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Find two integers whose product is 136 such that one of the integers is one more than twice the other integer.
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Let integers be x and y. So we know the product = 136 So x*y = 136 Then we know ONE of the integers is ONE MORE than TWICE the other integer so let's say: x = 2y + 1 Now substitute x = 2y + 1 into "x * y= 136"
Ok, I did that and I got \[2y ^{2}+1y-136=0\]
I tried to do the diamond method but I couldn't get an answer
yep now we want to solve for y so you're going to have to change that into the form (2y ) (y ) = 0
alright let's think of two numbers that multiply to give you 136. Hint: i'd use 17 and 8
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