Solve {l}{2x+y=11}{2x^2-y^2=23}

$$\left. \begin{array} { l } { 2 x + y = 11 } \\ { 2 x ^ { 2 } - y ^ { 2 } = 23 } \end{array} \right.$$

Solve for x, y

$$x=18\text{, }y=-25$$

$$x=4\text{, }y=3$$

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.

$$2x+y=11,-y^{2}+2x^{2}=23$$

Solve 2x+y=11 for x by isolating x on the left hand side of the equal sign.

$$2x+y=11$$

Subtract y from both sides of the equation.

$$2x=-y+11$$

Divide both sides by 2.

$$x=-\frac{1}{2}y+\frac{11}{2}$$

Substitute -\frac{1}{2}y+\frac{11}{2} for x in the other equation, -y^{2}+2x^{2}=23.

$$-y^{2}+2\left(-\frac{1}{2}y+\frac{11}{2}\right)^{2}=23$$

Square -\frac{1}{2}y+\frac{11}{2}.

$$-y^{2}+2\left(\frac{1}{4}y^{2}-\frac{11}{2}y+\frac{121}{4}\right)=23$$

Multiply 2 times \frac{1}{4}y^{2}-\frac{11}{2}y+\frac{121}{4}.

$$-y^{2}+\frac{1}{2}y^{2}-11y+\frac{121}{2}=23$$

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

$$-\frac{1}{2}y^{2}-11y+\frac{121}{2}=23$$

Subtract 23 from both sides of the equation.

$$-\frac{1}{2}y^{2}-11y+\frac{75}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+2\left(-\frac{1}{2}\right)^{2} for a, 2\times \frac{11}{2}\left(-\frac{1}{2}\right)\times 2 for b, and \frac{75}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-\frac{1}{2}\right)\times \frac{75}{2}}}{2\left(-\frac{1}{2}\right)}$$

Square 2\times \frac{11}{2}\left(-\frac{1}{2}\right)\times 2.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\left(-\frac{1}{2}\right)\times \frac{75}{2}}}{2\left(-\frac{1}{2}\right)}$$

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

$$y=\frac{-\left(-11\right)±\sqrt{121+2\times \frac{75}{2}}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times \frac{75}{2}.

$$y=\frac{-\left(-11\right)±\sqrt{121+75}}{2\left(-\frac{1}{2}\right)}$$

Add 121 to 75.

$$y=\frac{-\left(-11\right)±\sqrt{196}}{2\left(-\frac{1}{2}\right)}$$

Take the square root of 196.

$$y=\frac{-\left(-11\right)±14}{2\left(-\frac{1}{2}\right)}$$

The opposite of 2\times \frac{11}{2}\left(-\frac{1}{2}\right)\times 2 is 11.

$$y=\frac{11±14}{2\left(-\frac{1}{2}\right)}$$

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

$$y=\frac{11±14}{-1}$$

Now solve the equation y=\frac{11±14}{-1} when ± is plus. Add 11 to 14.

$$y=\frac{25}{-1}$$

Divide 25 by -1.

$$y=-25$$

Now solve the equation y=\frac{11±14}{-1} when ± is minus. Subtract 14 from 11.

$$y=-\frac{3}{-1}$$

Divide -3 by -1.

$$y=3$$

There are two solutions for y: -25 and 3. Substitute -25 for y in the equation x=-\frac{1}{2}y+\frac{11}{2} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\left(-25\right)+\frac{11}{2}$$

Multiply -\frac{1}{2} times -25.

$$x=\frac{25+11}{2}$$

Add -25\left(-\frac{1}{2}\right) to \frac{11}{2}.

$$x=18$$

Now substitute 3 for y in the equation x=-\frac{1}{2}y+\frac{11}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times 3+\frac{11}{2}$$

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

$$x=\frac{-3+11}{2}$$

Add -\frac{1}{2}\times 3 to \frac{11}{2}.

$$x=4$$

The system is now solved.

$$x=18,y=-25\text{ or }x=4,y=3$$