Solve {l}{4x+2y=8}{x^2+y^2=64}

$$\left. \begin{array} { l } { 4 x + 2 y = 8 } \\ { x ^ { 2 } + y ^ { 2 } = 64 } \end{array} \right.$$

Solve for x, y

$$x=\frac{4\sqrt{19}+8}{5}\approx 5.087119155\text{, }y=\frac{4-8\sqrt{19}}{5}\approx -6.17423831$$

$$x=\frac{8-4\sqrt{19}}{5}\approx -1.887119155\text{, }y=\frac{8\sqrt{19}+4}{5}\approx 7.77423831$$

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.

$$4x+2y=8,y^{2}+x^{2}=64$$

Solve 4x+2y=8 for x by isolating x on the left hand side of the equal sign.

$$4x+2y=8$$

Subtract 2y from both sides of the equation.

$$4x=-2y+8$$

Divide both sides by 4.

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

Substitute -\frac{1}{2}y+2 for x in the other equation, y^{2}+x^{2}=64.

$$y^{2}+\left(-\frac{1}{2}y+2\right)^{2}=64$$

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

$$y^{2}+\frac{1}{4}y^{2}-2y+4=64$$

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

$$\frac{5}{4}y^{2}-2y+4=64$$

Subtract 64 from both sides of the equation.

$$\frac{5}{4}y^{2}-2y-60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{1}{2}\right)^{2} for a, 1\times 2\left(-\frac{1}{2}\right)\times 2 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times \frac{5}{4}\left(-60\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-2\right)±\sqrt{4-4\times \frac{5}{4}\left(-60\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-2\right)±\sqrt{4-5\left(-60\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -60.

$$y=\frac{-\left(-2\right)±\sqrt{4+300}}{2\times \frac{5}{4}}$$

Add 4 to 300.

$$y=\frac{-\left(-2\right)±\sqrt{304}}{2\times \frac{5}{4}}$$

Take the square root of 304.

$$y=\frac{-\left(-2\right)±4\sqrt{19}}{2\times \frac{5}{4}}$$

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

$$y=\frac{2±4\sqrt{19}}{2\times \frac{5}{4}}$$

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

$$y=\frac{2±4\sqrt{19}}{\frac{5}{2}}$$

Now solve the equation y=\frac{2±4\sqrt{19}}{\frac{5}{2}} when ± is plus. Add 2 to 4\sqrt{19}.

$$y=\frac{4\sqrt{19}+2}{\frac{5}{2}}$$

Divide 2+4\sqrt{19} by \frac{5}{2} by multiplying 2+4\sqrt{19} by the reciprocal of \frac{5}{2}.

$$y=\frac{8\sqrt{19}+4}{5}$$

Now solve the equation y=\frac{2±4\sqrt{19}}{\frac{5}{2}} when ± is minus. Subtract 4\sqrt{19} from 2.

$$y=\frac{2-4\sqrt{19}}{\frac{5}{2}}$$

Divide 2-4\sqrt{19} by \frac{5}{2} by multiplying 2-4\sqrt{19} by the reciprocal of \frac{5}{2}.

$$y=\frac{4-8\sqrt{19}}{5}$$

There are two solutions for y: \frac{4+8\sqrt{19}}{5} and \frac{4-8\sqrt{19}}{5}. Substitute \frac{4+8\sqrt{19}}{5} for y in the equation x=-\frac{1}{2}y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times \frac{8\sqrt{19}+4}{5}+2$$

Multiply -\frac{1}{2} times \frac{4+8\sqrt{19}}{5}.

$$x=-\frac{8\sqrt{19}+4}{2\times 5}+2$$

Now substitute \frac{4-8\sqrt{19}}{5} for y in the equation x=-\frac{1}{2}y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times \frac{4-8\sqrt{19}}{5}+2$$

Multiply -\frac{1}{2} times \frac{4-8\sqrt{19}}{5}.

$$x=-\frac{4-8\sqrt{19}}{2\times 5}+2$$

The system is now solved.

$$x=-\frac{8\sqrt{19}+4}{2\times 5}+2,y=\frac{8\sqrt{19}+4}{5}\text{ or }x=-\frac{4-8\sqrt{19}}{2\times 5}+2,y=\frac{4-8\sqrt{19}}{5}$$