Solve {l}{2x-y=6}{2x^2+y^2=66}

$$\left\{ \begin{array} { l } { 2 x - y = 6 } \\ { 2 x ^ { 2 } + y ^ { 2 } = 66 } \end{array} \right.$$

Solve for x, y

$$x=5\text{, }y=4$$

$$x=-1\text{, }y=-8$$

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.

$$2x-y=6,y^{2}+2x^{2}=66$$

Solve 2x-y=6 for x by isolating x on the left hand side of the equal sign.

$$2x-y=6$$

Subtract -y from both sides of the equation.

$$2x=y+6$$

Divide both sides by 2.

$$x=\frac{1}{2}y+3$$

Substitute \frac{1}{2}y+3 for x in the other equation, y^{2}+2x^{2}=66.

$$y^{2}+2\left(\frac{1}{2}y+3\right)^{2}=66$$

Square \frac{1}{2}y+3.

$$y^{2}+2\left(\frac{1}{4}y^{2}+3y+9\right)=66$$

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

$$y^{2}+\frac{1}{2}y^{2}+6y+18=66$$

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

$$\frac{3}{2}y^{2}+6y+18=66$$

Subtract 66 from both sides of the equation.

$$\frac{3}{2}y^{2}+6y-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+2\times \left(\frac{1}{2}\right)^{2} for a, 2\times 3\times \frac{1}{2}\times 2 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\times \frac{3}{2}\left(-48\right)}}{2\times \frac{3}{2}}$$

Square 2\times 3\times \frac{1}{2}\times 2.

$$y=\frac{-6±\sqrt{36-4\times \frac{3}{2}\left(-48\right)}}{2\times \frac{3}{2}}$$

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

$$y=\frac{-6±\sqrt{36-6\left(-48\right)}}{2\times \frac{3}{2}}$$

Multiply -6 times -48.

$$y=\frac{-6±\sqrt{36+288}}{2\times \frac{3}{2}}$$

Add 36 to 288.

$$y=\frac{-6±\sqrt{324}}{2\times \frac{3}{2}}$$

Take the square root of 324.

$$y=\frac{-6±18}{2\times \frac{3}{2}}$$

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

$$y=\frac{-6±18}{3}$$

Now solve the equation y=\frac{-6±18}{3} when ± is plus. Add -6 to 18.

$$y=\frac{12}{3}$$

Divide 12 by 3.

$$y=4$$

Now solve the equation y=\frac{-6±18}{3} when ± is minus. Subtract 18 from -6.

$$y=-\frac{24}{3}$$

Divide -24 by 3.

$$y=-8$$

There are two solutions for y: 4 and -8. Substitute 4 for y in the equation x=\frac{1}{2}y+3 to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\times 4+3$$

Multiply \frac{1}{2} times 4.

$$x=2+3$$

Add \frac{1}{2}\times 4 to 3.

$$x=5$$

Now substitute -8 for y in the equation x=\frac{1}{2}y+3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\left(-8\right)+3$$

Multiply \frac{1}{2} times -8.

$$x=-4+3$$

Add -8\times \frac{1}{2} to 3.

$$x=-1$$

The system is now solved.

$$x=5,y=4\text{ or }x=-1,y=-8$$