Solve {l}{y=4/3x+3}{x^2+y^2=9}

$$\left. \begin{array} { l } { y = \frac { 4 } { 3 } x + 3 } \\ { x ^ { 2 } + y ^ { 2 } = 9 } \end{array} \right.$$

Solve for y, x

$$x=0\text{, }y=3$$

$$x=-\frac{72}{25}=-2.88\text{, }y=-\frac{21}{25}=-0.84$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract \frac{4}{3}x from both sides.

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

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.

$$y-\frac{4}{3}x=3,x^{2}+y^{2}=9$$

Solve y-\frac{4}{3}x=3 for y by isolating y on the left hand side of the equal sign.

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

Subtract -\frac{4}{3}x from both sides of the equation.

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

Substitute \frac{4}{3}x+3 for y in the other equation, x^{2}+y^{2}=9.

$$x^{2}+\left(\frac{4}{3}x+3\right)^{2}=9$$

Square \frac{4}{3}x+3.

$$x^{2}+\frac{16}{9}x^{2}+8x+9=9$$

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

$$\frac{25}{9}x^{2}+8x+9=9$$

Subtract 9 from both sides of the equation.

$$\frac{25}{9}x^{2}+8x=0$$

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

$$x=\frac{-8±\sqrt{8^{2}}}{2\times \frac{25}{9}}$$

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\times \frac{25}{9}}$$

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

$$x=\frac{-8±8}{\frac{50}{9}}$$

Now solve the equation x=\frac{-8±8}{\frac{50}{9}} when ± is plus. Add -8 to 8.

$$x=\frac{0}{\frac{50}{9}}$$

Divide 0 by \frac{50}{9} by multiplying 0 by the reciprocal of \frac{50}{9}.

$$x=0$$

Now solve the equation x=\frac{-8±8}{\frac{50}{9}} when ± is minus. Subtract 8 from -8.

$$x=-\frac{16}{\frac{50}{9}}$$

Divide -16 by \frac{50}{9} by multiplying -16 by the reciprocal of \frac{50}{9}.

$$x=-\frac{72}{25}$$

There are two solutions for x: 0 and -\frac{72}{25}. Substitute 0 for x in the equation y=\frac{4}{3}x+3 to find the corresponding solution for y that satisfies both equations.

$$y=3$$

Now substitute -\frac{72}{25} for x in the equation y=\frac{4}{3}x+3 and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{3}\left(-\frac{72}{25}\right)+3$$

Multiply \frac{4}{3} times -\frac{72}{25} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{96}{25}+3$$

Add -\frac{72}{25}\times \frac{4}{3} to 3.

$$y=-\frac{21}{25}$$

The system is now solved.

$$y=3,x=0\text{ or }y=-\frac{21}{25},x=-\frac{72}{25}$$