Question

Solve the system algebraically. Make sure that any points you name satisfy both equations.

Answer 1

Write out the two equations given

`\begin{gathered} y=-x^2+5=====(\text{equation 1)} \\ -x+y=3======(\text{equation 2)} \end{gathered}`

Make y the subject of equation 2

`\begin{gathered} -x+y=3 \\ y=3+x====(\text{equation 3)} \end{gathered}`

Since y is equal to y, then equations 1 and 3 are equal

`\begin{gathered} y=-x^2+5 \\ y=3+x \\ y=y \\ -x^2+5=3+x \\ x^2+x+3-5=0 \\ x^2+x-2=0 \end{gathered}`
`\begin{gathered} x^2-x+2x-2=0 \\ x(x-1)+2(x-1)=0 \\ (x-1)(x+2)=0 \\ x-1=0,x=1 \\ \text{or} \\ x+2=0,x=-2 \end{gathered}`

Substitute x into equation 3

`\begin{gathered} y=3+x \\ \text{when x=1} \\ y=3+1=4(1,4) \\ \text{when x=-2} \\ y=3+(-2) \\ y=3-2=1(-2,1) \end{gathered}`

Hence, the coordinates of the solution are (1,4) (-2,1)