LESSON NOTES ON LINEAR EQUATIONS IN ONE UNKNOWN

SUBJECT: Algebra

TOPIC: Linear Equations in one unknown

LESSON: Solving Linear Equations in one unknown

PREVIOUS KNOWLEDGE: Students can simplify linear expressions as well as calculate the numerical values of linear expressions.

TEACHING AIDS: A cardboard in which a river with a bridge made up of two poles drawn; a mechanical scale balance is also drawn below.

TEACHING METHODS: Demonstration, Question and Assignment methods.

LESSON OBJECTIVES: By the end of the lesson, students are expected to be able to: identify and solve linear equations in one unknown involving sum, products,quotient, brackets and word problems.

GRADE LEVEL: 7

DURATION: 45 minute

DATE: 11/17/2018

1) INTRODUCTION:

-What is a scale balance used for? . Idea here is to introduce the addition and subtraction method for solving linear equations by maintaining balance on both sides of the equality sign.

What is a bridge used for? . To connect two points ; the idea of the “equality sign”

2)PRESENTATION:

Definition of a Linear Equation: An equation which contains only the first power of one unknown quantity is called a linear equation.

Examples of linear equations are: 4x=20; 2a+4=8; 6=3y+2.

In the equation 4x=20, 4x is the left hand side and 20 is the right hand side. X is called the unknown quantity while 4 is called the numerical coefficient of x. From above, we can write 4x=x+x+x+x.

Examples of non-linear equations are: x²+2=6; y²-1=0 ; 2-u³=10.

2.1 SOLUTION OF SIMPLE LINEAR EQUATIONS: To find the solution of a linear equation , is to solve the equation. The element of the solution set is called the solution of the equation. In the process of solving an equation, we must maintain the equality of both sides of the equation.We try to replace the equation by an equivalent equation which is simple,until we reach the equation x=…

Intuitively, we can picture an equation as a scale balance where in order to maintain the balance, any value of the mass we add on one side must be added to the other side.

So if we multiply or divide both sides of an equation by the same quantity or add the same quantity or subtract the same quantity from both sides of the equation, they remain equal.

SOLVING EQUATIONS REQUIRING ADDITION AND SUBTRACTION

Example : Solve the following equations i) x-3=4 , ii) x+8=6,

Solution

x-3=4 (add 3 to both sides to maintain balance)

x-3+3=4+3

x-o=7

x=7 . answer

ii) x+8=6 (subtract 8 from both sides to maintain balance)

x+8-8=6-8

x+o=-2

x=-2. Answer

If you replace x by -2 in the equation above, -2+8=-6

-6=-6.

left hand side( LHS)= right hand side(RHS).

SOLVING EQUATIONS REQUIRING MULTIPLICATION AND DIVISION

Example 1: Solve the following equations i) x =5 ii) 3x=24

Solution

i) X =5 (multiply both sides by 3)

X×3=5×3

X×1=15

X=15. Answer.

ii) 3x=24 (divide both sides by 3)

3x =24

3 3

x = 8. Answer

Example 2: Solve the following equations i) 2x+2 =2 ii) 5x+4=19

SOLUTION

i) 2x+2=2 ( multiply both sides by 2)

(2x+2 ) ×2=2×2

2x+2=4 (subtract 2 from both sides)

2x+2-2=4-2

2x=2 (divide both sides 2)

2x =2

2 2

x=1. Answer

ii) 5x+4=19 (subtract 4 from both sides)

5x+4-4=19-4

5x+0=15

5x=15 (divide both sides by 5)

5x = 15

5 5

X=3 .Answer

Example 3: Solve the following equations i) 9x=10-x ii) 18-4x=2x+6, iii) 2(3p+6)=2p+2 iv) 4(x+2)+3(x+6)=3

SOLUTION

i) 9x=10-x (collect the unknown to LHS, ie add x to both sides)

9x+x=10-x+x

10x=10( divide both sides by 10)

10x=10

10 10

X=1. Answer

ii) 18-4x =2x+6( collect the unknown ie subtract 2x from both sides)

18-4x-2x=2x+6-2x

18-6x=0+6

18-6x=6(subtract 18 from both sides)

18-6x-18=6-18

0-6x=-12

-6x=-12( divide both sides by -6)

-6x =-12

-6 -6

X=2. Answer.

iii) 2(3p+6) =2p+2 ( remove the brackets from the LHS)

6p+18=2p+2 (collect the unknown to LHS ie subtract 2p from both sides)

6p+18-2p=2p+2-2p

4p+18=2

4p+18-18=2-18

4p+0=-16

4p=-16 ( divide both sides by 4)

4p = -16

4 4

P=-4 . Answer.

iv) 4(x+2)+3(x+6)=3 (remove the brackets from the LHS)

4x+8+3x+18=3

7x+26=3 (Subtract 26 from both sides)

7x+26-26=3-26

7x+0=-23

7x=-23 (divide both sides by 7)

7x = -23

7 7

X= -23

7. Answer

EVALUATION

Solve the following equations: i) 4x-7=1; ii) 12x=16-5x; iii) 6(n+12)=2(2n-4); iv) n=n-25 . v) 4t+3 = t+4

2 6 2

SOLUTION

i) 4x-7=1

4x-7+7=1+7

4x=8

4x =8

4 4

X=2. Answer

ii) 12x=16-5x

12x+5x=16-5x+5x

17x=16

17 17

X= 16

17. Answer

SUMMARY.

-To solve any equation of the form x+5=7, eliminate the number 5 from the LHS by subtracting 5 from both sides.

-To solve any equation of the form x-5=7, eliminate the number 5 from the RHS by adding 5 to both sides.

-T o solve any equation of the form 3x=12, eliminate 3 From the LHS by dividing both sides by 3.OR multiply through by 1.

-To solve any equation of the form x =2, multiply both sides 5, the denominator of the left hand side( LHS).

-To solve any euation of the form x+3=4=2x, collect the unknown term to the LHS and solve the resulting equation as prescribed above.

-To solve an equation of the form 2(x+5)=4(x-3); remove the brackets and collect the unknown to the Left hand side(LHS) and solve the resulting equation and solve as done above.

CONCLUSION.

Assignment.

Solve each of the following equations simplifying your answer as far as possible.

i) x+2=5

ii) x-5=4

iii) 2n+3=5

iv) 2(y+4) =y+3

v) 30+7(x-1)=3(2x+7)

vi) x +1=3

v) 22-4x =2x.

vi) 2m – m =8

4 3 5
-Get your mathematical equipment Here.