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

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

Solve for x, y

$$x=0\text{, }y=2$$

$$x=\frac{8}{5}=1.6\text{, }y=\frac{6}{5}=1.2$$

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.

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

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

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

Add 4 to both sides of the equation.

$$x+2y=4$$

Subtract 2y from both sides of the equation.

$$x=-2y+4$$

Substitute -2y+4 for x in the other equation, y^{2}+x^{2}=4.

$$y^{2}+\left(-2y+4\right)^{2}=4$$

Square -2y+4.

$$y^{2}+4y^{2}-16y+16=4$$

Add y^{2} to 4y^{2}.

$$5y^{2}-16y+16=4$$

Subtract 4 from both sides of the equation.

$$5y^{2}-16y+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-2\right)^{2} for a, 1\times 4\left(-2\right)\times 2 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 12}}{2\times 5}$$

Square 1\times 4\left(-2\right)\times 2.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 12}}{2\times 5}$$

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

$$y=\frac{-\left(-16\right)±\sqrt{256-20\times 12}}{2\times 5}$$

Multiply -20 times 12.

$$y=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 5}$$

Add 256 to -240.

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

Take the square root of 16.

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

The opposite of 1\times 4\left(-2\right)\times 2 is 16.

$$y=\frac{16±4}{2\times 5}$$

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

$$y=\frac{16±4}{10}$$

Now solve the equation y=\frac{16±4}{10} when ± is plus. Add 16 to 4.

$$y=\frac{20}{10}$$

Divide 20 by 10.

$$y=2$$

Now solve the equation y=\frac{16±4}{10} when ± is minus. Subtract 4 from 16.

$$y=\frac{12}{10}$$

Reduce the fraction \frac{12}{10} to lowest terms by extracting and canceling out 2.

$$y=\frac{6}{5}$$

There are two solutions for y: 2 and \frac{6}{5}. Substitute 2 for y in the equation x=-2y+4 to find the corresponding solution for x that satisfies both equations.

$$x=-2\times 2+4$$

Multiply -2 times 2.

$$x=-4+4$$

Add -2\times 2 to 4.

$$x=0$$

Now substitute \frac{6}{5} for y in the equation x=-2y+4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{6}{5}+4$$

Multiply -2 times \frac{6}{5}.

$$x=-\frac{12}{5}+4$$

Add -2\times \frac{6}{5} to 4.

$$x=\frac{8}{5}$$

The system is now solved.

$$x=0,y=2\text{ or }x=\frac{8}{5},y=\frac{6}{5}$$