Solve {l}{2x+y=10}{x^2+y^2=28}

$$\left\{ \begin{array} { l } { 2 x + y = 10 } \\ { x ^ { 2 } + y ^ { 2 } = 28 } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\sqrt{10}}{5}+4\approx 5.264911064\text{, }y=-\frac{4\sqrt{10}}{5}+2\approx -0.529822128$$

$$x=-\frac{2\sqrt{10}}{5}+4\approx 2.735088936\text{, }y=\frac{4\sqrt{10}}{5}+2\approx 4.529822128$$

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.

$$2x+y=10,y^{2}+x^{2}=28$$

Solve 2x+y=10 for x by isolating x on the left hand side of the equal sign.

$$2x+y=10$$

Subtract y from both sides of the equation.

$$2x=-y+10$$

Divide both sides by 2.

$$x=-\frac{1}{2}y+5$$

Substitute -\frac{1}{2}y+5 for x in the other equation, y^{2}+x^{2}=28.

$$y^{2}+\left(-\frac{1}{2}y+5\right)^{2}=28$$

Square -\frac{1}{2}y+5.

$$y^{2}+\frac{1}{4}y^{2}-5y+25=28$$

Add y^{2} to \frac{1}{4}y^{2}.

$$\frac{5}{4}y^{2}-5y+25=28$$

Subtract 28 from both sides of the equation.

$$\frac{5}{4}y^{2}-5y-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{1}{2}\right)^{2} for a, 1\times 5\left(-\frac{1}{2}\right)\times 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times \frac{5}{4}\left(-3\right)}}{2\times \frac{5}{4}}$$

Square 1\times 5\left(-\frac{1}{2}\right)\times 2.

$$y=\frac{-\left(-5\right)±\sqrt{25-4\times \frac{5}{4}\left(-3\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-5\right)±\sqrt{25-5\left(-3\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -3.

$$y=\frac{-\left(-5\right)±\sqrt{25+15}}{2\times \frac{5}{4}}$$

Add 25 to 15.

$$y=\frac{-\left(-5\right)±\sqrt{40}}{2\times \frac{5}{4}}$$

Take the square root of 40.

$$y=\frac{-\left(-5\right)±2\sqrt{10}}{2\times \frac{5}{4}}$$

The opposite of 1\times 5\left(-\frac{1}{2}\right)\times 2 is 5.

$$y=\frac{5±2\sqrt{10}}{2\times \frac{5}{4}}$$

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

$$y=\frac{5±2\sqrt{10}}{\frac{5}{2}}$$

Now solve the equation y=\frac{5±2\sqrt{10}}{\frac{5}{2}} when ± is plus. Add 5 to 2\sqrt{10}.

$$y=\frac{2\sqrt{10}+5}{\frac{5}{2}}$$

Divide 5+2\sqrt{10} by \frac{5}{2} by multiplying 5+2\sqrt{10} by the reciprocal of \frac{5}{2}.

$$y=\frac{4\sqrt{10}}{5}+2$$

Now solve the equation y=\frac{5±2\sqrt{10}}{\frac{5}{2}} when ± is minus. Subtract 2\sqrt{10} from 5.

$$y=\frac{5-2\sqrt{10}}{\frac{5}{2}}$$

Divide 5-2\sqrt{10} by \frac{5}{2} by multiplying 5-2\sqrt{10} by the reciprocal of \frac{5}{2}.

$$y=-\frac{4\sqrt{10}}{5}+2$$

There are two solutions for y: 2+\frac{4\sqrt{10}}{5} and 2-\frac{4\sqrt{10}}{5}. Substitute 2+\frac{4\sqrt{10}}{5} for y in the equation x=-\frac{1}{2}y+5 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\left(\frac{4\sqrt{10}}{5}+2\right)+5$$

Multiply -\frac{1}{2} times 2+\frac{4\sqrt{10}}{5}.

$$x=-\frac{\frac{4\sqrt{10}}{5}+2}{2}+5$$

Now substitute 2-\frac{4\sqrt{10}}{5} for y in the equation x=-\frac{1}{2}y+5 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\left(-\frac{4\sqrt{10}}{5}+2\right)+5$$

Multiply -\frac{1}{2} times 2-\frac{4\sqrt{10}}{5}.

$$x=-\frac{-\frac{4\sqrt{10}}{5}+2}{2}+5$$

The system is now solved.

$$x=-\frac{\frac{4\sqrt{10}}{5}+2}{2}+5,y=\frac{4\sqrt{10}}{5}+2\text{ or }x=-\frac{-\frac{4\sqrt{10}}{5}+2}{2}+5,y=-\frac{4\sqrt{10}}{5}+2$$