Question

What are the solutions for the following system of equations?–2x^2 + y = 64x^2 + 3y = 28 (0, 6) and (1, 8) (0, 6) and (–1, 8) (–1, 4) and (1, 4) (–1, 8) and (1, 8)

Answer 1

Step 1

Given;

`\begin{gathered} -2x^2+y=6---(1) \\ 4x^2+3y=28---(2) \end{gathered}`

Step 2

Make x² the subject of the formula in equation 1

`\begin{gathered} x^2=(6-y)/(-2) \\ x^2=(-6+y)/(2) \\ Substitute\text{ for x}^2\text{ in equation 2} \\ 4((-6+y)/(2))+3y=28 \end{gathered}`
`\begin{gathered} -12+2y+3y=28 \\ 5y=28+12 \\ 5y=40 \\ y=8 \end{gathered}`
`\begin{gathered} From\text{ equation 1} \\ x^2=(-6+y)/(2) \\ x^2=(-6+8)/(2) \\ x^2=1 \\ x=\pm1 \\ x=-1\text{ or 1} \end{gathered}`

The second value of y will be;

`8`

Answer; in (x,y)

`(-1,8)\text{ and \lparen1,8\rparen}`