6 2x y x y

Decide to eliminate Make coefficients of in equations identical

1 3: 6 9 15 3

2 2 : 6 4 12 4

34 3 : 5 3 5

5

9 16 8Substitute 5 into 1 : 2 5 2

5 5 5

x x

x y

x y

y y

x x x

Check or test correctness of solution:

8 3 16 92 3 5 Solution satisfies 1

5 5 5 5

8 3 24 63 2 6 Solution satisfies 2

5 5 5 5

8 3 and is the solution of the system.

5 5x y

Tutorial 3: Solving Simultaneous Linear Equations

2 3 8 Given the 2 2 system of linear equations

3 4 39

Solve the given system by using the:

1 Graphical method

2 Subtitution method

3 Elimination meth

1

2

od

x y

x y

PAUSE VIDEO

• Do Tutorial 3



1 Solve graphically the 2 2 system of linear equations:

2 3 8 1

2

3 4 39

x y

x y

Use any method from Grade 9

to sketch the two lines.

5 and 6 satisfies

both equations 1 and 2

Left as an exercise

x y


2 Solve the 2 2 system of

linear equations by means of

substitution

2 3 8

1

23 4 39

x y

x y

3From 1 it follows that 4 3

2

Substitute 3 into 2 :

912 4 39 4

2

Solve equation 4 :

9 24 8 78 17 102 6 5

yx

yy

y y y y

Back-substitute 5 into 3 :

3 64 9 4 5

2x

Show that 5 and 6

is a solution of the system

(Left as an exercise)

x y


3 Solve by means of elimination

the 2 2 system of linear equations:

2 3 8 1

2

3 4 39

x y

x y

Decide to eliminate :

1 4 : 8 12 32 3

2 3: 9 12 117 4

3 4 : 17 85 5 5

Substitute 5 into 4 :

45 12 117 12 72 6

y

x y

x y

x x

y y y

Check (Test) that

5 and 6

is the solution of

the system (Exercises!)

x y

REMEMBER!

•Consult text-books for additional examples.

•Attempt as many as possible other similar

examples on your own.

•Compare your methods with those that were

discussed in the Video.

•Repeat this procedure until you are confident.

•Do not forget:

Practice makes perfect!

End of Video on Solving Linear and

Quadratic Equations

Documents

Solving Linear and Quadratic Equations resources... · In the linear equation 3 2 we showed and 4 2 3 6 82 tested that 2 is the solution or root for this equation. 33 x x x x §·