Hi, can you please help me solve the simultanious equation with the steps of how to do it.

\(x^2 + y^2 = 25\quad(1)\\ 3x + y = 5 \quad(2)\)

\(x^2 + y^2 = 25\) This is a circle centre (0,0) with radius 5

\(y = -3x+5 \) and this line with gradient -3 and y intercept 5

So straight off I can see that one point of intersection is (0,5)

I know that there is one more.

I'm going square the second equ and then the first from it

\((3x+y)^2=5^2\\ 9x^2+y^2+6xy=25\qquad (2a)\\ (2a)-(1)\\ 9x^2+y^2+6xy-x^2-y^2=25-25\qquad \\ 8x^2+6xy=0\qquad \\ 2x(4x+3y)=0\qquad \\ x=0 \qquad or \qquad 4x+3y=0\\ When\;\; x=0 \quad y=5 \quad \text{We know about that one}\\ 4x+3y=0\\ 4x=-3y\\ y=\frac{-4x}{3}\qquad (3) \\ \text{Now I can solve (2) and (3) simultaneously}\\ -3x+5=\frac{-4x}{3}\\ -9x+15=-4x\\ 15=5x\\ x=3\\ y=-3*3+5=-4\\ (3,-4) \)

So the points of intersection are (0,5) and (3,-4)

And here I have checked my answer :)