SOLUTION: Set up an equation and solve the problem. Find two integers whose product is 65 such that one of the integers is three more than one of the integers is three more than twice the o

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Set up an equation and solve the problem. Find two integers whose product is 65 such that one of the integers is three more than one of the integers is three more than twice the o      Log On


   

Question 846596: Set up an equation and solve the problem.
Find two integers whose product is 65 such that one of the integers is three more than one of the integers is three more than twice the other integer.
I have:
x(y)=65
x=2y+3
(2y+3) y=65
2y^2+3y-65=0
(2y^2+13y)+(-10y-65)=0
y(2y+13)-5(2y+13)
(2y+13) (y-5)
2y+13=0 =-13/2
y-5=0. =5
Webassign told me it was wrong and I just want to know how to do it.
Thank you so much!!
Answer by josh_jordan(263) About Me  (Show Source):
You can put this solution on YOUR website!
Hello. You were correct up until you stopped at y = -13/2 and 5. The word problem says that both numbers have to be integers, and -13/2 is not an integer. So, 5 would be one of the integers. To find the other integer, substitute 5 for y in x = 2y + 3:

x = 2(5) + 3 -----> x = 13

So, your equation would be set up as xy=65 AND x=2y%2B3 and your two integers would be 5 and 13