Verified Answer :

we are given

\(y=-2x^2\)

\(y=x-2\)

we can eliminate x from second equation

\(x=y+2\)

now, we can plug back in first equation

\(y=-2(y+2)^2\)

now, we can expand it

\(y=-2(y^2+4y+4)\)

\(y=-2y^2-8y-8\)

so, option-C.....................Answer

Answer:

Option C - \(y=-2y^2-8y-8;\ x=y+2\)

Step-by-step explanation:

Given : The system of equation \(y=-2x^2\ ;y=x-2\)

To find : Which system is equivalent to given system?

Solution :

Let Equation 1- \(y=-2x^2\)

Equation 2 - \(y=x-2\)

We can re-write equation 2 in terms of x as  \(x=y+2\)

Substitute the value of x in equation 1,

\(y=-2(y+2)^2\)

Solve,

\(y=-2(y^2+2^2+2(y)(2))\)

\(y=-2(y^2+4+4y)\)

\(y=-2y^2-8-8y\)

\(y=-2y^2-8y-8\)

The required equations are \(y=-2y^2-8y-8;\ x=y+2\)

Therefore, option C is correct.

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