Solve {l}{x^2+y^2=89}{2x+y=4}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 89 } \\ { 2 x + y = 4 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{429}+8}{5}\approx 5.742463035\text{, }y=\frac{4-2\sqrt{429}}{5}\approx -7.484926071$$

$$x=\frac{8-\sqrt{429}}{5}\approx -2.542463035\text{, }y=\frac{2\sqrt{429}+4}{5}\approx 9.084926071$$

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.

$$2x+y=4,y^{2}+x^{2}=89$$

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

$$2x+y=4$$

Subtract y from both sides of the equation.

$$2x=-y+4$$

Divide both sides by 2.

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

Substitute -\frac{1}{2}y+2 for x in the other equation, y^{2}+x^{2}=89.

$$y^{2}+\left(-\frac{1}{2}y+2\right)^{2}=89$$

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

$$y^{2}+\frac{1}{4}y^{2}-2y+4=89$$

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

$$\frac{5}{4}y^{2}-2y+4=89$$

Subtract 89 from both sides of the equation.

$$\frac{5}{4}y^{2}-2y-85=0$$

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 -85 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times \frac{5}{4}\left(-85\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-2\right)±\sqrt{4-4\times \frac{5}{4}\left(-85\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-2\right)±\sqrt{4-5\left(-85\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -85.

$$y=\frac{-\left(-2\right)±\sqrt{4+425}}{2\times \frac{5}{4}}$$

Add 4 to 425.

$$y=\frac{-\left(-2\right)±\sqrt{429}}{2\times \frac{5}{4}}$$

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

$$y=\frac{2±\sqrt{429}}{2\times \frac{5}{4}}$$

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

$$y=\frac{2±\sqrt{429}}{\frac{5}{2}}$$

Now solve the equation y=\frac{2±\sqrt{429}}{\frac{5}{2}} when ± is plus. Add 2 to \sqrt{429}.

$$y=\frac{\sqrt{429}+2}{\frac{5}{2}}$$

Divide 2+\sqrt{429} by \frac{5}{2} by multiplying 2+\sqrt{429} by the reciprocal of \frac{5}{2}.

$$y=\frac{2\sqrt{429}+4}{5}$$

Now solve the equation y=\frac{2±\sqrt{429}}{\frac{5}{2}} when ± is minus. Subtract \sqrt{429} from 2.

$$y=\frac{2-\sqrt{429}}{\frac{5}{2}}$$

Divide 2-\sqrt{429} by \frac{5}{2} by multiplying 2-\sqrt{429} by the reciprocal of \frac{5}{2}.

$$y=\frac{4-2\sqrt{429}}{5}$$

There are two solutions for y: \frac{4+2\sqrt{429}}{5} and \frac{4-2\sqrt{429}}{5}. Substitute \frac{4+2\sqrt{429}}{5} for y in the equation x=-\frac{1}{2}y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times \frac{2\sqrt{429}+4}{5}+2$$

Multiply -\frac{1}{2} times \frac{4+2\sqrt{429}}{5}.

$$x=-\frac{2\sqrt{429}+4}{2\times 5}+2$$

Now substitute \frac{4-2\sqrt{429}}{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.

$$x=-\frac{1}{2}\times \frac{4-2\sqrt{429}}{5}+2$$

Multiply -\frac{1}{2} times \frac{4-2\sqrt{429}}{5}.

$$x=-\frac{4-2\sqrt{429}}{2\times 5}+2$$

The system is now solved.

$$x=-\frac{2\sqrt{429}+4}{2\times 5}+2,y=\frac{2\sqrt{429}+4}{5}\text{ or }x=-\frac{4-2\sqrt{429}}{2\times 5}+2,y=\frac{4-2\sqrt{429}}{5}$$