Solve {l}{x^2+y^2=5}{x-y=-1}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 5 } \\ { x - y = - 1 } \end{array} \right.$$

Solve for x, y

$$x=1\text{, }y=2$$

$$x=-2\text{, }y=-1$$

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.

$$x-y=-1,y^{2}+x^{2}=5$$

Solve x-y=-1 for x by isolating x on the left hand side of the equal sign.

$$x-y=-1$$

Subtract -y from both sides of the equation.

$$x=y-1$$

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

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

Square y-1.

$$y^{2}+y^{2}-2y+1=5$$

Add y^{2} to y^{2}.

$$2y^{2}-2y+1=5$$

Subtract 5 from both sides of the equation.

$$2y^{2}-2y-4=0$$

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

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

Square 1\left(-1\right)\times 1\times 2.

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

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

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

Multiply -8 times -4.

$$y=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times 2}$$

Add 4 to 32.

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

Take the square root of 36.

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

The opposite of 1\left(-1\right)\times 1\times 2 is 2.

$$y=\frac{2±6}{2\times 2}$$

Multiply 2 times 1+1\times 1^{2}.

$$y=\frac{2±6}{4}$$

Now solve the equation y=\frac{2±6}{4} when ± is plus. Add 2 to 6.

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

Now solve the equation y=\frac{2±6}{4} when ± is minus. Subtract 6 from 2.

$$y=-\frac{4}{4}$$

Divide -4 by 4.

$$y=-1$$

There are two solutions for y: 2 and -1. Substitute 2 for y in the equation x=y-1 to find the corresponding solution for x that satisfies both equations.

$$x=2-1$$

Add 1\times 2 to -1.

$$x=1$$

Now substitute -1 for y in the equation x=y-1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-1-1$$

Add -1 to -1.

$$x=-2$$

The system is now solved.

$$x=1,y=2\text{ or }x=-2,y=-1$$