SOLUTION: I need help with this systems problem. 7x^2 + y^2 = 64 x + y = 4 Thanks

Algebra ->  Systems-of-equations -> SOLUTION: I need help with this systems problem. 7x^2 + y^2 = 64 x + y = 4 Thanks      Log On


   

Question 618242: I need help with this systems problem.
7x^2 + y^2 = 64
x + y = 4
Thanks
Answer by Maths68(1474) About Me  (Show Source):
You can put this solution on YOUR website!
7x%5E2+%2B+y%5E2+=+64...........(1)
x+%2B+y+=+4.................(2)
Take (2)
x+%2B+y+=+4
x=4-y.....................(3)
Put the value of x from (3) to (1)
7x%5E2+%2B+y%5E2+=+64...........(1)
7%284-y%29%5E2+%2B+y%5E2+=+64
7%2816-8y%2By%5E2%29+%2B+y%5E2+=+64
112-56y%2B7y%5E2+%2B+y%5E2+=+64
112-56y%2B8y%5E2=+64
8%2814-7y%2By%5E2%29=+64
Divide by 8 both sides of the above equation
8%2814-7y%2By%5E2%29%2F8=+64%2F8
cross%288%29%2814-7y%2By%5E2%29%2Fcross%288%29=+cross%2864%29%2Fcross%288%29
14-7y%2By%5E2=+8
14-7y%2By%5E2-8=0
y%5E2-7y%2B14-8=0
y%5E2-7y%2B6=0
Now solve for y
y%5E2-6y-y%2B6=0
y%28y-6%29-1%28y-6%29=0
%28y-6%29%28y-1%29=0
y-6=0 Or y-1=0
y=6 Or y=1
Put the values of y in (3)
x=4-y.....................(3)
y=6
x=4-6
x=-2
y=1
x=4-1
x=3
Solution set (x,y)=(-2,6),(3,1)




Check
======
Put the values of x and y in either equation (1) or (2)
When(x,y)=(-2,6)
7x%5E2+%2B+y%5E2+=+64...........(1)
7%28-2%29%5E2+%2B+%286%29%5E2+=+64
7%284%29+%2B+36+=+64
28+%2B+36+=+64
64=64
When(x,y)=(3,1)
7x%5E2+%2B+y%5E2+=+64...........(1)
7%283%29%5E2+%2B+%281%29%5E2+=+64
7%289%29+%2B+1+=+64
63+%2B+1+=+64
64=64