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Q: Find two consecutive integers such that the larger is nine more than twice the smaller?

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Nice little problem !Call the consecutive integers ' x ' and ' x+1 '.The sum is [ x + 2(x+1) ] = Ax + 2x + 2 = A3x + 2 = ASubtract 2 from each side:3x = A - 2Divide each side by 3:x = (A-2)/3 and that's the smaller of the two integers.The larger one is (x+1) = (A+1)/3

integers are x and x + 2, so x + 2 = 2x - 10 ie x = 12. Integers are 12 and 14

no one wants to know the answer. its freaking math

(x)(x+2)(x+4)

There are two consecutive odd numbers such that five times the smaller plus twice the greater is 25. What is the smaller number?

The sum of two consecutive integers can not be even unless both are even or both are odd.x+y=70y=4+2xx+(4+2x)=704+3x=703x=70-43x=66x=22y=70-22y=48The two numbers are 22 and 48. 22 is the smaller.

5872345098234783904672083946728390752430689723409687298290843 theres your answer

They are (14, 15, 16).

9 and 10 9 + 2(10) = 29

18, 20 and 22

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