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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
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Expansion of simple expressions.

To expand the bracket means to remove the bracket. To do this, multiply each term inside the bracket by the term outside the bracket. Then simplify by adding the like terms together.

Same signs give positive

\( \scriptsize + \)\( \scriptsize \times \)\( \scriptsize + \)=\( \scriptsize + \)
\( \scriptsize \times \)=\( \scriptsize + \)

Different signs give negative

\( \scriptsize \times \)\( \scriptsize + \)=
\( \scriptsize + \)\( \scriptsize \times \)=

Expand the following

a. 3(a + b) 

=  3a  + 3b

b. 8(3  – y)

=  24 x – 8y

c. 5(6 – z)

= 30 – 5z

d. 4(m + n)

= 4m + 4n

e. \( \scriptsize 4x \left( 2y \: – \: z \right) \\ = \scriptsize 4x \left( 2y \right) \: – \: 4x \left(z \right) \\ = \scriptsize 8xy \; – \; 4xz \)

f. 3a ( x – 4y)

= 3ax – 12ay

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