Solve {l}{x^2+y^2=1}{4x-3y=5}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { 4 x - 3 y = 5 } \end{array} \right.$$

Solve for x, y

$$x=\frac{4}{5}=0.8$$

$$y=-\frac{3}{5}=-0.6$$

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.

$$4x-3y=5,y^{2}+x^{2}=1$$

Solve 4x-3y=5 for x by isolating x on the left hand side of the equal sign.

$$4x-3y=5$$

Subtract -3y from both sides of the equation.

$$4x=3y+5$$

Divide both sides by 4.

$$x=\frac{3}{4}y+\frac{5}{4}$$

Substitute \frac{3}{4}y+\frac{5}{4} for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(\frac{3}{4}y+\frac{5}{4}\right)^{2}=1$$

Square \frac{3}{4}y+\frac{5}{4}.

$$y^{2}+\frac{9}{16}y^{2}+\frac{15}{8}y+\frac{25}{16}=1$$

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

$$\frac{25}{16}y^{2}+\frac{15}{8}y+\frac{25}{16}=1$$

Subtract 1 from both sides of the equation.

$$\frac{25}{16}y^{2}+\frac{15}{8}y+\frac{9}{16}=0$$

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 \frac{5}{4}\times \frac{3}{4}\times 2 for b, and \frac{9}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{15}{8}±\sqrt{\left(\frac{15}{8}\right)^{2}-4\times \frac{25}{16}\times \frac{9}{16}}}{2\times \frac{25}{16}}$$

Square 1\times \frac{5}{4}\times \frac{3}{4}\times 2.

$$y=\frac{-\frac{15}{8}±\sqrt{\frac{225}{64}-4\times \frac{25}{16}\times \frac{9}{16}}}{2\times \frac{25}{16}}$$

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

$$y=\frac{-\frac{15}{8}±\sqrt{\frac{225}{64}-\frac{25}{4}\times \frac{9}{16}}}{2\times \frac{25}{16}}$$

Multiply -\frac{25}{4} times \frac{9}{16} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{15}{8}±\sqrt{\frac{225-225}{64}}}{2\times \frac{25}{16}}$$

Add \frac{225}{64} to -\frac{225}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{15}{8}±\sqrt{0}}{2\times \frac{25}{16}}$$

Take the square root of 0.

$$y=-\frac{\frac{15}{8}}{2\times \frac{25}{16}}$$

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

$$y=-\frac{\frac{15}{8}}{\frac{25}{8}}$$

Divide -\frac{15}{8} by \frac{25}{8} by multiplying -\frac{15}{8} by the reciprocal of \frac{25}{8}.

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

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

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

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

$$x=-\frac{9}{20}+\frac{5}{4}$$

Add -\frac{3}{5}\times \frac{3}{4} to \frac{5}{4}.

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

The system is now solved.

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