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

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

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 second equation. Subtract x from both sides.

$$y-x=a$$

Subtract -x from both sides of the equation.

$$y=x+a$$

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

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

Square x+a.

$$x^{2}+x^{2}+2ax+a^{2}=9$$

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

$$2x^{2}+2ax+a^{2}=9$$

Subtract 9 from both sides of the equation.

$$2x^{2}+2ax+a^{2}-9=0$$

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

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

Square 1\times 1\times 2a.

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

Multiply -4 times 1+1\times 1^{2}.

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

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

$$x=\frac{-2a±\sqrt{4a^{2}+72-8a^{2}}}{2\times 2}$$

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

$$x=\frac{-2a±\sqrt{72-4a^{2}}}{2\times 2}$$

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

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

Multiply 2 times 1+1\times 1^{2}.

$$x=\frac{-2a±2\sqrt{18-a^{2}}}{4}$$

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

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

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

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

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

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

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

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

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

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

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

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

The system is now solved.

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

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 second equation. Subtract x from both sides.

$$y-x=a$$

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.

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

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

$$y-x=a$$

Subtract -x from both sides of the equation.

$$y=x+a$$

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

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

Square x+a.

$$x^{2}+x^{2}+2ax+a^{2}=9$$

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

$$2x^{2}+2ax+a^{2}=9$$

Subtract 9 from both sides of the equation.

$$2x^{2}+2ax+a^{2}-9=0$$

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

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

Square 1\times 1\times 2a.

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

Multiply -4 times 1+1\times 1^{2}.

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

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

$$x=\frac{-2a±\sqrt{4a^{2}+72-8a^{2}}}{2\times 2}$$

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

$$x=\frac{-2a±\sqrt{72-4a^{2}}}{2\times 2}$$

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

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

Multiply 2 times 1+1\times 1^{2}.

$$x=\frac{-2a±2\sqrt{18-a^{2}}}{4}$$

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

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

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

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

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

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

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

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

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

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

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

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

The system is now solved.

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