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JSS2: MATHEMATICS - 2ND TERM

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  1. Transactions in the Homes and Offices | Week 1
    8 Topics
    |
    1 Quiz
  2. Expansion and Factorization of Algebraic Expressions | Week 2
    4 Topics
    |
    1 Quiz
  3. Algebraic Expansion and Factorization of Algebraic Expression | Week 3
    4 Topics
    |
    1 Quiz
  4. Algebraic Fractions I | Week 4
    4 Topics
    |
    1 Quiz
  5. Addition and Subtraction of Algebraic Fractions | Week 5
    2 Topics
    |
    1 Quiz
  6. Solving Simple Equations | Week 6
    4 Topics
    |
    1 Quiz
  7. Linear Inequalities I | Week 7
    4 Topics
    |
    1 Quiz
  8. Linear Inequalities II | Week 8
    2 Topics
    |
    1 Quiz
  9. Quadrilaterals | Week 9
    2 Topics
    |
    1 Quiz
  10. Angles in a Polygon | Week 10
    4 Topics
    |
    1 Quiz
  11. The Cartesian Plane Co-ordinate System I | Week 11
    3 Topics
    |
    1 Quiz
  12. The Cartesian Plane Co-ordinate System II | Week 12
    1 Topic
    |
    1 Quiz
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Topic Content:

  • Balancing Method

Solving an equation means finding the value of the unknown, which makes the equation true.

To use the balancing method to solve a simple equation, add, subtract, divide or multiply both sides of the equation with the same quantity. In other words, whatever operation is done to one side of the equation needs to also be done to the other.

Example 6.1.1:

Solve the following equation using the balancing method:

a.  \( \scriptsize x \: + \: 8 \: = \: 19\)

b. \( \scriptsize x \: – \: 20 \: = \: 35\)

c. \( \scriptsize 3x \: = \: 27\)

d. \( \scriptsize 3x \: – \: 7 \: = \: 8\)

e. \( \frac{x}{6} \scriptsize \: = \: 7\)

Solution 

a.  \( \scriptsize x \: + \: 8 \: = \: 19\)

Additive inverse of +8 = -8 

We are going to add (-8) to both sides

\( \scriptsize x \: + \: 8 \: + \: (-8) \: = \: 19 \: + \: (-8) \\ \scriptsize x \: + \: 8 \: – \: 8 \: = \: 19 \: – \: 8 \\ \scriptsize x \: + \: 0 \: = \: 19 \: – \: 8 \\ \scriptsize x \: = \: 19 \: – \: 8 \\ \scriptsize x \: = \: 11 \)

b. \( \scriptsize x \: – \: 20 \: = \: 35\)

Additive inverse of -20 = +20

Add +20 to both sides

\( \scriptsize x \: – \: 20 \: + \: (+20) \: = \: 35 \: + \: (+20)\)

\( \scriptsize x \: – \: 20 \: + \: 20 \: = \: 35 \: + \: 20\)

\( \scriptsize x \: = \: 35 \: + \: 20\)

\( \scriptsize x \: = \: 55\)

check: 55 – 20 = 35  

c. \( \scriptsize 3x \: = \: 27\)

Multiplicative inverse of 3 = \( \frac{1}{3} \)

Multiply both sides by \( \frac{1}{3} \)

\( \scriptsize 3x \: \times \: \normalsize \frac{1}{3} \scriptsize = \: 27 \: \times \: \normalsize \frac{1}{3}\)

\( \scriptsize \not{\! 3}x \: \times \: \normalsize \frac{1}{\not{3}} \scriptsize = \: 27 \: \times \: \normalsize \frac{1}{3}\)

\( \scriptsize x = \normalsize \frac{27}{3}\)

\( \scriptsize x = 9\)

d. \( \scriptsize 3x \: – \: 7 \: = \: 8\)

Additive inverse of -7 = +7

Add (+7) to both sides

\( \scriptsize 3x \: – \: 7 \: + \: (+7) \: = \: 8 + \: (+7) \)

\( \scriptsize 3x = \: 8 \: + \: 7 \)

\( \scriptsize 3x = 15 \)

To find x, Divide both sides by the coefficient of x, which is 3

\( \frac{3x}{3} = \frac{15}{3} \)

\( \frac{\not{3}x}{\not{3}} = \frac{15}{3} \)

\( \scriptsize x = \normalsize \frac{15}{3} \)

\( \scriptsize x = 5 \)

e. \( \frac{x}{6} \scriptsize \: = \: 7\)

