Solve {l}{y=4/9x+4/3}{x^2/25+y^2/16=1}

$$\left\{ \begin{array} { l } { y = \frac { 4 } { 9 } x + \frac { 4 } { 3 } } \\ { \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1 } \end{array} \right.$$

Solve for y, x

$$x=\frac{-45\sqrt{97}-75}{106}\approx -4.888666048\text{, }y=\frac{54-10\sqrt{97}}{53}\approx -0.839407132$$

$$x=\frac{45\sqrt{97}-75}{106}\approx 3.473571708\text{, }y=\frac{10\sqrt{97}+54}{53}\approx 2.877142981$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract \frac{4}{9}x from both sides.

$$y-\frac{4}{9}x=\frac{4}{3}$$

Consider the second equation. Multiply both sides of the equation by 400, the least common multiple of 25,16.

$$16x^{2}+25y^{2}=400$$

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.

$$y-\frac{4}{9}x=\frac{4}{3},16x^{2}+25y^{2}=400$$

Solve y-\frac{4}{9}x=\frac{4}{3} for y by isolating y on the left hand side of the equal sign.

$$y-\frac{4}{9}x=\frac{4}{3}$$

Subtract -\frac{4}{9}x from both sides of the equation.

$$y=\frac{4}{9}x+\frac{4}{3}$$

Substitute \frac{4}{9}x+\frac{4}{3} for y in the other equation, 16x^{2}+25y^{2}=400.

$$16x^{2}+25\left(\frac{4}{9}x+\frac{4}{3}\right)^{2}=400$$

Square \frac{4}{9}x+\frac{4}{3}.

$$16x^{2}+25\left(\frac{16}{81}x^{2}+\frac{32}{27}x+\frac{16}{9}\right)=400$$

Multiply 25 times \frac{16}{81}x^{2}+\frac{32}{27}x+\frac{16}{9}.

$$16x^{2}+\frac{400}{81}x^{2}+\frac{800}{27}x+\frac{400}{9}=400$$

Add 16x^{2} to \frac{400}{81}x^{2}.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x+\frac{400}{9}=400$$

Subtract 400 from both sides of the equation.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x-\frac{3200}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16+25\times \left(\frac{4}{9}\right)^{2} for a, 25\times \frac{4}{3}\times \frac{4}{9}\times 2 for b, and -\frac{3200}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{800}{27}±\sqrt{\left(\frac{800}{27}\right)^{2}-4\times \frac{1696}{81}\left(-\frac{3200}{9}\right)}}{2\times \frac{1696}{81}}$$

Square 25\times \frac{4}{3}\times \frac{4}{9}\times 2.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{640000}{729}-4\times \frac{1696}{81}\left(-\frac{3200}{9}\right)}}{2\times \frac{1696}{81}}$$

Multiply -4 times 16+25\times \left(\frac{4}{9}\right)^{2}.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{640000}{729}-\frac{6784}{81}\left(-\frac{3200}{9}\right)}}{2\times \frac{1696}{81}}$$

Multiply -\frac{6784}{81} times -\frac{3200}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{640000+21708800}{729}}}{2\times \frac{1696}{81}}$$

Add \frac{640000}{729} to \frac{21708800}{729} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{2483200}{81}}}{2\times \frac{1696}{81}}$$

Take the square root of \frac{2483200}{81}.

$$x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{2\times \frac{1696}{81}}$$

Multiply 2 times 16+25\times \left(\frac{4}{9}\right)^{2}.

$$x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{\frac{3392}{81}}$$

Now solve the equation x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{\frac{3392}{81}} when ± is plus. Add -\frac{800}{27} to \frac{160\sqrt{97}}{9}.

$$x=\frac{\frac{160\sqrt{97}}{9}-\frac{800}{27}}{\frac{3392}{81}}$$

Divide -\frac{800}{27}+\frac{160\sqrt{97}}{9} by \frac{3392}{81} by multiplying -\frac{800}{27}+\frac{160\sqrt{97}}{9} by the reciprocal of \frac{3392}{81}.

$$x=\frac{45\sqrt{97}-75}{106}$$

Now solve the equation x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{\frac{3392}{81}} when ± is minus. Subtract \frac{160\sqrt{97}}{9} from -\frac{800}{27}.

$$x=\frac{-\frac{160\sqrt{97}}{9}-\frac{800}{27}}{\frac{3392}{81}}$$

Divide -\frac{800}{27}-\frac{160\sqrt{97}}{9} by \frac{3392}{81} by multiplying -\frac{800}{27}-\frac{160\sqrt{97}}{9} by the reciprocal of \frac{3392}{81}.

$$x=\frac{-45\sqrt{97}-75}{106}$$

There are two solutions for x: \frac{-75+45\sqrt{97}}{106} and \frac{-75-45\sqrt{97}}{106}. Substitute \frac{-75+45\sqrt{97}}{106} for x in the equation y=\frac{4}{9}x+\frac{4}{3} to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{9}\times \frac{45\sqrt{97}-75}{106}+\frac{4}{3}$$

Multiply \frac{4}{9} times \frac{-75+45\sqrt{97}}{106}.

$$y=\frac{4\times \frac{45\sqrt{97}-75}{106}}{9}+\frac{4}{3}$$

Now substitute \frac{-75-45\sqrt{97}}{106} for x in the equation y=\frac{4}{9}x+\frac{4}{3} and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{9}\times \frac{-45\sqrt{97}-75}{106}+\frac{4}{3}$$

Multiply \frac{4}{9} times \frac{-75-45\sqrt{97}}{106}.

$$y=\frac{4\times \frac{-45\sqrt{97}-75}{106}}{9}+\frac{4}{3}$$

The system is now solved.

$$y=\frac{4\times \frac{45\sqrt{97}-75}{106}}{9}+\frac{4}{3},x=\frac{45\sqrt{97}-75}{106}\text{ or }y=\frac{4\times \frac{-45\sqrt{97}-75}{106}}{9}+\frac{4}{3},x=\frac{-45\sqrt{97}-75}{106}$$