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

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

Solve for y, x

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

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

Steps Using Substitution and Quadratic Formula

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

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

Consider the second equation. Add x^{2} to both sides.

$$y^{2}+x^{2}=9$$

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.

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

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

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

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

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

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

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

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

$$x^{2}+\frac{16}{9}x^{2}-\frac{32}{3}x+16=9$$

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

$$\frac{25}{9}x^{2}-\frac{32}{3}x+16=9$$

Subtract 9 from both sides of the equation.

$$\frac{25}{9}x^{2}-\frac{32}{3}x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{4}{3}\right)^{2} for a, 1\left(-4\right)\times \frac{4}{3}\times 2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{32}{3}\right)±\sqrt{\left(-\frac{32}{3}\right)^{2}-4\times \frac{25}{9}\times 7}}{2\times \frac{25}{9}}$$

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

$$x=\frac{-\left(-\frac{32}{3}\right)±\sqrt{\frac{1024}{9}-4\times \frac{25}{9}\times 7}}{2\times \frac{25}{9}}$$

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

$$x=\frac{-\left(-\frac{32}{3}\right)±\sqrt{\frac{1024}{9}-\frac{100}{9}\times 7}}{2\times \frac{25}{9}}$$

Multiply -\frac{100}{9} times 7.

$$x=\frac{-\left(-\frac{32}{3}\right)±\sqrt{\frac{1024-700}{9}}}{2\times \frac{25}{9}}$$

Add \frac{1024}{9} to -\frac{700}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{32}{3}\right)±\sqrt{36}}{2\times \frac{25}{9}}$$

Take the square root of 36.

$$x=\frac{-\left(-\frac{32}{3}\right)±6}{2\times \frac{25}{9}}$$

The opposite of 1\left(-4\right)\times \frac{4}{3}\times 2 is \frac{32}{3}.

$$x=\frac{\frac{32}{3}±6}{2\times \frac{25}{9}}$$

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

$$x=\frac{\frac{32}{3}±6}{\frac{50}{9}}$$

Now solve the equation x=\frac{\frac{32}{3}±6}{\frac{50}{9}} when ± is plus. Add \frac{32}{3} to 6.

$$x=\frac{\frac{50}{3}}{\frac{50}{9}}$$

Divide \frac{50}{3} by \frac{50}{9} by multiplying \frac{50}{3} by the reciprocal of \frac{50}{9}.

$$x=3$$

Now solve the equation x=\frac{\frac{32}{3}±6}{\frac{50}{9}} when ± is minus. Subtract 6 from \frac{32}{3}.

$$x=\frac{\frac{14}{3}}{\frac{50}{9}}$$

Divide \frac{14}{3} by \frac{50}{9} by multiplying \frac{14}{3} by the reciprocal of \frac{50}{9}.

$$x=\frac{21}{25}$$

There are two solutions for x: 3 and \frac{21}{25}. Substitute 3 for x in the equation y=\frac{4}{3}x-4 to find the corresponding solution for y that satisfies both equations.

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

Multiply \frac{4}{3} times 3.

$$y=4-4$$

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

$$y=0$$

Now substitute \frac{21}{25} for x in the equation y=\frac{4}{3}x-4 and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{3}\times \frac{21}{25}-4$$

Multiply \frac{4}{3} times \frac{21}{25} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{28}{25}-4$$

Add \frac{21}{25}\times \frac{4}{3} to -4.

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

The system is now solved.

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