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प्रश्न
Subtract – 5a2 – 3a + 1 from the sum of 4a2 + 3 – 8a and 9a – 7.
बेरीज
उत्तर
sum of 4a2 + 3 – 8a and 9a – 7
= 4a2 + 3 - 8a + 9a - 7 = 4a2 + a - 4
∴ (4a2 + a - 4) - (- 5a2 - 3a + 1)
= 4a2 + a - 4 + 5a2 + 3a - 1
= 4a2 + 5a2 + a + 3a - 4 - 1
= 9a2 + 4a - 5
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