Respuesta :
we are given
[tex] y=-2x^2 [/tex]
[tex] y=x-2 [/tex]
we can eliminate x from second equation
[tex] x=y+2 [/tex]
now, we can plug back in first equation
[tex] y=-2(y+2)^2 [/tex]
now, we can expand it
[tex] y=-2(y^2+4y+4) [/tex]
[tex] y=-2y^2-8y-8 [/tex]
so, option-C.....................Answer
Answer:
Option C - [tex]y=-2y^2-8y-8;\ x=y+2[/tex]
Step-by-step explanation:
Given : The system of equation [tex]y=-2x^2\ ;y=x-2[/tex]
To find : Which system is equivalent to given system?
Solution :
Let Equation 1- [tex]y=-2x^2[/tex]
Equation 2 - [tex]y=x-2[/tex]
We can re-write equation 2 in terms of x as [tex]x=y+2[/tex]
Substitute the value of x in equation 1,
[tex]y=-2(y+2)^2[/tex]
Solve,
[tex]y=-2(y^2+2^2+2(y)(2))[/tex]
[tex]y=-2(y^2+4+4y)[/tex]
[tex]y=-2y^2-8-8y[/tex]
[tex]y=-2y^2-8y-8[/tex]
The required equations are [tex]y=-2y^2-8y-8;\ x=y+2[/tex]
Therefore, option C is correct.
