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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
3
Solution
i)
X =5 (multiply both sides by 3)
3
X×3=5×3
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
8
SOLUTION
i)
2x+2=2 ( multiply both sides by 2)
2
(2x+2 ) ×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
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.
3
-To solve any
equation of the form x =2,
multiply both sides 5, the denominator of the left hand side( LHS).
5
-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
4
v)
22-4x =2x.
5
vi)
2m – m =8
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