Question

Solve the system by the substitution method.
xy = 12
x^2 + y^2 = 40

A. {( 2, 6), ( 6, 2), ( 2, -6), ( 6, -2)}
B. {( 2, 6), ( -2, -6), ( 2, -6), ( -2, 6)}
C. {( 2, 6), ( -2, -6), ( 6, 2), ( -6, -2)}
D. {( -2, -6), ( -6, -2), ( -2, 6), ( -6, 2)}

Answer 1

On the bottom equation, they are being squared. It doesn't matter if their sign is positive or negative.
However, in the top equation they are being multiplied and the result must be positive, therefore their signs must be the same. So, -2 with -6 or +2 with +6.
Just look for the list where the 2 and the 6 always have the same sign.

Answer 2

so C?

Answer 3

Yes, C is correct.

Answer 4

thank you

Answer 5

You're welcome.

Answer 6

from the first equation, xy=12 ---> y=12/x
substitute y=12/x to the 2nd equation, gives
x^2 + (12/x)^2 = 40
x^2 + 144/x^2 = 40
multiply by x^2 to both sides
x^4 +144=40x^2
x^4-40x^2+144=0
(x^2-36)(x^2-4)=0
(x+6)(x-6)(x+2)(x-2)=0
for zeroes, x=-6, x=6, x=-2, x=2
subtitute of them to y=12/x again, u will get values of y,
and u will get order pairs that solutions