Question

What is one solution of the following system? StartLayout Enlarged left-brace 1st row 2 y minus 2 x = 12 2nd row x squared + y squared = 36 EndLayout (–6, 0) (–2, 4) (0, –6) (4, –2)

Answer 1

Answer:

A on edge

Step-by-step explanation:

Answer 2

Answer:

Option A.

Step-by-step explanation:

The given equations are

$2y-2x=12$     ...(1)

$x^2+y^2=36$    ...(2)

From equation (1), we get

$2y=12+2x$

$y=\dfrac{12+2x}{2}$

$y=6+x$         ...(3)

Substitute $y=6+x$ in equation (2).

$x^2+(6+x)^2=36$

$x^2+36+12x+x^2=36$

$2x^2+12x=0$

$2x(x+6)=0$

$x=0,-6$

Put x=0, in equation (3).

$y=6+0=6$

Put x=-6, in equation (3).

$y=6-6=0$

It means, (0,6) and (-6,0) are two solutions of the given system of equations.

Therefore, the correct option is A.