Twice the sum of two consecutive integers is 10 less than 5 times the smaller integer. Find the greater integer.

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Call the smaller consecutive integer X

Then the greater is X+1

The sum of these integers is X+(X+1) = 2X +1

Twice the sum is 2(2X+1) = 4X+2

10 less than 5 times smaller = 5X-10

So 5X-10 = 4X +2

X = 12

Therefore the greater integer X+1 = 13

Let x be the smaller integer.

Since they're consecutive, the larger one must be x + 1.

"Twice the sum of two consecutive integers" is 2(x + x + 1) = 2(2x + 1) = 4x + 2

"10 less than 5 times the smaller integer" is 5x - 10

So...

4x + 2 = 5x - 10

2 = x - 10

12 = x

So 12 is the smaller integer, which means that 13 is the larger one.

And, checking,

Twice the sum of 12 and 13 (2*25 = 50) is 10 less than 5*12, which is 60-10=50

they match!

Twice the sum of 12 and 13 (2*25 = 50) is 10 less than 5*12, which is 60-10=50

they match!

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