Solve {l}{x^2+y^2=8}{x+y=m}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 8 } \\ { x + y = m } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{\sqrt{16-m^{2}}+m}{2}\text{, }y=\frac{-\sqrt{16-m^{2}}+m}{2}$$

$$x=\frac{-\sqrt{16-m^{2}}+m}{2}\text{, }y=\frac{\sqrt{16-m^{2}}+m}{2}$$

Steps Using Substitution and Quadratic Formula

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

$$x+y=m$$

Subtract y from both sides of the equation.

$$x=-y+m$$

Substitute -y+m for x in the other equation, y^{2}+x^{2}=8.

$$y^{2}+\left(-y+m\right)^{2}=8$$

Square -y+m.

$$y^{2}+y^{2}+\left(-2m\right)y+m^{2}=8$$

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

$$2y^{2}+\left(-2m\right)y+m^{2}=8$$

Subtract 8 from both sides of the equation.

$$2y^{2}+\left(-2m\right)y+m^{2}-8=0$$

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 2m for b, and m^{2}-8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

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

Multiply -8 times m^{2}-8.

$$y=\frac{-\left(-2m\right)±\sqrt{4m^{2}+64-8m^{2}}}{2\times 2}$$

Add 4m^{2} to -8m^{2}+64.

$$y=\frac{-\left(-2m\right)±\sqrt{64-4m^{2}}}{2\times 2}$$

Take the square root of -4m^{2}+64.

$$y=\frac{-\left(-2m\right)±2\sqrt{\left(4-m\right)\left(m+4\right)}}{2\times 2}$$

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

$$y=\frac{2m±2\sqrt{\left(4-m\right)\left(m+4\right)}}{4}$$

Now solve the equation y=\frac{2m±2\sqrt{\left(4-m\right)\left(m+4\right)}}{4} when ± is plus. Add 2m to 2\sqrt{\left(4+m\right)\left(4-m\right)}.

$$y=\frac{2\sqrt{16-m^{2}}+2m}{4}$$

Divide 2m+2\sqrt{16-m^{2}} by 4.

$$y=\frac{\sqrt{16-m^{2}}+m}{2}$$

Now solve the equation y=\frac{2m±2\sqrt{\left(4-m\right)\left(m+4\right)}}{4} when ± is minus. Subtract 2\sqrt{\left(4+m\right)\left(4-m\right)} from 2m.

$$y=\frac{-2\sqrt{16-m^{2}}+2m}{4}$$

Divide 2m-2\sqrt{-m^{2}+16} by 4.

$$y=\frac{-\sqrt{16-m^{2}}+m}{2}$$

There are two solutions for y: \frac{m+\sqrt{16-m^{2}}}{2} and \frac{m-\sqrt{16-m^{2}}}{2}. Substitute \frac{m+\sqrt{16-m^{2}}}{2} for y in the equation x=-y+m to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{16-m^{2}}+m}{2}+m$$

Now substitute \frac{m-\sqrt{16-m^{2}}}{2} for y in the equation x=-y+m and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{16-m^{2}}+m}{2}+m$$

The system is now solved.

$$x=-\frac{\sqrt{16-m^{2}}+m}{2}+m,y=\frac{\sqrt{16-m^{2}}+m}{2}\text{ or }x=-\frac{-\sqrt{16-m^{2}}+m}{2}+m,y=\frac{-\sqrt{16-m^{2}}+m}{2}$$

Solve for x, y

$$x=\frac{\sqrt{16-m^{2}}+m}{2}\text{, }y=\frac{-\sqrt{16-m^{2}}+m}{2}$$

$$x=\frac{-\sqrt{16-m^{2}}+m}{2}\text{, }y=\frac{\sqrt{16-m^{2}}+m}{2}\text{, }|m|\leq 4$$

Steps Using Substitution and Quadratic Formula

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.

$$x+y=m,y^{2}+x^{2}=8$$

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

$$x+y=m$$

Subtract y from both sides of the equation.

$$x=-y+m$$

Substitute -y+m for x in the other equation, y^{2}+x^{2}=8.

$$y^{2}+\left(-y+m\right)^{2}=8$$

Square -y+m.

$$y^{2}+y^{2}+\left(-2m\right)y+m^{2}=8$$

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

$$2y^{2}+\left(-2m\right)y+m^{2}=8$$

Subtract 8 from both sides of the equation.

$$2y^{2}+\left(-2m\right)y+m^{2}-8=0$$

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 2m for b, and m^{2}-8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

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

Multiply -8 times m^{2}-8.

$$y=\frac{-\left(-2m\right)±\sqrt{4m^{2}+64-8m^{2}}}{2\times 2}$$

Add 4m^{2} to -8m^{2}+64.

$$y=\frac{-\left(-2m\right)±\sqrt{64-4m^{2}}}{2\times 2}$$

Take the square root of -4m^{2}+64.

$$y=\frac{-\left(-2m\right)±2\sqrt{16-m^{2}}}{2\times 2}$$

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

$$y=\frac{2m±2\sqrt{16-m^{2}}}{4}$$

Now solve the equation y=\frac{2m±2\sqrt{16-m^{2}}}{4} when ± is plus. Add 2m to 2\sqrt{16-m^{2}}.

$$y=\frac{2\sqrt{16-m^{2}}+2m}{4}$$

Divide 2m+2\sqrt{16-m^{2}} by 4.

$$y=\frac{\sqrt{16-m^{2}}+m}{2}$$

Now solve the equation y=\frac{2m±2\sqrt{16-m^{2}}}{4} when ± is minus. Subtract 2\sqrt{16-m^{2}} from 2m.

$$y=\frac{-2\sqrt{16-m^{2}}+2m}{4}$$

Divide 2m-2\sqrt{16-m^{2}} by 4.

$$y=\frac{-\sqrt{16-m^{2}}+m}{2}$$

There are two solutions for y: \frac{m+\sqrt{16-m^{2}}}{2} and \frac{m-\sqrt{16-m^{2}}}{2}. Substitute \frac{m+\sqrt{16-m^{2}}}{2} for y in the equation x=-y+m to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{16-m^{2}}+m}{2}+m$$

Now substitute \frac{m-\sqrt{16-m^{2}}}{2} for y in the equation x=-y+m and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{16-m^{2}}+m}{2}+m$$

The system is now solved.

$$x=-\frac{\sqrt{16-m^{2}}+m}{2}+m,y=\frac{\sqrt{16-m^{2}}+m}{2}\text{ or }x=-\frac{-\sqrt{16-m^{2}}+m}{2}+m,y=\frac{-\sqrt{16-m^{2}}+m}{2}$$