Question

What are the solutions to the following system? StartLayout Enlarged left-brace 1st row negative 2 x squared + y = negative 5 2nd row y = negative 3 x squared + 5 EndLayout (0, 2) (1, –2) (StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1) (StartRoot 5 EndRoot, negative 10) and (negative StartRoot 5 EndRoot, negative 10)

Answer 1

Answer:

$(\sqrt{2},-1),(-\sqrt{2},-1)$

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

Step-by-step explanation:

we have

$-2x^{2} +y=-5$ ----> equation A

$y=-3x^{2} +5$ -----> equation B

solve by substitution

substitute equation B in equation A

$-2x^{2} +(-3x^{2} +5)=-5$

solve for x

$-5x^{2} +5=-5$

$-5x^{2}=-10$

$x^{2}=2$

$x=\pm\sqrt{2}$

Find the value of y

$y=-3x^{2} +5$

For $x=\sqrt{2}$ ----> $y=-3(\sqrt{2})^{2} +5=-1$

For $x=-\sqrt{2}$ ----> $y=-3(-\sqrt{2})^{2} +5=-1$

therefore

The solutions are

$(\sqrt{2},-1),(-\sqrt{2},-1)$

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

Answer 2

Answer:

answer is C. on edgenuty. Good luck. I might fail and repeat junior year bc im so behind

Step-by-step explanation: