Question 204618: a positive number is 1 more than twice another number.if the difference of their squares is 40, find the larger of the two numbers.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a positive number is 1 more than twice another number.
if the difference of their squares is 40, find the larger of the two numbers.
:
let x = the larger number
let y = the other number
:
"a positive number is 1 more than twice another number."
x = 2y + 1
:
"if the difference of their squares is 40,"
x^2 - y^2 = 40
:
Replace x with (2y+1)
(2y+1)^2 - y^2 = 40
FOIL
4y^2 + 4y + 1 - y^2 = 40
:
3y^2 + 4y + 1 - 40 = 0
:
3y^2 + 4y - 39 = 0
Factor
(3y + 13)(y - 3) = 0
the positive integer solution
y = 3
:
Find x:
x = 2(3) + 1
x = 7
:
Nos. are 7 and 3
:
:
Check: 49 - 9 = 40
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