Solve {l}{3x+4y=10}{x^2-y^2=25}

$$\left. \begin{array} { l } { 3 x + 4 y = 10 } \\ { x ^ { 2 } - y ^ { 2 } = 25 } \end{array} \right.$$

Solve for x, y

$$x=\frac{20\sqrt{11}-30}{7}\approx 5.190356544\text{, }y=\frac{40-15\sqrt{11}}{7}\approx -1.392767408$$

$$x=\frac{-20\sqrt{11}-30}{7}\approx -13.761785115\text{, }y=\frac{15\sqrt{11}+40}{7}\approx 12.821338836$$

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.

$$3x+4y=10,-y^{2}+x^{2}=25$$

Solve 3x+4y=10 for x by isolating x on the left hand side of the equal sign.

$$3x+4y=10$$

Subtract 4y from both sides of the equation.

$$3x=-4y+10$$

Divide both sides by 3.

$$x=-\frac{4}{3}y+\frac{10}{3}$$

Substitute -\frac{4}{3}y+\frac{10}{3} for x in the other equation, -y^{2}+x^{2}=25.

$$-y^{2}+\left(-\frac{4}{3}y+\frac{10}{3}\right)^{2}=25$$

Square -\frac{4}{3}y+\frac{10}{3}.

$$-y^{2}+\frac{16}{9}y^{2}-\frac{80}{9}y+\frac{100}{9}=25$$

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

$$\frac{7}{9}y^{2}-\frac{80}{9}y+\frac{100}{9}=25$$

Subtract 25 from both sides of the equation.

$$\frac{7}{9}y^{2}-\frac{80}{9}y-\frac{125}{9}=0$$

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{10}{3}\left(-\frac{4}{3}\right)\times 2 for b, and -\frac{125}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{80}{9}\right)±\sqrt{\left(-\frac{80}{9}\right)^{2}-4\times \frac{7}{9}\left(-\frac{125}{9}\right)}}{2\times \frac{7}{9}}$$

Square 1\times \frac{10}{3}\left(-\frac{4}{3}\right)\times 2.

$$y=\frac{-\left(-\frac{80}{9}\right)±\sqrt{\frac{6400}{81}-4\times \frac{7}{9}\left(-\frac{125}{9}\right)}}{2\times \frac{7}{9}}$$

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

$$y=\frac{-\left(-\frac{80}{9}\right)±\sqrt{\frac{6400}{81}-\frac{28}{9}\left(-\frac{125}{9}\right)}}{2\times \frac{7}{9}}$$

Multiply -\frac{28}{9} times -\frac{125}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{80}{9}\right)±\sqrt{\frac{6400+3500}{81}}}{2\times \frac{7}{9}}$$

Add \frac{6400}{81} to \frac{3500}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{80}{9}\right)±\sqrt{\frac{1100}{9}}}{2\times \frac{7}{9}}$$

Take the square root of \frac{1100}{9}.

$$y=\frac{-\left(-\frac{80}{9}\right)±\frac{10\sqrt{11}}{3}}{2\times \frac{7}{9}}$$

The opposite of 1\times \frac{10}{3}\left(-\frac{4}{3}\right)\times 2 is \frac{80}{9}.

$$y=\frac{\frac{80}{9}±\frac{10\sqrt{11}}{3}}{2\times \frac{7}{9}}$$

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

$$y=\frac{\frac{80}{9}±\frac{10\sqrt{11}}{3}}{\frac{14}{9}}$$

Now solve the equation y=\frac{\frac{80}{9}±\frac{10\sqrt{11}}{3}}{\frac{14}{9}} when ± is plus. Add \frac{80}{9} to \frac{10\sqrt{11}}{3}.

$$y=\frac{\frac{10\sqrt{11}}{3}+\frac{80}{9}}{\frac{14}{9}}$$

Divide \frac{80}{9}+\frac{10\sqrt{11}}{3} by \frac{14}{9} by multiplying \frac{80}{9}+\frac{10\sqrt{11}}{3} by the reciprocal of \frac{14}{9}.

$$y=\frac{15\sqrt{11}+40}{7}$$

Now solve the equation y=\frac{\frac{80}{9}±\frac{10\sqrt{11}}{3}}{\frac{14}{9}} when ± is minus. Subtract \frac{10\sqrt{11}}{3} from \frac{80}{9}.

$$y=\frac{-\frac{10\sqrt{11}}{3}+\frac{80}{9}}{\frac{14}{9}}$$

Divide \frac{80}{9}-\frac{10\sqrt{11}}{3} by \frac{14}{9} by multiplying \frac{80}{9}-\frac{10\sqrt{11}}{3} by the reciprocal of \frac{14}{9}.

$$y=\frac{40-15\sqrt{11}}{7}$$

There are two solutions for y: \frac{40+15\sqrt{11}}{7} and \frac{40-15\sqrt{11}}{7}. Substitute \frac{40+15\sqrt{11}}{7} for y in the equation x=-\frac{4}{3}y+\frac{10}{3} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{4}{3}\times \frac{15\sqrt{11}+40}{7}+\frac{10}{3}$$

Multiply -\frac{4}{3} times \frac{40+15\sqrt{11}}{7}.

$$x=\frac{-4\times \frac{15\sqrt{11}+40}{7}+10}{3}$$

Now substitute \frac{40-15\sqrt{11}}{7} for y in the equation x=-\frac{4}{3}y+\frac{10}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{4}{3}\times \frac{40-15\sqrt{11}}{7}+\frac{10}{3}$$

Multiply -\frac{4}{3} times \frac{40-15\sqrt{11}}{7}.

$$x=\frac{-4\times \frac{40-15\sqrt{11}}{7}+10}{3}$$

The system is now solved.

$$x=\frac{-4\times \frac{15\sqrt{11}+40}{7}+10}{3},y=\frac{15\sqrt{11}+40}{7}\text{ or }x=\frac{-4\times \frac{40-15\sqrt{11}}{7}+10}{3},y=\frac{40-15\sqrt{11}}{7}$$