Solve {c}{x^2+y^2=25}{x+y=7}

$$\left. \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 25 } \\ { x + y = 7 } \end{array} \right.$$

Solve for x, y

$$x=4\text{, }y=3$$

$$x=3\text{, }y=4$$

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.

$$x+y=7,y^{2}+x^{2}=25$$

Solve x+y=7 for x by isolating x on the left hand side of the equal sign.

$$x+y=7$$

Subtract y from both sides of the equation.

$$x=-y+7$$

Substitute -y+7 for x in the other equation, y^{2}+x^{2}=25.

$$y^{2}+\left(-y+7\right)^{2}=25$$

Square -y+7.

$$y^{2}+y^{2}-14y+49=25$$

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

$$2y^{2}-14y+49=25$$

Subtract 25 from both sides of the equation.

$$2y^{2}-14y+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 7\left(-1\right)\times 2 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square 1\times 7\left(-1\right)\times 2.

$$y=\frac{-\left(-14\right)±\sqrt{196-4\times 2\times 24}}{2\times 2}$$

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

$$y=\frac{-\left(-14\right)±\sqrt{196-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$y=\frac{-\left(-14\right)±\sqrt{196-192}}{2\times 2}$$

Add 196 to -192.

$$y=\frac{-\left(-14\right)±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$y=\frac{-\left(-14\right)±2}{2\times 2}$$

The opposite of 1\times 7\left(-1\right)\times 2 is 14.

$$y=\frac{14±2}{2\times 2}$$

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

$$y=\frac{14±2}{4}$$

Now solve the equation y=\frac{14±2}{4} when ± is plus. Add 14 to 2.

$$y=\frac{16}{4}$$

Divide 16 by 4.

$$y=4$$

Now solve the equation y=\frac{14±2}{4} when ± is minus. Subtract 2 from 14.

$$y=\frac{12}{4}$$

Divide 12 by 4.

$$y=3$$

There are two solutions for y: 4 and 3. Substitute 4 for y in the equation x=-y+7 to find the corresponding solution for x that satisfies both equations.

$$x=-4+7$$

Add -4 to 7.

$$x=3$$

Now substitute 3 for y in the equation x=-y+7 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-3+7$$

Add -3 to 7.

$$x=4$$

The system is now solved.

$$x=3,y=4\text{ or }x=4,y=3$$