Question

HELP PLEASE!!!!!!!!!!!!!
Use the method of substitution to solve the system. (Enter your answer(s) as ordered pair(s). If there is more than one ordered pair, use commas between ordered pairs. If there is no solution, enter NO SOLUTION.)

Answer 1

\[y^{2}-11.y+28=0\]

Answer 2

y = 28/x^2 (1)
y = 11 - x^2 (2)

looking at equation (2) x^2 = 11 - y

substitute into equation (1)

y = 28/(11 -y)

so

y(11 - y) = 28

expand and simplify

11y - y^2 = 28

y^2 - 11y + 28 = 8

(y - 7)(y - 4) = 0

then y = 4 or 7

substitute into equation (2) to find x

x^2 = 11 - y


x^2 = 11 - 7

\[x = \pm 2 \]

x^2 = 11 - 4

\[x = \pm \sqrt{7}\]

so the ordered pairs are

(-2, 7), (2, 7) \[(-\sqrt{7}, 4)\] \[(\sqrt{7}, 4)\]

Answer 3

Thank you))