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JSS3: MATHEMATICS - 1ST TERM

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  1. Binary Number System I | Week 1
    5Topics
    |
    1 Quiz
  2. Binary Number System II | Week 2
    6Topics
    |
    1 Quiz
  3. Word Problems I | Week 3
    4Topics
    |
    1 Quiz
  4. Word Problems with Fractions II | Week 4
    1Topic
    |
    1 Quiz
  5. Factorisation I | Week 5
    4Topics
    |
    1 Quiz
  6. Factorisation II | Week 6
    3Topics
    |
    1 Quiz
  7. Factorisation III | Week 7
    3Topics
    |
    1 Quiz
  8. Substitution & Change of Subject of Formulae | Week 8
    2Topics
    |
    1 Quiz
  9. Simple Equations Involving Fractions | Week 9
    3Topics
    |
    1 Quiz
  10. Word Problems | Week 10
    1Topic
    |
    1 Quiz
Lesson Progress
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To solve simple equations containing fractions, first eliminate the fractions by multiplying each term in the equation by the LCM of the denominator.

i. \( \frac{x}{12} = \scriptsize 5 \)

\( \frac{x}{12}\scriptsize \: \times \: 12 = 5 \: \times \: 12\)

\( \scriptsize x = 12 \: \times \: 5 \\ \scriptsize = 60\)

ii. \( \frac{x}{9} = \frac{4}{3}\)

Cross multiply

\( \frac{x\: \searrow}{9 \: \nearrow} = \frac{ \swarrow \: 4}{\nwarrow \: 3 }\)

\( \scriptsize 3 \: \times \: x = 9 \: \times \: 4 \)

\( \scriptsize 3x = 36 \)

Divide by 3

\( \frac{3x}{3} = \frac{36}{3}\)

\( \frac{\not{3}x}{\not{3}} = \frac{36}{3}\)

x = 12

iii. \( \frac{y \: – \: 5}{4} = \scriptsize 8 \)

Multiply through by 4

\( \scriptsize 4 \: \times \: \normalsize \frac{y \: – \: 5}{4} = \scriptsize 4 \: \times \: 8 \)

\( \scriptsize y \: – \: 5 = 32 \)

Add 5 to both sides

\( \scriptsize y \: – \: 5 \: + \: 5 = 32 \: + \: 5 \)

\( \scriptsize y = 37 \)

iv. \( \scriptsize 2x \: – \: \normalsize \frac{x}{8} = \scriptsize 12 \frac{1}{2} \)

\( \frac{2x}{1} \: – \: \frac{x}{8}= \frac{25}{2}\)

Take the L.C.M

\( \frac{8(2x)\: – \: 1(x)}{8} = \frac{25}{2}\)

\( \frac{16x\: – \: x}{8} = \frac{25}{2}\)

\( \frac{15x}{8} = \frac{25}{2}\)

\( \scriptsize 15x \: \times \: 2 =25 \: \times \: 8 \)

\( \scriptsize 30x =200 \)

x = \( \frac{200}{30} \)

x = \(\scriptsize 6 \frac{2}{3} \: or \: 6.67 \)

v. \( \frac{5x \: – \: 1}{4} = \frac{2x \: – \: 1}{2}\)

Cross multiply

\( \scriptsize 2(5x \: – \: 1) = 4(2x \: – \: 1) \)

\( \scriptsize 10x \: – \: 2 = 8x \: – \: 4 \)

Take like terms

\( \scriptsize 10x \: – \: 8x = -4 \: + \: 2 \)

\( \scriptsize 2x = -2 \)

Divide both sides by 2

\( \frac {2x}{2} =\frac{ -2}{2} \)

x = -1

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