Question 109323: one positive integer is 7 less than twice another. the sum of their squares is 346. find the integers
i dont understand how to set this up
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! You set it up by writing exactly what it says
(think of the word "is" as meaning equal "=")
:
Let x = "one positive integer
Let y = "another positive integer
:
One positive integer is 7 less than twice another
x = 2y - 7
:
the sum of their squares is 346.
x^2 + y^2 = 346
:
Find the integers
:
From the 1st statement, we know that x = (2y-7),
x^2 + y^2 = 346
:
Substitute (2y-7) for x in the above equation
(2y-7)^2 + y^2 = 346
:
FOIL (2y-7)(2y-7)
4y^2 - 28y + 49 + y^2 = 346
:
5y^2 - 28y + 49 - 346 = 0
:
5y^2 - 28y - 297 = 0
:
We probably should use the quadratic formula, however,
we know the positive solution has to be an integer,
let's play around with the factors of 297, we come up with:
(5y + 27) (y - 11) = 0
:
y = +11
:
x = 2(11) - 7
x = 22 - 7
x = 15
:
Check our solutions:
11^2 + 15^2 =
121 + 225 = 346
|
|
|
