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Question
Multiply and then evaluate:
(x – 2y + z) and (x – 3z); when x = − 2, y = − 1 and z = 1.
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Solution
(x − 2y + z) × (x − 3z)
= x (x − 3z) − 2y (x − 3z) + z (x − 3z)
= x2 − 2zx − 2xy + 6yz − 3z2
Verification:
When x = − 2, y = − 1, z = 1
L.H.S. = (x − 2y + z) × (x − 3z)
= [− 2 − 2 × (− 1) × 1] × [− 2 − 3 × 1]
= (− 2 + 2 + 1) × (− 2 − 3)
= 1 × (− 5)
= − 5
R.H.S. = x2 − 2zx − 2xy + 6yz − 3z2
= (− 2)2 − 2 (1) (− 2) − 2 (− 2) (− 1) + 6 (− 1) (1) − 3(1)2
= 4 + 4 − 4 − 6 − 3
= 8 − 13
= − 5
∴ L.H.S. = R.H.S.
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