Solve {l}{x^2+y^2=251}{x+y=7}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 251 } \\ { x + y = 7 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{453}+7}{2}\approx 14.141898327\text{, }y=\frac{7-\sqrt{453}}{2}\approx -7.141898327$$

$$x=\frac{7-\sqrt{453}}{2}\approx -7.141898327\text{, }y=\frac{\sqrt{453}+7}{2}\approx 14.141898327$$

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.

$$x+y=7,y^{2}+x^{2}=251$$

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}=251.

$$y^{2}+\left(-y+7\right)^{2}=251$$

Square -y+7.

$$y^{2}+y^{2}-14y+49=251$$

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

$$2y^{2}-14y+49=251$$

Subtract 251 from both sides of the equation.

$$2y^{2}-14y-202=0$$

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 -202 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\left(-202\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-14\right)±\sqrt{196-4\times 2\left(-202\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-14\right)±\sqrt{196-8\left(-202\right)}}{2\times 2}$$

Multiply -8 times -202.

$$y=\frac{-\left(-14\right)±\sqrt{196+1616}}{2\times 2}$$

Add 196 to 1616.

$$y=\frac{-\left(-14\right)±\sqrt{1812}}{2\times 2}$$

Take the square root of 1812.

$$y=\frac{-\left(-14\right)±2\sqrt{453}}{2\times 2}$$

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

$$y=\frac{14±2\sqrt{453}}{2\times 2}$$

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

$$y=\frac{14±2\sqrt{453}}{4}$$

Now solve the equation y=\frac{14±2\sqrt{453}}{4} when ± is plus. Add 14 to 2\sqrt{453}.

$$y=\frac{2\sqrt{453}+14}{4}$$

Divide 14+2\sqrt{453} by 4.

$$y=\frac{\sqrt{453}+7}{2}$$

Now solve the equation y=\frac{14±2\sqrt{453}}{4} when ± is minus. Subtract 2\sqrt{453} from 14.

$$y=\frac{14-2\sqrt{453}}{4}$$

Divide 14-2\sqrt{453} by 4.

$$y=\frac{7-\sqrt{453}}{2}$$

There are two solutions for y: \frac{7+\sqrt{453}}{2} and \frac{7-\sqrt{453}}{2}. Substitute \frac{7+\sqrt{453}}{2} for y in the equation x=-y+7 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{453}+7}{2}+7$$

Now substitute \frac{7-\sqrt{453}}{2} for y in the equation x=-y+7 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{7-\sqrt{453}}{2}+7$$

The system is now solved.

$$x=-\frac{\sqrt{453}+7}{2}+7,y=\frac{\sqrt{453}+7}{2}\text{ or }x=-\frac{7-\sqrt{453}}{2}+7,y=\frac{7-\sqrt{453}}{2}$$