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## JSS1: MATHEMATICS - 2ND TERM

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#### Quizzes

Lesson 3, Topic 3
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# Substitution into Algebraic Expressions

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In algebra, substitution means replacing letters with numbers. The operation sign is important when carrying out the substitution.

### Example

If a = 2, b = -5, c = 10, d = 3

Evaluate:

i. $$\scriptsize a \: + \: b \: + \: c$$
ii. $$\scriptsize c \: \div \: b$$
iii. $$\scriptsize a \: \times\: d$$
iv. $$\frac{d \: \times \: c}{a}$$
v. $$\frac{2b \: \times \: c}{10}$$

Solution

i. $$\scriptsize a \: + \: b \: + \: c$$

From the question the values given are a = 2, b = -5, c = 10

Substituting the values for a, b and c into the equation we get

= $$\scriptsize 2 \: + \: (-5) \: + \: 10$$

Note:

$$\scriptsize +(-) \; = \; –$$

or

$$\scriptsize \; + \; \times \; \; – \; = \; –$$

= $$\scriptsize 2 \; – \: 5 \: + \: 10$$

Using Bodmas rule, Addition comes before subtraction

Add 2 + 10 before subtracting 5

$$\scriptsize 2\: + \: 10 \; – \: 5$$

= $$\scriptsize 12 \; – \: 5$$

= $$\scriptsize 7$$

ii. $$\scriptsize c \: \div \: b$$

From the question, the values given are c = 10, b = -5

Substituting the values for c and b into the equation we get

$$\scriptsize 10 \: \div \: (-5)$$

= $$\frac {10}{-5}$$

= $$\scriptsize \: -2$$

iii. $$\scriptsize a \: \times\: d$$

From the question, the values given are a = 2, d = 3

Substituting the values for a and d into the equation we get

$$\scriptsize 2 \: \times\: 3 = 6$$

iv. $$\frac{d \: \times \: c}{a}$$

From the question, the values given are a = 2, c = 10, d = 3

Substituting the values for a, c and d into the equation we get

$$\frac{3 \: \times \: 10}{2} \\ \frac{30}{2} \\ = \scriptsize 15$$

v. $$\frac{2b \: \times \: c}{10}$$

From the question, the values given are b = -5, c = 10

Substituting the values for b and c into the equation we get

$$\frac{2(-5) \: \times \: 10}{10} \\ = \frac{-10 \: \times \: 10}{10} \\ = \frac{-100}{10} \\ = \scriptsize -10$$