Solve {l}{-4x+2y+20=0}{x^2+y^2-20=0}

$$\left\{ \begin{array} { l } { - 4 x + 2 y + 20 = 0 } \\ { x ^ { 2 } + y ^ { 2 } - 20 = 0 } \end{array} \right.$$

Solve for x, y

$$x=4$$

$$y=-2$$

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.

$$-4x+2y+20=0,y^{2}+x^{2}-20=0$$

Solve -4x+2y+20=0 for x by isolating x on the left hand side of the equal sign.

$$-4x+2y+20=0$$

Subtract 20 from both sides of the equation.

$$-4x+2y=-20$$

Subtract 2y from both sides of the equation.

$$-4x=-2y-20$$

Divide both sides by -4.

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

Substitute \frac{1}{2}y+5 for x in the other equation, y^{2}+x^{2}-20=0.

$$y^{2}+\left(\frac{1}{2}y+5\right)^{2}-20=0$$

Square \frac{1}{2}y+5.

$$y^{2}+\frac{1}{4}y^{2}+5y+25-20=0$$

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

$$\frac{5}{4}y^{2}+5y+25-20=0$$

Add 1\times 5^{2} to -20.

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

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

$$y=\frac{-5±\sqrt{5^{2}-4\times \frac{5}{4}\times 5}}{2\times \frac{5}{4}}$$

Square 1\times 5\times \frac{1}{2}\times 2.

$$y=\frac{-5±\sqrt{25-4\times \frac{5}{4}\times 5}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-5±\sqrt{25-5\times 5}}{2\times \frac{5}{4}}$$

Multiply -5 times 5.

$$y=\frac{-5±\sqrt{25-25}}{2\times \frac{5}{4}}$$

Add 25 to -25.

$$y=\frac{-5±\sqrt{0}}{2\times \frac{5}{4}}$$

Take the square root of 0.

$$y=-\frac{5}{2\times \frac{5}{4}}$$

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

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

Divide -5 by \frac{5}{2} by multiplying -5 by the reciprocal of \frac{5}{2}.

$$y=-2$$

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

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

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

$$x=-1+5$$

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

$$x=4$$

The system is now solved.

$$x=4,y=-2\text{ or }x=4,y=-2$$