Solve {l}{x^2/25+y^2/16=1}{y=4/5x-12/5}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1 } \\ { y = \frac { 4 } { 5 } x - \frac { 12 } { 5 } } \end{array} \right.$$

Solve for x, y

$$x=\frac{3-\sqrt{41}}{2}\approx -1.701562119\text{, }y=\frac{-2\sqrt{41}-6}{5}\approx -3.761249695$$

$$x=\frac{\sqrt{41}+3}{2}\approx 4.701562119\text{, }y=\frac{2\sqrt{41}-6}{5}\approx 1.361249695$$

Steps Using Substitution and Quadratic Formula

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

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

Consider the second equation. Subtract \frac{4}{5}x from both sides.

$$y-\frac{4}{5}x=-\frac{12}{5}$$

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.

$$y-\frac{4}{5}x=-\frac{12}{5},16x^{2}+25y^{2}=400$$

Solve y-\frac{4}{5}x=-\frac{12}{5} for y by isolating y on the left hand side of the equal sign.

$$y-\frac{4}{5}x=-\frac{12}{5}$$

Subtract -\frac{4}{5}x from both sides of the equation.

$$y=\frac{4}{5}x-\frac{12}{5}$$

Substitute \frac{4}{5}x-\frac{12}{5} for y in the other equation, 16x^{2}+25y^{2}=400.

$$16x^{2}+25\left(\frac{4}{5}x-\frac{12}{5}\right)^{2}=400$$

Square \frac{4}{5}x-\frac{12}{5}.

$$16x^{2}+25\left(\frac{16}{25}x^{2}-\frac{96}{25}x+\frac{144}{25}\right)=400$$

Multiply 25 times \frac{16}{25}x^{2}-\frac{96}{25}x+\frac{144}{25}.

$$16x^{2}+16x^{2}-96x+144=400$$

Add 16x^{2} to 16x^{2}.

$$32x^{2}-96x+144=400$$

Subtract 400 from both sides of the equation.

$$32x^{2}-96x-256=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16+25\times \left(\frac{4}{5}\right)^{2} for a, 25\left(-\frac{12}{5}\right)\times \frac{4}{5}\times 2 for b, and -256 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-96\right)±\sqrt{\left(-96\right)^{2}-4\times 32\left(-256\right)}}{2\times 32}$$

Square 25\left(-\frac{12}{5}\right)\times \frac{4}{5}\times 2.

$$x=\frac{-\left(-96\right)±\sqrt{9216-4\times 32\left(-256\right)}}{2\times 32}$$

Multiply -4 times 16+25\times \left(\frac{4}{5}\right)^{2}.

$$x=\frac{-\left(-96\right)±\sqrt{9216-128\left(-256\right)}}{2\times 32}$$

Multiply -128 times -256.

$$x=\frac{-\left(-96\right)±\sqrt{9216+32768}}{2\times 32}$$

Add 9216 to 32768.

$$x=\frac{-\left(-96\right)±\sqrt{41984}}{2\times 32}$$

Take the square root of 41984.

$$x=\frac{-\left(-96\right)±32\sqrt{41}}{2\times 32}$$

The opposite of 25\left(-\frac{12}{5}\right)\times \frac{4}{5}\times 2 is 96.

$$x=\frac{96±32\sqrt{41}}{2\times 32}$$

Multiply 2 times 16+25\times \left(\frac{4}{5}\right)^{2}.

$$x=\frac{96±32\sqrt{41}}{64}$$

Now solve the equation x=\frac{96±32\sqrt{41}}{64} when ± is plus. Add 96 to 32\sqrt{41}.

$$x=\frac{32\sqrt{41}+96}{64}$$

Divide 96+32\sqrt{41} by 64.

$$x=\frac{\sqrt{41}+3}{2}$$

Now solve the equation x=\frac{96±32\sqrt{41}}{64} when ± is minus. Subtract 32\sqrt{41} from 96.

$$x=\frac{96-32\sqrt{41}}{64}$$

Divide 96-32\sqrt{41} by 64.

$$x=\frac{3-\sqrt{41}}{2}$$

There are two solutions for x: \frac{3+\sqrt{41}}{2} and \frac{3-\sqrt{41}}{2}. Substitute \frac{3+\sqrt{41}}{2} for x in the equation y=\frac{4}{5}x-\frac{12}{5} to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{5}\times \frac{\sqrt{41}+3}{2}-\frac{12}{5}$$

Multiply \frac{4}{5} times \frac{3+\sqrt{41}}{2}.

$$y=\frac{4\times \frac{\sqrt{41}+3}{2}-12}{5}$$

Now substitute \frac{3-\sqrt{41}}{2} for x in the equation y=\frac{4}{5}x-\frac{12}{5} and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{5}\times \frac{3-\sqrt{41}}{2}-\frac{12}{5}$$

Multiply \frac{4}{5} times \frac{3-\sqrt{41}}{2}.

$$y=\frac{4\times \frac{3-\sqrt{41}}{2}-12}{5}$$

The system is now solved.

$$y=\frac{4\times \frac{\sqrt{41}+3}{2}-12}{5},x=\frac{\sqrt{41}+3}{2}\text{ or }y=\frac{4\times \frac{3-\sqrt{41}}{2}-12}{5},x=\frac{3-\sqrt{41}}{2}$$