Solve {c}{2x-y=2}{x^2+2y^2=12}

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

Solve for x, y

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

$$x=-\frac{2}{9}\approx -0.222222222\text{, }y=-\frac{22}{9}\approx -2.444444444$$

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.

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

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

$$2x-y=2$$

Subtract -y from both sides of the equation.

$$2x=y+2$$

Divide both sides by 2.

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

Substitute \frac{1}{2}y+1 for x in the other equation, 2y^{2}+x^{2}=12.

$$2y^{2}+\left(\frac{1}{2}y+1\right)^{2}=12$$

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

$$2y^{2}+\frac{1}{4}y^{2}+y+1=12$$

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

$$\frac{9}{4}y^{2}+y+1=12$$

Subtract 12 from both sides of the equation.

$$\frac{9}{4}y^{2}+y-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2+1\times \left(\frac{1}{2}\right)^{2} for a, 1\times 1\times \frac{1}{2}\times 2 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\times \frac{9}{4}\left(-11\right)}}{2\times \frac{9}{4}}$$

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

$$y=\frac{-1±\sqrt{1-4\times \frac{9}{4}\left(-11\right)}}{2\times \frac{9}{4}}$$

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

$$y=\frac{-1±\sqrt{1-9\left(-11\right)}}{2\times \frac{9}{4}}$$

Multiply -9 times -11.

$$y=\frac{-1±\sqrt{1+99}}{2\times \frac{9}{4}}$$

Add 1 to 99.

$$y=\frac{-1±\sqrt{100}}{2\times \frac{9}{4}}$$

Take the square root of 100.

$$y=\frac{-1±10}{2\times \frac{9}{4}}$$

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

$$y=\frac{-1±10}{\frac{9}{2}}$$

Now solve the equation y=\frac{-1±10}{\frac{9}{2}} when ± is plus. Add -1 to 10.

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

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

$$y=2$$

Now solve the equation y=\frac{-1±10}{\frac{9}{2}} when ± is minus. Subtract 10 from -1.

$$y=-\frac{11}{\frac{9}{2}}$$

Divide -11 by \frac{9}{2} by multiplying -11 by the reciprocal of \frac{9}{2}.

$$y=-\frac{22}{9}$$

There are two solutions for y: 2 and -\frac{22}{9}. Substitute 2 for y in the equation x=\frac{1}{2}y+1 to find the corresponding solution for x that satisfies both equations.

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

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

$$x=1+1$$

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

$$x=2$$

Now substitute -\frac{22}{9} for y in the equation x=\frac{1}{2}y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\left(-\frac{22}{9}\right)+1$$

Multiply \frac{1}{2} times -\frac{22}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{11}{9}+1$$

Add -\frac{22}{9}\times \frac{1}{2} to 1.

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

The system is now solved.

$$x=2,y=2\text{ or }x=-\frac{2}{9},y=-\frac{22}{9}$$