Solve {l}{x^2+y^2=1}{3x+4y-5=0}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { 3 x + 4 y - 5 = 0 } \end{array} \right.$$

Solve for x, y

$$x=\frac{3}{5}=0.6$$

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

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.

$$3x+4y-5=0,y^{2}+x^{2}=1$$

Solve 3x+4y-5=0 for x by isolating x on the left hand side of the equal sign.

$$3x+4y-5=0$$

Add 5 to both sides of the equation.

$$3x+4y=5$$

Subtract 4y from both sides of the equation.

$$3x=-4y+5$$

Divide both sides by 3.

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

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

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

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

$$y^{2}+\frac{16}{9}y^{2}-\frac{40}{9}y+\frac{25}{9}=1$$

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

$$\frac{25}{9}y^{2}-\frac{40}{9}y+\frac{25}{9}=1$$

Subtract 1 from both sides of the equation.

$$\frac{25}{9}y^{2}-\frac{40}{9}y+\frac{16}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{4}{3}\right)^{2} for a, 1\times \frac{5}{3}\left(-\frac{4}{3}\right)\times 2 for b, and \frac{16}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{40}{9}\right)±\sqrt{\left(-\frac{40}{9}\right)^{2}-4\times \frac{25}{9}\times \frac{16}{9}}}{2\times \frac{25}{9}}$$

Square 1\times \frac{5}{3}\left(-\frac{4}{3}\right)\times 2.

$$y=\frac{-\left(-\frac{40}{9}\right)±\sqrt{\frac{1600}{81}-4\times \frac{25}{9}\times \frac{16}{9}}}{2\times \frac{25}{9}}$$

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

$$y=\frac{-\left(-\frac{40}{9}\right)±\sqrt{\frac{1600}{81}-\frac{100}{9}\times \frac{16}{9}}}{2\times \frac{25}{9}}$$

Multiply -\frac{100}{9} times \frac{16}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{40}{9}\right)±\sqrt{\frac{1600-1600}{81}}}{2\times \frac{25}{9}}$$

Add \frac{1600}{81} to -\frac{1600}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{40}{9}\right)±\sqrt{0}}{2\times \frac{25}{9}}$$

Take the square root of 0.

$$y=-\frac{-\frac{40}{9}}{2\times \frac{25}{9}}$$

The opposite of 1\times \frac{5}{3}\left(-\frac{4}{3}\right)\times 2 is \frac{40}{9}.

$$y=\frac{\frac{40}{9}}{2\times \frac{25}{9}}$$

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

$$y=\frac{\frac{40}{9}}{\frac{50}{9}}$$

Divide \frac{40}{9} by \frac{50}{9} by multiplying \frac{40}{9} by the reciprocal of \frac{50}{9}.

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

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

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

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

$$x=-\frac{16}{15}+\frac{5}{3}$$

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

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

The system is now solved.

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