Solve {l}{y^2-x^2=5}{3x+2y=12}

$$\left. \begin{array} { l } { y ^ { 2 } - x ^ { 2 } = 5 } \\ { 3 x + 2 y = 12 } \end{array} \right.$$

Solve for y, x

$$x=\frac{62}{5}=12.4\text{, }y=-\frac{63}{5}=-12.6$$

$$x=2\text{, }y=3$$

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.

$$3x+2y=12,y^{2}-x^{2}=5$$

Solve 3x+2y=12 for x by isolating x on the left hand side of the equal sign.

$$3x+2y=12$$

Subtract 2y from both sides of the equation.

$$3x=-2y+12$$

Divide both sides by 3.

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

Substitute -\frac{2}{3}y+4 for x in the other equation, y^{2}-x^{2}=5.

$$y^{2}-\left(-\frac{2}{3}y+4\right)^{2}=5$$

Square -\frac{2}{3}y+4.

$$y^{2}-\left(\frac{4}{9}y^{2}-\frac{16}{3}y+16\right)=5$$

Multiply -1 times \frac{4}{9}y^{2}-\frac{16}{3}y+16.

$$y^{2}-\frac{4}{9}y^{2}+\frac{16}{3}y-16=5$$

Add y^{2} to -\frac{4}{9}y^{2}.

$$\frac{5}{9}y^{2}+\frac{16}{3}y-16=5$$

Subtract 5 from both sides of the equation.

$$\frac{5}{9}y^{2}+\frac{16}{3}y-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1-\left(-\frac{2}{3}\right)^{2} for a, -4\left(-\frac{2}{3}\right)\times 2 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{16}{3}±\sqrt{\left(\frac{16}{3}\right)^{2}-4\times \frac{5}{9}\left(-21\right)}}{2\times \frac{5}{9}}$$

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

$$y=\frac{-\frac{16}{3}±\sqrt{\frac{256}{9}-4\times \frac{5}{9}\left(-21\right)}}{2\times \frac{5}{9}}$$

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

$$y=\frac{-\frac{16}{3}±\sqrt{\frac{256}{9}-\frac{20}{9}\left(-21\right)}}{2\times \frac{5}{9}}$$

Multiply -\frac{20}{9} times -21.

$$y=\frac{-\frac{16}{3}±\sqrt{\frac{256}{9}+\frac{140}{3}}}{2\times \frac{5}{9}}$$

Add \frac{256}{9} to \frac{140}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{16}{3}±\sqrt{\frac{676}{9}}}{2\times \frac{5}{9}}$$

Take the square root of \frac{676}{9}.

$$y=\frac{-\frac{16}{3}±\frac{26}{3}}{2\times \frac{5}{9}}$$

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

$$y=\frac{-\frac{16}{3}±\frac{26}{3}}{\frac{10}{9}}$$

Now solve the equation y=\frac{-\frac{16}{3}±\frac{26}{3}}{\frac{10}{9}} when ± is plus. Add -\frac{16}{3} to \frac{26}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{\frac{10}{3}}{\frac{10}{9}}$$

Divide \frac{10}{3} by \frac{10}{9} by multiplying \frac{10}{3} by the reciprocal of \frac{10}{9}.

$$y=3$$

Now solve the equation y=\frac{-\frac{16}{3}±\frac{26}{3}}{\frac{10}{9}} when ± is minus. Subtract \frac{26}{3} from -\frac{16}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{14}{\frac{10}{9}}$$

Divide -14 by \frac{10}{9} by multiplying -14 by the reciprocal of \frac{10}{9}.

$$y=-\frac{63}{5}$$

There are two solutions for y: 3 and -\frac{63}{5}. Substitute 3 for y in the equation x=-\frac{2}{3}y+4 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{2}{3}\times 3+4$$

Multiply -\frac{2}{3} times 3.

$$x=-2+4$$

Add -\frac{2}{3}\times 3 to 4.

$$x=2$$

Now substitute -\frac{63}{5} for y in the equation x=-\frac{2}{3}y+4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{2}{3}\left(-\frac{63}{5}\right)+4$$

Multiply -\frac{2}{3} times -\frac{63}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{42}{5}+4$$

Add -\frac{63}{5}\left(-\frac{2}{3}\right) to 4.

$$x=\frac{62}{5}$$

The system is now solved.

$$x=2,y=3\text{ or }x=\frac{62}{5},y=-\frac{63}{5}$$