Solve {l}{y=3-x}{x^2+y^2=9}

$$\left\{ \begin{array} { l } { y = 3 - x } \\ { x ^ { 2 } + y ^ { 2 } = 9 } \end{array} \right.$$

Solve for y, x

$$x=0\text{, }y=3$$

$$x=3\text{, }y=0$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Add x to both sides.

$$y+x=3$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$y+x=3,x^{2}+y^{2}=9$$

Solve y+x=3 for y by isolating y on the left hand side of the equal sign.

$$y+x=3$$

Subtract x from both sides of the equation.

$$y=-x+3$$

Substitute -x+3 for y in the other equation, x^{2}+y^{2}=9.

$$x^{2}+\left(-x+3\right)^{2}=9$$

Square -x+3.

$$x^{2}+x^{2}-6x+9=9$$

Add x^{2} to x^{2}.

$$2x^{2}-6x+9=9$$

Subtract 9 from both sides of the equation.

$$2x^{2}-6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 3\left(-1\right)\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}}}{2\times 2}$$

Take the square root of \left(-6\right)^{2}.

$$x=\frac{-\left(-6\right)±6}{2\times 2}$$

The opposite of 1\times 3\left(-1\right)\times 2 is 6.

$$x=\frac{6±6}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$x=\frac{6±6}{4}$$

Now solve the equation x=\frac{6±6}{4} when ± is plus. Add 6 to 6.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{6±6}{4} when ± is minus. Subtract 6 from 6.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

There are two solutions for x: 3 and 0. Substitute 3 for x in the equation y=-x+3 to find the corresponding solution for y that satisfies both equations.

$$y=-3+3$$

Add -3 to 3.

$$y=0$$

Now substitute 0 for x in the equation y=-x+3 and solve to find the corresponding solution for y that satisfies both equations.

$$y=3$$

The system is now solved.

$$y=0,x=3\text{ or }y=3,x=0$$