Solve x+y=3,x^2+y^2=5

$$x + y = 3 , x ^ { 2 } + y ^ { 2 } = 5$$

Solve for x, y

$$x=2\text{, }y=1$$

$$x=1\text{, }y=2$$

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.

$$x+y=3,y^{2}+x^{2}=5$$

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

$$x+y=3$$

Subtract y from both sides of the equation.

$$x=-y+3$$

Substitute -y+3 for x in the other equation, y^{2}+x^{2}=5.

$$y^{2}+\left(-y+3\right)^{2}=5$$

Square -y+3.

$$y^{2}+y^{2}-6y+9=5$$

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

$$2y^{2}-6y+9=5$$

Subtract 5 from both sides of the equation.

$$2y^{2}-6y+4=0$$

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

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\times 4}}{2\times 2}$$

Square 1\times 3\left(-1\right)\times 2.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 2\times 4}}{2\times 2}$$

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

$$y=\frac{-\left(-6\right)±\sqrt{36-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$y=\frac{-\left(-6\right)±\sqrt{36-32}}{2\times 2}$$

Add 36 to -32.

$$y=\frac{-\left(-6\right)±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

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

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

$$y=\frac{6±2}{2\times 2}$$

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

$$y=\frac{6±2}{4}$$

Now solve the equation y=\frac{6±2}{4} when ± is plus. Add 6 to 2.

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

Now solve the equation y=\frac{6±2}{4} when ± is minus. Subtract 2 from 6.

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

Divide 4 by 4.

$$y=1$$

There are two solutions for y: 2 and 1. Substitute 2 for y in the equation x=-y+3 to find the corresponding solution for x that satisfies both equations.

$$x=-2+3$$

Add -2 to 3.

$$x=1$$

Now substitute 1 for y in the equation x=-y+3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-1+3$$

Add -1 to 3.

$$x=2$$

The system is now solved.

$$x=1,y=2\text{ or }x=2,y=1$$