Solve {l}{x-y=1}{4(x^2+y^2)=20}

$$\left\{ \begin{array} { l } { x - y = 1 } \\ { 4 ( x ^ { 2 } + y ^ { 2 } ) = 20 } \end{array} \right.$$

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

Consider the second equation. Divide both sides by 4.

$$x^{2}+y^{2}=\frac{20}{4}$$

Divide 20 by 4 to get 5.

$$x^{2}+y^{2}=5$$

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\times 1\times 1\times 2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1\times 1\times 1\times 2.

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

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

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

Multiply -8 times -4.

$$y=\frac{-2±\sqrt{4+32}}{2\times 2}$$

Add 4 to 32.

$$y=\frac{-2±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$y=\frac{-2±6}{2\times 2}$$

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.

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

Divide 4 by 4.

$$y=1$$

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

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

Divide -8 by 4.

$$y=-2$$

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

$$x=1+1$$

Add 1\times 1 to 1.

$$x=2$$

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

$$x=-2+1$$

Add -2 to 1.

$$x=-1$$

The system is now solved.

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