SOLUTION: Set up the necessary equation(s) and solve algebraically. two numbers which differ by 3, have a product of 88. Find the numbers.

SOLUTION: Set up the necessary equation(s) and solve algebraically. two numbers which differ by 3, have a product of 88. Find the numbers.

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Question 465893: Set up the necessary equation(s) and solve algebraically.
two numbers which differ by 3, have a product of 88. Find the numbers.
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
p(p+3)=88
p2+3p=88
p2+3p-88=0
(p+11)(p-8)=0
p=-11 or 8
Throwing out the negative answer, we get the two numbers to be 8 and 11..
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