Solve {l}{x^2-y^2=7}{2y=2+x}

$$\left. \begin{array} { l } { x ^ { 2 } - y ^ { 2 } = 7 } \\ { 2 y = 2 + x } \end{array} \right.$$

Solve for x, y

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

$$x=-\frac{8}{3}\approx -2.666666667\text{, }y=-\frac{1}{3}\approx -0.333333333$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract x from both sides.

$$2y-x=2$$

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.

$$2y-x=2,x^{2}-y^{2}=7$$

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

$$2y-x=2$$

Subtract -x from both sides of the equation.

$$2y=x+2$$

Divide both sides by 2.

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

Substitute \frac{1}{2}x+1 for y in the other equation, x^{2}-y^{2}=7.

$$x^{2}-\left(\frac{1}{2}x+1\right)^{2}=7$$

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

$$x^{2}-\left(\frac{1}{4}x^{2}+x+1\right)=7$$

Multiply -1 times \frac{1}{4}x^{2}+x+1.

$$x^{2}-\frac{1}{4}x^{2}-x-1=7$$

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

$$\frac{3}{4}x^{2}-x-1=7$$

Subtract 7 from both sides of the equation.

$$\frac{3}{4}x^{2}-x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1-\left(\frac{1}{2}\right)^{2} for a, -\frac{1}{2}\times 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{3}{4}\left(-8\right)}}{2\times \frac{3}{4}}$$

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

$$x=\frac{-\left(-1\right)±\sqrt{1-3\left(-8\right)}}{2\times \frac{3}{4}}$$

Multiply -3 times -8.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times \frac{3}{4}}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\times \frac{3}{4}}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\times \frac{3}{4}}$$

The opposite of -\frac{1}{2}\times 2 is 1.

$$x=\frac{1±5}{2\times \frac{3}{4}}$$

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

$$x=\frac{1±5}{\frac{3}{2}}$$

Now solve the equation x=\frac{1±5}{\frac{3}{2}} when ± is plus. Add 1 to 5.

$$x=\frac{6}{\frac{3}{2}}$$

Divide 6 by \frac{3}{2} by multiplying 6 by the reciprocal of \frac{3}{2}.

$$x=4$$

Now solve the equation x=\frac{1±5}{\frac{3}{2}} when ± is minus. Subtract 5 from 1.

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

Divide -4 by \frac{3}{2} by multiplying -4 by the reciprocal of \frac{3}{2}.

$$x=-\frac{8}{3}$$

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

$$y=\frac{1}{2}\times 4+1$$

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

$$y=2+1$$

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

$$y=3$$

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

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

Multiply \frac{1}{2} times -\frac{8}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{4}{3}+1$$

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

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

The system is now solved.

$$y=3,x=4\text{ or }y=-\frac{1}{3},x=-\frac{8}{3}$$