Solve the System of Equations x^2+(y+3)^2=25 x-y=4

Replace all occurrences of ${\displaystyle x}$ in ${\displaystyle x^2+{(y+3)}^2=25}$ with ${\displaystyle 4+y}$.

Simplify the left side.

Subtract ${\displaystyle 25}$ from both sides of the equation.

Combine the opposite terms in ${\displaystyle 2y^2+14y+25-25}$.

Factor ${\displaystyle 2y}$ out of ${\displaystyle 2y^2+14y}$.

If any individual factor on the left side of the equation is equal to ${\displaystyle 0}$, the entire expression will be equal to ${\displaystyle 0}$.

Set ${\displaystyle y}$ equal to ${\displaystyle 0}$.

Set ${\displaystyle y+7}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle y}$.

The final solution is all the values that make ${\displaystyle 2y(y+7)=0}$ true.

Replace all occurrences of ${\displaystyle y}$ in ${\displaystyle x=4+y}$ with ${\displaystyle 0}$.

Simplify ${\displaystyle x=4+0}$.

Replace all occurrences of ${\displaystyle y}$ in ${\displaystyle x=4+y}$ with ${\displaystyle -7}$.

Simplify ${\displaystyle x=4-7}$.