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

Explanation:

Answer 2

`(√(2),-1),(-√(2),-1)`

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

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√(2)`

Find the value of y

`y=-3x^(2) +5`

For
`x=√(2)` ---->
`y=-3(√(2))^(2) +5=-1`

For
`x=-√(2)` ---->
`y=-3(-√(2))^(2) +5=-1`

therefore

The solutions are

`(√(2),-1),(-√(2),-1)`

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