Back to: MATHEMATICS JSS 2
Welcome to JSS2 Second Term!
We are building on our work from the First term as we learn more about Mathematics.
In today’s class, We will be discussing Solving Algebraic Equations. We hope you enjoy the class!
CONTENT
- Using directed number
- Unknown on both sides
- Equations with brackets
- Equations with fractions
Using Directed Numbers
Example
(1) Solve 25 – 9x = 2 and 12 = 9 – 3a
25 – 9x = 2
Subtract 25 from both sides
25 – 25 – 9x = 2 – 25
-9x = -23.
Divide both sides by – 9
-9x = – 23
– 9 -9
X = 23/9 = 2 5/9
Check
when x = 23/9
LHS = 25 – 9x 23/9 = 25 – 23 = 2 RHS
(2) 12 = 9 – 31
12 = 9 – 31
Subtract 9 – from both sides
12 – 9 = -9 – 9 – 3a
3 = -31
Divide both sides by -3
3/-3 = -31/-3
-1 = a
:. A = -1
Check when x = -1
LHS = 12 = 9 – 3 ( -1) = RHS
LHS = 12 = 9 + 3 + RHS
LHS = 12 = 12 = RHS
SELF EVALUATION
Solve the following equations and check the Solutions.
- 5 – 5n = 8
- 12 + 5a = 23
- 20 – 5t = 5
- Unknown on both sides.
If an equation has unknown terms on both sides of the equal sign, collect the unknown terms on one side and the number terms on the other side.
Worked Examples
Solve the equation.
- 5x – 4 = 2x + 11
Solution
5x – 4 = 2x + 11
Subtract 2x from both sides of ( 1)
5x – 2x – 4 = 2x – 2x + 11
3x – 4 = 11 ………………. ( 2 )
Add 4 to both sides of …..(2)
3x – 4 + 4 = 11 + 4
3x = 15 ………………… ( 3)
3x/3 = 15/3
X = 5.
Check: when x = 5
LHS = 5 x 5 – 4 = 25 – 4 = 21
RHS = 2 x 5 + 11 = 10 + 11 = 21 = LHS
SELF EVALUATION
Solve the following equations
- 18 – 5f = 2f + 4
- 5a + 6 = 2a + 20
Equations with brackets
Always remove brackets before collecting terms.
Examples.
Solve
(1) 3 ( 3x – 1) = 4 ( x + 3 )
(2) 5 (x +11) + 2 (2x – 5) = 0
Solution
3 ( 3x -1) = 4 ( x + 3)…………..(1)
Remove brackets
9x – 3 = 4x + 12 …………….. ( 2)
Subtracts 4x from both sides and add 3 to both sides.
9x – 4x – 3 + 3 = 4x – 4x + 12 + 3
5x = 15 …………………………(3)
Divide both sides by 5
X = 3
Check : When x = 3
LHS = 3 ( 3 x 3 – 1 ) = 3 ( 9-1) = 3 x 8 = 24
LHS = 4 ( 3 + 3) = 4 x 6 =24 = LHS
(2) .5 ( x + 11) + 2 ( 2x – 5 ) = 0
Remove brackets
5x + 55 + 4x – 10 = 0
Collect like terms
9x + 45 = 0
Subtract 45 from both sides
9x = -45
Divide both sides by 9
X = -5
Check : when x = -5
LHS = 5(-5 + 11) + 2 (2x (-5) – 5 )
= 5 x 6 + 2 ( -10 -5 )
=30 + 2x ( -15) = 30 – 30 = 0 = RHS
Self Evaluation
Solve the following equations and check the Solutions
- 2 (x + 5) = 18 2.) 5(a + 2) = 4 ( a -1) 3.) 2(y – 2) + 3 ( y – 7 ) = 0
Equation with fractions
Always clear fractions before collecting terms. To clear fractions, multiply both sides of the equation by the LCM of the denomination of the fractions.
Example
Solve the equations
Solutions
The LCM of 5 and 3 is 15. Multiply every term by 15.
EVALUATION
Solve the following equations.
GENERAL SELF EVALUATION
- solve 7a + 2 = 5 ( a – 4)
- Solve the equation
- Solve the equation
REVISION QUESTION
Solve the following equations
- 3(x + 5) – 35 = -4(2x -6)
- 4(2x + 5) -2 (x – 3 ) = 8(2x + 4 )
READING ASSIGNMENT
New General Mathematics, UBE Edition chapter 13 pgs. 110-113.
Essential Mathematics by AJS Oluwasanmi, chapter 19pgs 200-205
WEEKEND ASSIGNMENT
- Solve the equation 4b +24 = 0 A. -3 B. -4 C. -5 D. -6
- Solve 7c – 6 = c A. 1 B. 2 C. 3 D. 4
- Solve 15 = 3(x – 3) A. 6 B. 7 C. 8 D. 4
- Solve 0 = 7 (x – 3 ) A. 1 B. 2 C. 3 D. 4
- Solve 3m + 8 = m A. -1 B. -2 C. -3 D. -4
THEORY
Solve the following equations.
- 3(2x-1) =39 – 2(y + 1)
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
In our next class, we will be talking about Word Problems On Algebraic Fraction. We are eager to meet you there.
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