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

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Set up the necessary equation(s) and solve algebraically. two numbers which differ by 3, have a product of 88. Find the numbers.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

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) About Me  (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..