The product of two consecutive odd integers is equal to 27 plus six times the sum of the integers. What is the larger integer?

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product of 2 consecutive odd integers

let the numbers be a and b

so:

a x b = 27 + 2(a+b)

a=b+2

Replace a with b+2

b(b+2)= 27 +2((b+2)+b)

b^{2}+2b=27 + 2b +4+2b

Group like terms together

b^{2} +2b-2b-2b=27 +4

b^{2} -2b= 31

factorize

Because of the timing, let me finish up in five minutes, I will Post the complete answer now

b(b+2)=27+6(b+b+2)

b^{2}+2b=27+6(2b+2)

b^{2}+2b=27+12b+12

factoring gives

(bâˆ’13)(b+3)=0

b=13 or -3

Sorry for the delay, I had experienced some network problems.. Thanks

Not satisfied. In my book it says the correct answer is 15 or -1.. Did you mess up somewhere along the way?

Nothing is wrong!

The value is b is 13

the value of a which is equal to b +2 is 15

So your book is right too

The numbers are 13 and 15

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