Page 143 - ICSE Math 6

(c) x – 2 −   {–5y + ( 2 –10 8x− )}  

= x – [2 – {–5y + (2 – 10 + 8x)}] (Removing bar bracket)
= x – [2 – {–5y + (–8 + 8x)}] (Simplifying terms in small bracket)
= x – [2 – {–5y – 8 + 8x}] (Removing small bracket)

= x – [2 + 5y + 8 – 8x] (Removing curly bracket)
= x – [10 + 5y – 8x] (Simplifying terms in big bracket)
= x – 10 – 5y + 8x (Removing big bracket)
= 9x – 5y – 10

Substitution

The value of an algebraic expression having variables depends on the value of its variables. Consider
the expression 3x + 20. Here x is the variable. So, the value of the expression will depend on x.
For example,
(a) When x = 3, then 3x + 20 = 3 × 3 + 20 = 9 + 20 = 29
(b) When x = –2, then 3x + 20 = 3 × (–2) + 20 = –6 + 20 = 14
Thus, to find the value of an expression, the variables are substituted by their values. This method is
known as substitution.

Example 5: Find the value of 2y + 20 when:
(a) y = 2 (b) y = 10 (c) y = –15

Solution: (a) Substitute y = 2 to get:
2y + 20 = 2 × 2 + 20 = 4 + 20 = 24
(b) Substitute y = 10 to get:
2y + 20 = 2 × 10 + 20

= 20 + 20 = 40
(c) Substitute y = –15 to get:
2y + 20 = 2 × (–15) + 20
= –30 + 20 = –10

2
Example 6: Find the value of 2x + 3y – 10 when:
(a) x = 2, y = 3 (b) x = –3, y = 4 (c) x = –5, y = –15

Solution: (a) Substitute x = 2 and y = 3 to get:
2
2
2x + 3y – 10 = 2 × (2) + 3 × 3 – 10
= 2 × 4 + 9 – 10 = 8 + 9 – 10 = 7
(b) Substitute x = –3 and y = 4 to get:
2
2
2x + 3y – 10 = 2 × (–3) + 3 × (4) – 10
= 2 × 9 + 12 – 10 = 18 + 12 – 10 = 20
(c) Substitute x = –5 and y = –15 to get:
2
2
2x + 3y – 10 = 2 × (–5) + 3 × (–15) – 10
= 50 – 45 – 10 = –5

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