Solve {l}{x/y=1/4}{x^2+y^2=1}

$$\left. \begin{array} { l } { \frac { x } { y } = \frac { 1 } { 4 } } \\ { x ^ { 2 } + y ^ { 2 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{\sqrt{17}}{17}\approx -0.242535625\text{, }y=-\frac{4\sqrt{17}}{17}\approx -0.9701425$$

$$x=\frac{\sqrt{17}}{17}\approx 0.242535625\text{, }y=\frac{4\sqrt{17}}{17}\approx 0.9701425$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4y, the least common multiple of y,4.

$$4x=y$$

Divide both sides by 4.

$$x=\frac{1}{4}y$$

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

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

Square \frac{1}{4}y.

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

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

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

Subtract 1 from both sides of the equation.

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

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

$$y=\frac{0±\sqrt{0^{2}-4\times \frac{17}{16}\left(-1\right)}}{2\times \frac{17}{16}}$$

Square 1\times 0\times \frac{1}{4}\times 2.

$$y=\frac{0±\sqrt{-4\times \frac{17}{16}\left(-1\right)}}{2\times \frac{17}{16}}$$

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

$$y=\frac{0±\sqrt{-\frac{17}{4}\left(-1\right)}}{2\times \frac{17}{16}}$$

Multiply -\frac{17}{4} times -1.

$$y=\frac{0±\sqrt{\frac{17}{4}}}{2\times \frac{17}{16}}$$

Take the square root of \frac{17}{4}.

$$y=\frac{0±\frac{\sqrt{17}}{2}}{2\times \frac{17}{16}}$$

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

$$y=\frac{0±\frac{\sqrt{17}}{2}}{\frac{17}{8}}$$

Now solve the equation y=\frac{0±\frac{\sqrt{17}}{2}}{\frac{17}{8}} when ± is plus.

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

Now solve the equation y=\frac{0±\frac{\sqrt{17}}{2}}{\frac{17}{8}} when ± is minus.

$$y=-\frac{4\sqrt{17}}{17}$$

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

$$x=\frac{1}{4}\times \frac{4\sqrt{17}}{17}$$

Multiply \frac{1}{4} times \frac{4\sqrt{17}}{17}.

$$x=\frac{4\sqrt{17}}{4\times 17}$$

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

$$x=\frac{1}{4}\left(-\frac{4\sqrt{17}}{17}\right)$$

Multiply \frac{1}{4} times -\frac{4\sqrt{17}}{17}.

$$x=-\frac{4\sqrt{17}}{4\times 17}$$

The system is now solved.

$$x=\frac{4\sqrt{17}}{4\times 17},y=\frac{4\sqrt{17}}{17}\text{ or }x=-\frac{4\sqrt{17}}{4\times 17},y=-\frac{4\sqrt{17}}{17}$$