Solve {l}{2x+y=9}{x^2-y^2=0}

$$\left. \begin{array} { l } { 2 x + y = 9 } \\ { x ^ { 2 } - y ^ { 2 } = 0 } \end{array} \right.$$

Solve for x, y

$$x=9\text{, }y=-9$$

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

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.

$$2x+y=9,-y^{2}+x^{2}=0$$

Solve 2x+y=9 for x by isolating x on the left hand side of the equal sign.

$$2x+y=9$$

Subtract y from both sides of the equation.

$$2x=-y+9$$

Divide both sides by 2.

$$x=-\frac{1}{2}y+\frac{9}{2}$$

Substitute -\frac{1}{2}y+\frac{9}{2} for x in the other equation, -y^{2}+x^{2}=0.

$$-y^{2}+\left(-\frac{1}{2}y+\frac{9}{2}\right)^{2}=0$$

Square -\frac{1}{2}y+\frac{9}{2}.

$$-y^{2}+\frac{1}{4}y^{2}-\frac{9}{2}y+\frac{81}{4}=0$$

Add -y^{2} to \frac{1}{4}y^{2}.

$$-\frac{3}{4}y^{2}-\frac{9}{2}y+\frac{81}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+1\left(-\frac{1}{2}\right)^{2} for a, 1\times \frac{9}{2}\left(-\frac{1}{2}\right)\times 2 for b, and \frac{81}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\left(-\frac{9}{2}\right)^{2}-4\left(-\frac{3}{4}\right)\times \frac{81}{4}}}{2\left(-\frac{3}{4}\right)}$$

Square 1\times \frac{9}{2}\left(-\frac{1}{2}\right)\times 2.

$$y=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-4\left(-\frac{3}{4}\right)\times \frac{81}{4}}}{2\left(-\frac{3}{4}\right)}$$

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

$$y=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}+3\times \frac{81}{4}}}{2\left(-\frac{3}{4}\right)}$$

Multiply 3 times \frac{81}{4}.

$$y=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81+243}{4}}}{2\left(-\frac{3}{4}\right)}$$

Add \frac{81}{4} to \frac{243}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{9}{2}\right)±\sqrt{81}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of 81.

$$y=\frac{-\left(-\frac{9}{2}\right)±9}{2\left(-\frac{3}{4}\right)}$$

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

$$y=\frac{\frac{9}{2}±9}{2\left(-\frac{3}{4}\right)}$$

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

$$y=\frac{\frac{9}{2}±9}{-\frac{3}{2}}$$

Now solve the equation y=\frac{\frac{9}{2}±9}{-\frac{3}{2}} when ± is plus. Add \frac{9}{2} to 9.

$$y=\frac{\frac{27}{2}}{-\frac{3}{2}}$$

Divide \frac{27}{2} by -\frac{3}{2} by multiplying \frac{27}{2} by the reciprocal of -\frac{3}{2}.

$$y=-9$$

Now solve the equation y=\frac{\frac{9}{2}±9}{-\frac{3}{2}} when ± is minus. Subtract 9 from \frac{9}{2}.

$$y=-\frac{\frac{9}{2}}{-\frac{3}{2}}$$

Divide -\frac{9}{2} by -\frac{3}{2} by multiplying -\frac{9}{2} by the reciprocal of -\frac{3}{2}.

$$y=3$$

There are two solutions for y: -9 and 3. Substitute -9 for y in the equation x=-\frac{1}{2}y+\frac{9}{2} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\left(-9\right)+\frac{9}{2}$$

Multiply -\frac{1}{2} times -9.

$$x=\frac{9+9}{2}$$

Add -9\left(-\frac{1}{2}\right) to \frac{9}{2}.

$$x=9$$

Now substitute 3 for y in the equation x=-\frac{1}{2}y+\frac{9}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times 3+\frac{9}{2}$$

Multiply -\frac{1}{2} times 3.

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

Add -\frac{1}{2}\times 3 to \frac{9}{2}.

$$x=3$$

The system is now solved.

$$x=9,y=-9\text{ or }x=3,y=3$$