Solve {c}{x^2+y^2=25}{x+y=5}

$$\left\{ \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 25 } \\ { x + y = 5 } \end{array} \right.$$

Solve for x, y

$$x=0\text{, }y=5$$

$$x=5\text{, }y=0$$

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.

$$x+y=5,y^{2}+x^{2}=25$$

Solve x+y=5 for x by isolating x on the left hand side of the equal sign.

$$x+y=5$$

Subtract y from both sides of the equation.

$$x=-y+5$$

Substitute -y+5 for x in the other equation, y^{2}+x^{2}=25.

$$y^{2}+\left(-y+5\right)^{2}=25$$

Square -y+5.

$$y^{2}+y^{2}-10y+25=25$$

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

$$2y^{2}-10y+25=25$$

Subtract 25 from both sides of the equation.

$$2y^{2}-10y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 5\left(-1\right)\times 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times 2}$$

Take the square root of \left(-10\right)^{2}.

$$y=\frac{-\left(-10\right)±10}{2\times 2}$$

The opposite of 1\times 5\left(-1\right)\times 2 is 10.

$$y=\frac{10±10}{2\times 2}$$

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

$$y=\frac{10±10}{4}$$

Now solve the equation y=\frac{10±10}{4} when ± is plus. Add 10 to 10.

$$y=\frac{20}{4}$$

Divide 20 by 4.

$$y=5$$

Now solve the equation y=\frac{10±10}{4} when ± is minus. Subtract 10 from 10.

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

Divide 0 by 4.

$$y=0$$

There are two solutions for y: 5 and 0. Substitute 5 for y in the equation x=-y+5 to find the corresponding solution for x that satisfies both equations.

$$x=-5+5$$

Add -5 to 5.

$$x=0$$

Now substitute 0 for y in the equation x=-y+5 and solve to find the corresponding solution for x that satisfies both equations.

$$x=5$$

The system is now solved.

$$x=0,y=5\text{ or }x=5,y=0$$