Solve {l}{4x-3y=0}{x^2+y^2=1}

$$\left\{ \begin{array} { l } { 4 x - 3 y = 0 } \\ { x ^ { 2 } + y ^ { 2 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{3}{5}=-0.6\text{, }y=-\frac{4}{5}=-0.8$$

$$x=\frac{3}{5}=0.6\text{, }y=\frac{4}{5}=0.8$$

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.

$$4x-3y=0,y^{2}+x^{2}=1$$

Solve 4x-3y=0 for x by isolating x on the left hand side of the equal sign.

$$4x-3y=0$$

Subtract -3y from both sides of the equation.

$$4x=3y$$

Divide both sides by 4.

$$x=\frac{3}{4}y$$

Substitute \frac{3}{4}y for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(\frac{3}{4}y\right)^{2}=1$$

Square \frac{3}{4}y.

$$y^{2}+\frac{9}{16}y^{2}=1$$

Add y^{2} to \frac{9}{16}y^{2}.

$$\frac{25}{16}y^{2}=1$$

Subtract 1 from both sides of the equation.

$$\frac{25}{16}y^{2}-1=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\times \frac{25}{16}\left(-1\right)}}{2\times \frac{25}{16}}$$

Square 1\times 0\times \frac{3}{4}\times 2.

$$y=\frac{0±\sqrt{-4\times \frac{25}{16}\left(-1\right)}}{2\times \frac{25}{16}}$$

Multiply -4 times 1+1\times \left(\frac{3}{4}\right)^{2}.

$$y=\frac{0±\sqrt{-\frac{25}{4}\left(-1\right)}}{2\times \frac{25}{16}}$$

Multiply -\frac{25}{4} times -1.

$$y=\frac{0±\sqrt{\frac{25}{4}}}{2\times \frac{25}{16}}$$

Take the square root of \frac{25}{4}.

$$y=\frac{0±\frac{5}{2}}{2\times \frac{25}{16}}$$

Multiply 2 times 1+1\times \left(\frac{3}{4}\right)^{2}.

$$y=\frac{0±\frac{5}{2}}{\frac{25}{8}}$$

Now solve the equation y=\frac{0±\frac{5}{2}}{\frac{25}{8}} when ± is plus.

$$y=\frac{4}{5}$$

Now solve the equation y=\frac{0±\frac{5}{2}}{\frac{25}{8}} when ± is minus.

$$y=-\frac{4}{5}$$

There are two solutions for y: \frac{4}{5} and -\frac{4}{5}. Substitute \frac{4}{5} for y in the equation x=\frac{3}{4}y to find the corresponding solution for x that satisfies both equations.

$$x=\frac{3}{4}\times \frac{4}{5}$$

Multiply \frac{3}{4} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{3}{5}$$

Now substitute -\frac{4}{5} for y in the equation x=\frac{3}{4}y and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{3}{4}\left(-\frac{4}{5}\right)$$

Multiply \frac{3}{4} times -\frac{4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{3}{5}$$

The system is now solved.

$$x=\frac{3}{5},y=\frac{4}{5}\text{ or }x=-\frac{3}{5},y=-\frac{4}{5}$$