Solve for: x=-2\text{,#}y=1
x=-14/3 \approx-4.666666667\text{,#}y=-13/3 \approx-4.333333333

Solve for x, y: $x=-2\text{, }y=1$
$x=-\frac{14}{3}\approx -4.666666667\text{, }y=-\frac{13}{3}\approx -4.333333333$

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.

$$2x-y+5=0,-y^{2}+x^{2}=3$$

Solve $2x-y+5=0$ for $x$ by isolating $x$ on the left hand side of the equal sign.

$$2x-y+5=0$$

Subtract $5$ from both sides of the equation.

$$2x-y=-5$$

Subtract $-y$ from both sides of the equation.

$$2x=y-5$$

Divide both sides by $2$.

$$x=\frac{1}{2}y-\frac{5}{2}$$

Substitute $\frac{1}{2}y-\frac{5}{2}$ for $x$ in the other equation, $-y^{2}+x^{2}=3$.

$$-y^{2}+\left(\frac{1}{2}y-\frac{5}{2}\right)^{2}=3$$

Square $\frac{1}{2}y-\frac{5}{2}$.

$$-y^{2}+\frac{1}{4}y^{2}-\frac{5}{2}y+\frac{25}{4}=3$$

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

$$-\frac{3}{4}y^{2}-\frac{5}{2}y+\frac{25}{4}=3$$

Subtract $3$ from both sides of the equation.

$$-\frac{3}{4}y^{2}-\frac{5}{2}y+\frac{13}{4}=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1+1\times \left(\frac{1}{2}\right)^{2}$ for $a$, $1\left(-\frac{5}{2}\right)\times \left(\frac{1}{2}\right)\times 2$ for $b$, and $\frac{13}{4}$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\left(-\frac{5}{2}\right)^{2}-4\left(-\frac{3}{4}\right)\times \left(\frac{13}{4}\right)}}{2\left(-\frac{3}{4}\right)}$$

Square $1\left(-\frac{5}{2}\right)\times \left(\frac{1}{2}\right)\times 2$.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-4\left(-\frac{3}{4}\right)\times \left(\frac{13}{4}\right)}}{2\left(-\frac{3}{4}\right)}$$

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

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}+3\times \left(\frac{13}{4}\right)}}{2\left(-\frac{3}{4}\right)}$$

Multiply $3$ times $\frac{13}{4}$.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25+39}{4}}}{2\left(-\frac{3}{4}\right)}$$

Add $\frac{25}{4}$ to $\frac{39}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{16}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of $16$.

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

The opposite of $1\left(-\frac{5}{2}\right)\times \left(\frac{1}{2}\right)\times 2$ is $\frac{5}{2}$.

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

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

$$y=\frac{\frac{5}{2}±4}{-\frac{3}{2}}$$

Now solve the equation $y=\frac{\frac{5}{2}±4}{-\frac{3}{2}}$ when $±$ is plus. Add $\frac{5}{2}$ to $4$.

$$y=\frac{\frac{13}{2}}{-\frac{3}{2}}$$

Divide $\frac{13}{2}$ by $-\frac{3}{2}$ by multiplying $\frac{13}{2}$ by the reciprocal of $-\frac{3}{2}$.

$$y=-\frac{13}{3}$$

Now solve the equation $y=\frac{\frac{5}{2}±4}{-\frac{3}{2}}$ when $±$ is minus. Subtract $4$ from $\frac{5}{2}$.

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

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

$$y=1$$

There are two solutions for $y$: $-\frac{13}{3}$ and $1$. Substitute $-\frac{13}{3}$ for $y$ in the equation $x=\frac{1}{2}y-\frac{5}{2}$ to find the corresponding solution for $x$ that satisfies both equations.

$$x=\frac{1}{2}\left(-\frac{13}{3}\right)-\frac{5}{2}$$

Multiply $\frac{1}{2}$ times $-\frac{13}{3}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{13}{6}-\frac{5}{2}$$

Add $-\frac{13}{3}\times \left(\frac{1}{2}\right)$ to $-\frac{5}{2}$.

$$x=-\frac{14}{3}$$

Now substitute $1$ for $y$ in the equation $x=\frac{1}{2}y-\frac{5}{2}$ and solve to find the corresponding solution for $x$ that satisfies both equations.

$$x=\frac{1-5}{2}$$

Add $\frac{1}{2}\times 1$ to $-\frac{5}{2}$.

$$x=-2$$

The system is now solved.

$$x=-\frac{14}{3},y=-\frac{13}{3}\text{ or }x=-2,y=1$$