Multiplicative inverse of \( \frac{1}{6} = \scriptsize 6 \)

Multiply both sides by 6 

\( \frac{x}{6} \scriptsize \: \times \: 6 = 7 \: \times \: 6 \)

\( \frac{x}{\not{6}} \scriptsize \: \times \: \not{\! 6} = 7 \: \times \: 6 \)

\( \scriptsize x = 42 \)

Check: \( \frac{x}{6} \scriptsize \: = \: 7 \\ \frac{42}{6} \scriptsize \: = \: 7 \)

Evaluation Questions:

Solve the following equation using the balancing method 

(1) x – 12 = 20 

(2) y – 4 = 9 

(3) 2x = 10 

(4) 3x – 6 = 33

(5) –x + 8 = 24 

(6) \( \frac{y}{9} \scriptsize = 3 \)

(7) \( \frac{-2x}{8} \scriptsize = 5 \)

(8) -30b  = 15 

Answers 

(1)  x = 32 

(2)  y = 13 

(3)  x = 5 

(4)  x = 13

(5)  x  = -16 

(6)  y = 27

(7)  x = -20 

(8) \( \scriptsize b = \normalsize \frac{-1}{2} \)

Example 6.1.2:

Solve the following equations 

a. \(\scriptsize 8x \: – \: 14 = x \: – \: 7 \)

b. \(\scriptsize 4a \: – \: 3 = 2a \: + \: 6 \)

c. \(\scriptsize 5x \: – \: 10 \: – \: 3x = 2x \: + \: 20 \: -\: 3x \)

Solution  

a. \(\scriptsize 8x \: – \: 14 = x \: – \: 7 \)

Collect like terms 

\( \scriptsize 8 {\color{Blue} x} \: – \: {\color{Green} 1{\color{Green} 4}}= {\color{Blue} x} \: – \: {\color{Green} 7} \)

\(\scriptsize 8x \: – \: x = \: – \: 7 \: + \: 14\)

x changed to –x when it crossed to the LHS 
-14 changed to +14 when it crossed to the RHS 

\(\scriptsize 8x \: – \: x = \: – \: 7 \: + \: 14\)

\(\scriptsize 7x = 7\)

Divide both sides by 7

\( \frac{7x}{7} = \frac{7}{7} \)

\( \frac{\not{7}x}{\not{7}} = \frac{7}{7} \)

\( \scriptsize x = \normalsize \frac{7}{7} \)

\( \scriptsize x = 1 \)

b. \(\scriptsize 4a \: – \: 3 = 2a \: + \: 6 \)

Collect like terms

\(\scriptsize 4 {\color{Blue} a} \: – \: {\color{Green} 3} = 2{\color{Blue} a} \: + \: {\color{Green} 6}\)

\(\scriptsize 4a \: – \: 2a = +6 \: + \: 3 \)

\(\scriptsize 2a = 9 \)

Divide both sides by 2

\( \frac{2a}{2} = \frac{9}{2} \)

\( \frac{\not{2}a}{\not{2}} = \frac{9}{2} \)

\( \scriptsize a = \normalsize \frac{9}{2} \)

\( \scriptsize a = 4 \frac{1}{2} \)

c. \(\scriptsize 5x \: – \: 10 \: – \: 3x = 2x \: + \: 20 \: -\: 3x \)

 Re-arrange

\(\scriptsize 5x \: – \: 3x \: – \:10 = 2x \: -\: 3x \: + \: 20 \)

\(\scriptsize 2x \: – \:10 = -x \: + \: 20 \)

Collect like terms 

\(\scriptsize 2x \: + \:x = 20 \: + \: 10 \)

\(\scriptsize 3x = 20 \: + \: 10 \)

\(\scriptsize 3x = 30\)

Divide both sides by 3

\(\frac {3x}{3} = \frac{30}{3}\)

\(\frac {\not{3}x}{\not{3}} = \frac{30}{3}\)

\(\scriptsize x = \normalsize \frac{30}{3}\)

\(\scriptsize x = 10 \)

Evaluation Questions:

Solve the following equations:

  1. 5h – 12 = 20 – 3h 
  2. 10a – 20 = 9a + 1 
  3. 4x = 18 – 2x
  4. 12 – x – 5 = -5x – 3 
  5. 10 -2x = 2 – 6x

Answer

  1. h = 4
  2. a = 21 
  3. x = 3 
  4. x = -2½ 
  5. x = -2 
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