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

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

Solve for x, y

$$x=0\text{, }y=2$$

$$x=\frac{16}{17}\approx 0.941176471\text{, }y=-\frac{30}{17}\approx -1.764705882$$

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.

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

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

$$4x+y=2$$

Subtract y from both sides of the equation.

$$4x=-y+2$$

Divide both sides by 4.

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

Substitute -\frac{1}{4}y+\frac{1}{2} for x in the other equation, y^{2}+x^{2}=4.

$$y^{2}+\left(-\frac{1}{4}y+\frac{1}{2}\right)^{2}=4$$

Square -\frac{1}{4}y+\frac{1}{2}.

$$y^{2}+\frac{1}{16}y^{2}-\frac{1}{4}y+\frac{1}{4}=4$$

Add y^{2} to \frac{1}{16}y^{2}.

$$\frac{17}{16}y^{2}-\frac{1}{4}y+\frac{1}{4}=4$$

Subtract 4 from both sides of the equation.

$$\frac{17}{16}y^{2}-\frac{1}{4}y-\frac{15}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{1}{4}\right)^{2} for a, 1\times \frac{1}{2}\left(-\frac{1}{4}\right)\times 2 for b, and -\frac{15}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\left(-\frac{1}{4}\right)^{2}-4\times \frac{17}{16}\left(-\frac{15}{4}\right)}}{2\times \frac{17}{16}}$$

Square 1\times \frac{1}{2}\left(-\frac{1}{4}\right)\times 2.

$$y=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\frac{1}{16}-4\times \frac{17}{16}\left(-\frac{15}{4}\right)}}{2\times \frac{17}{16}}$$

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

$$y=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\frac{1}{16}-\frac{17}{4}\left(-\frac{15}{4}\right)}}{2\times \frac{17}{16}}$$

Multiply -\frac{17}{4} times -\frac{15}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\frac{1+255}{16}}}{2\times \frac{17}{16}}$$

Add \frac{1}{16} to \frac{255}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{4}\right)±\sqrt{16}}{2\times \frac{17}{16}}$$

Take the square root of 16.

$$y=\frac{-\left(-\frac{1}{4}\right)±4}{2\times \frac{17}{16}}$$

The opposite of 1\times \frac{1}{2}\left(-\frac{1}{4}\right)\times 2 is \frac{1}{4}.

$$y=\frac{\frac{1}{4}±4}{2\times \frac{17}{16}}$$

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

$$y=\frac{\frac{1}{4}±4}{\frac{17}{8}}$$

Now solve the equation y=\frac{\frac{1}{4}±4}{\frac{17}{8}} when ± is plus. Add \frac{1}{4} to 4.

$$y=\frac{\frac{17}{4}}{\frac{17}{8}}$$

Divide \frac{17}{4} by \frac{17}{8} by multiplying \frac{17}{4} by the reciprocal of \frac{17}{8}.

$$y=2$$

Now solve the equation y=\frac{\frac{1}{4}±4}{\frac{17}{8}} when ± is minus. Subtract 4 from \frac{1}{4}.

$$y=-\frac{\frac{15}{4}}{\frac{17}{8}}$$

Divide -\frac{15}{4} by \frac{17}{8} by multiplying -\frac{15}{4} by the reciprocal of \frac{17}{8}.

$$y=-\frac{30}{17}$$

There are two solutions for y: 2 and -\frac{30}{17}. Substitute 2 for y in the equation x=-\frac{1}{4}y+\frac{1}{2} to find the corresponding solution for x that satisfies both equations.

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

Multiply -\frac{1}{4} times 2.

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

Add -\frac{1}{4}\times 2 to \frac{1}{2}.

$$x=0$$

Now substitute -\frac{30}{17} for y in the equation x=-\frac{1}{4}y+\frac{1}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{4}\left(-\frac{30}{17}\right)+\frac{1}{2}$$

Multiply -\frac{1}{4} times -\frac{30}{17} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{15}{34}+\frac{1}{2}$$

Add -\frac{30}{17}\left(-\frac{1}{4}\right) to \frac{1}{2}.

$$x=\frac{16}{17}$$

The system is now solved.

$$x=0,y=2\text{ or }x=\frac{16}{17},y=-\frac{30}{17}$$