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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
Grade 10 CAPS Mathematics
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
Tutorial 1: Problem 2: Suggested Solution
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
Tutorial 1: Problem 3: Suggested Solution
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
Tutorial 1: Problem 4: Suggested Solution
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
Grade 10 CAPS Mathematics
Video Series
Lesson 2
Outcomes for Lesson 2
In this lesson we will focus on:
• 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
Example 2: Solving Quadratic Equations
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
Example 2: Graphical Solution
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
Example 3: Solving Quadratic Equations
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
Example 3: Graphical Solution
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
• Then View Solutions
Tutorial 2: Problem 1: Suggested Solution
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) :
Tutorial 2: Problem 2: Suggested Solution
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
Tutorial 2: Problem 3: Suggested Solution
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
Grade 10 CAPS Mathematics
Video Series
Lesson 3
Outcomes for Lesson 3
In this lesson we will focus on:
• 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
• Then View Solutions
Tutorial 3: Problem 1: Suggested Solution
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
Tutorial 3: Problem 2: Suggested Solution
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
Tutorial 3: Problem 3: Suggested Solution
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