Subtract the sum of 3a²-2a+5 and a²-5a-7 from the sum of 5a²-9a+3 and 2a-a²-1
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Step-by-step explanation:
Sum of 3a2 – 2a + 5 and a2 – 5a – 7
= 3a2 – 2a + 5 + a2 – 5a – 7
= 3a2 + a2 – 2a – 5a + 5 – 7
= 4a2 – 7a – 2
And sum of 5a2 – 9a + 3 and 2a – a2 - 1
= 5a2 – 9a + 3 + 2a – a2 – 1
= 5a2 – a2 – 9a + 2a + 3 – 1
= 4a2 – 7a + 2
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Answered by
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Answer:
Sum of 3a2 – 2a + 5 and a2 – 5a – 7
= 3a2 – 2a + 5 + a2 – 5a – 7
= 3a2 + a2 – 2a – 5a + 5 – 7
= 4a2 – 7a – 2
And sum of 5a2 – 9a + 3 and 2a – a2 - 1
= 5a2 – 9a + 3 + 2a – a2 – 1
= 5a2 – a2 – 9a + 2a + 3 – 1
= 4a2 – 7a + 2
Now, (4a2 – 7a + 2) – (4a2 – 7a – 2)
= 4a2 – 7a + 2 – 4a2 + 7a + 2
= 4a2 – 4a2 – 7a + 7a + 2 + 2
= 0 + 0 + 4 = 4
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