CONTENT
Solving equation by balance method
Equation with bracket
Equation with fraction.
Solving Equation by Balance Method
To solve an equation means to find the values of the unknown in the equation that makes it true.
For example: 2x – 9 = 15.
2x – 9 is on the left hand side (LHS) and 15 is on the right hand side (RHS) of the equals signs.
Worked examples
- solve 3x = 12
- 2x – 9 = 15.
Solution
1.3x = 12
Divide both sides by 3
3x = 12
3 3
x = 4.
2. 2x – 9 = 5
add 9 to both sides since +9 is the additive inverse of (-9)
2x – 9 + 9 = 15 + 9
2x = 24
x = 24/2 = 12
EVALUATION
Use the balance method to solve the following
(a) 3x – 8 = 10 (b) 20 = 9x + 11 (c ) 10y – 7 = 27 (d) 9 + 2x = 16.
Equation with bracket
Worked example
1. solve 3(3x – 1) = 4 ( x + 3)
2. Solve 5 ( x + 11) + 2 ( 2x – 5) = 0
Soluton
1. 3 (3x – 1) = 4 ( x + 3)
9x – 3 = 4x + 12
Collect like terms:
9x – 4x = 12 + 3
5x = 15
x = 15/3
x = 3
2. 5 (x + 11) + 2 ( 2x – 5) = 0
5x + 55 + 4x – 10 = 0
collect like terms
5x + 5x = 10 – 55
9x = -45
x = -45/9
x = -5.
EVALUATION
Solve the following :
1. 2 (x + 5) = 18 2. 6 (2s – 7) = 5s 3. 3x + 1 = 2(3x+5)
4. 8 (2d – 3) = 3 (4d – 7) 5. (y + 8 ) + 2 (y + 1) = 0.
READING ASSIGNMENT
New General Mathematics chapt. 13 Ex 13d nos 1-20.
WEEKEND ASSIGNMENT
1. Solve 3x + 9 = 117
(a) 38 (b) 36 (c) -36 (d) -38
2. If -2r = 18 what is 4?
(a) -9 (b) 20 (c) 9 (d) -20
3. solve 2 (x + 5) = 16
(a) 13 (b) 10 (c) 8 (d) 3
4. Solve x = 5 (a) -15 (b) 15 (c) 10 (d) -10
3
5. If x/5= ½ What is x? (a) 2 ½ (b) 2 2/3 (c) -2 ½ (d ) 2.
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