Question
Multiply and then evaluate :
$(4x + y)$ and $(x – 2y);$ when $x = 2$ and $y = 1.$
$(4x + y)$ and $(x – 2y);$ when $x = 2$ and $y = 1.$
✓
Answer
$(4x + y) \times (x − 2y)$
$= 4x (x − 2y) + y (x − 2y)$
$= 4x^2− 8xy + xy − 2y^2$
$= 4x^2 − 7xy − 2y^2$
Verification:
When $x = 2, y = 1$
$\text{L.H.S.} = (4x + y) (x − 2y)$
$= (4 \times 2 + 1) (2 − 2 \times 1)$
$= (8 + 1) (2 − 2)$
$= 9 \times 0$
$= 0$
$\text{R.H.S.} = 4x^2 − 7xy − 2y^2$
$= 4 (2)^2− 7 \times 2 \times 1 − 2 (1)^2$
$= 4 \times 4 − 14 − 2$
$= 16 − 16$
$= 0$
$\therefore \text{L.H.S.} = \text{R.H.S.}$
$= 4x (x − 2y) + y (x − 2y)$
$= 4x^2− 8xy + xy − 2y^2$
$= 4x^2 − 7xy − 2y^2$
Verification:
When $x = 2, y = 1$
$\text{L.H.S.} = (4x + y) (x − 2y)$
$= (4 \times 2 + 1) (2 − 2 \times 1)$
$= (8 + 1) (2 − 2)$
$= 9 \times 0$
$= 0$
$\text{R.H.S.} = 4x^2 − 7xy − 2y^2$
$= 4 (2)^2− 7 \times 2 \times 1 − 2 (1)^2$
$= 4 \times 4 − 14 − 2$
$= 16 − 16$
$= 0$
$\therefore \text{L.H.S.} = \text{R.H.S.}$
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