Question

Solve the system for all solutionsx^2 + (y - 1)^2 = 13x + y = -4

Answer 1

Given the system of equations :

`\begin{gathered} x^2+(y-1)^2=13 \\ x+y=-4 \end{gathered}`

From the second equation : x = -y - 4

Substitute with x at the first equation

`\begin{gathered} (-y-4)^2+(y-1)^2=13 \\ y^2+8y+16+y^2-2y+1=13 \\ 2y^2+6y+4=0 \\ y^2+3y+2=0 \\ (y+1)(y+2)=0 \end{gathered}`

So, there are two solutions

The first when : y + 1 = 0

`\begin{gathered} y+1=0 \\ y=-1 \\ x=-y-4=-(-1)-4=-3 \end{gathered}`

The second when : y + 2 = 0

`\begin{gathered} y+2=0 \\ y=-2 \\ x=-y-4=-(-2)-4=-2 \end{gathered}`

So, the answer is : the solution as order pair ( x , y )

`\begin{gathered} (x,y)=(-1,-3) \\ or,(x,y)=(-2,-2) \end{gathered}`