Solve {l}{a=x+y}{9=x^2+y^2}

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

Solve for x, y (complex solution)

$$x=\frac{\sqrt{18-a^{2}}+a}{2}\text{, }y=\frac{-\sqrt{18-a^{2}}+a}{2}$$

$$x=\frac{-\sqrt{18-a^{2}}+a}{2}\text{, }y=\frac{\sqrt{18-a^{2}}+a}{2}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$x+y=a$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

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

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

$$x+y=a$$

Subtract y from both sides of the equation.

$$x=-y+a$$

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

$$y^{2}+\left(-y+a\right)^{2}=9$$

Square -y+a.

$$y^{2}+y^{2}+\left(-2a\right)y+a^{2}=9$$

Add y^{2} to y^{2}.

$$2y^{2}+\left(-2a\right)y+a^{2}=9$$

Subtract 9 from both sides of the equation.

$$2y^{2}+\left(-2a\right)y+a^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\left(-1\right)\times 2a for b, and a^{2}-9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2a\right)±\sqrt{\left(-2a\right)^{2}-4\times 2\left(a^{2}-9\right)}}{2\times 2}$$

Square 1\left(-1\right)\times 2a.

$$y=\frac{-\left(-2a\right)±\sqrt{4a^{2}-4\times 2\left(a^{2}-9\right)}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$y=\frac{-\left(-2a\right)±\sqrt{4a^{2}-8\left(a^{2}-9\right)}}{2\times 2}$$

Multiply -8 times a^{2}-9.

$$y=\frac{-\left(-2a\right)±\sqrt{4a^{2}+72-8a^{2}}}{2\times 2}$$

Add 4a^{2} to -8a^{2}+72.

$$y=\frac{-\left(-2a\right)±\sqrt{72-4a^{2}}}{2\times 2}$$

Take the square root of -4a^{2}+72.

$$y=\frac{-\left(-2a\right)±2\sqrt{18-a^{2}}}{2\times 2}$$

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

$$y=\frac{2a±2\sqrt{18-a^{2}}}{4}$$

Now solve the equation y=\frac{2a±2\sqrt{18-a^{2}}}{4} when ± is plus. Add 2a to 2\sqrt{-a^{2}+18}.

$$y=\frac{2\sqrt{18-a^{2}}+2a}{4}$$

Divide 2a+2\sqrt{-a^{2}+18} by 4.

$$y=\frac{\sqrt{18-a^{2}}+a}{2}$$

Now solve the equation y=\frac{2a±2\sqrt{18-a^{2}}}{4} when ± is minus. Subtract 2\sqrt{-a^{2}+18} from 2a.

$$y=\frac{-2\sqrt{18-a^{2}}+2a}{4}$$

Divide 2a-2\sqrt{-a^{2}+18} by 4.

$$y=\frac{-\sqrt{18-a^{2}}+a}{2}$$

There are two solutions for y: \frac{a+\sqrt{-a^{2}+18}}{2} and \frac{a-\sqrt{-a^{2}+18}}{2}. Substitute \frac{a+\sqrt{-a^{2}+18}}{2} for y in the equation x=-y+a to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{18-a^{2}}+a}{2}+a$$

Now substitute \frac{a-\sqrt{-a^{2}+18}}{2} for y in the equation x=-y+a and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{18-a^{2}}+a}{2}+a$$

The system is now solved.

$$x=-\frac{\sqrt{18-a^{2}}+a}{2}+a,y=\frac{\sqrt{18-a^{2}}+a}{2}\text{ or }x=-\frac{-\sqrt{18-a^{2}}+a}{2}+a,y=\frac{-\sqrt{18-a^{2}}+a}{2}$$

Solve for x, y

$$x=\frac{\sqrt{18-a^{2}}+a}{2}\text{, }y=\frac{-\sqrt{18-a^{2}}+a}{2}$$

$$x=\frac{-\sqrt{18-a^{2}}+a}{2}\text{, }y=\frac{\sqrt{18-a^{2}}+a}{2}\text{, }|a|\leq 3\sqrt{2}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$x+y=a$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

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

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.

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

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

$$x+y=a$$

Subtract y from both sides of the equation.

$$x=-y+a$$

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

$$y^{2}+\left(-y+a\right)^{2}=9$$

Square -y+a.

$$y^{2}+y^{2}+\left(-2a\right)y+a^{2}=9$$

Add y^{2} to y^{2}.

$$2y^{2}+\left(-2a\right)y+a^{2}=9$$

Subtract 9 from both sides of the equation.

$$2y^{2}+\left(-2a\right)y+a^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\left(-1\right)\times 2a for b, and a^{2}-9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2a\right)±\sqrt{\left(-2a\right)^{2}-4\times 2\left(a^{2}-9\right)}}{2\times 2}$$

Square 1\left(-1\right)\times 2a.

$$y=\frac{-\left(-2a\right)±\sqrt{4a^{2}-4\times 2\left(a^{2}-9\right)}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$y=\frac{-\left(-2a\right)±\sqrt{4a^{2}-8\left(a^{2}-9\right)}}{2\times 2}$$

Multiply -8 times a^{2}-9.

$$y=\frac{-\left(-2a\right)±\sqrt{4a^{2}+72-8a^{2}}}{2\times 2}$$

Add 4a^{2} to -8a^{2}+72.

$$y=\frac{-\left(-2a\right)±\sqrt{72-4a^{2}}}{2\times 2}$$

Take the square root of -4a^{2}+72.

$$y=\frac{-\left(-2a\right)±2\sqrt{18-a^{2}}}{2\times 2}$$

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

$$y=\frac{2a±2\sqrt{18-a^{2}}}{4}$$

Now solve the equation y=\frac{2a±2\sqrt{18-a^{2}}}{4} when ± is plus. Add 2a to 2\sqrt{-a^{2}+18}.

$$y=\frac{2\sqrt{18-a^{2}}+2a}{4}$$

Divide 2a+2\sqrt{-a^{2}+18} by 4.

$$y=\frac{\sqrt{18-a^{2}}+a}{2}$$

Now solve the equation y=\frac{2a±2\sqrt{18-a^{2}}}{4} when ± is minus. Subtract 2\sqrt{-a^{2}+18} from 2a.

$$y=\frac{-2\sqrt{18-a^{2}}+2a}{4}$$

Divide 2a-2\sqrt{-a^{2}+18} by 4.

$$y=\frac{-\sqrt{18-a^{2}}+a}{2}$$

There are two solutions for y: \frac{a+\sqrt{-a^{2}+18}}{2} and \frac{a-\sqrt{-a^{2}+18}}{2}. Substitute \frac{a+\sqrt{-a^{2}+18}}{2} for y in the equation x=-y+a to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{18-a^{2}}+a}{2}+a$$

Now substitute \frac{a-\sqrt{-a^{2}+18}}{2} for y in the equation x=-y+a and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{18-a^{2}}+a}{2}+a$$

The system is now solved.

$$x=-\frac{\sqrt{18-a^{2}}+a}{2}+a,y=\frac{\sqrt{18-a^{2}}+a}{2}\text{ or }x=-\frac{-\sqrt{18-a^{2}}+a}{2}+a,y=\frac{-\sqrt{18-a^{2}}+a}{2}$$