Solve {l}{x^2/16+y^2/4=1}{y=-1/2x+2}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1 } \\ { y = - \frac { 1 } { 2 } x + 2 } \end{array} \right.$$

Solve for x, y

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

$$x=4\text{, }y=0$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 16, the least common multiple of 16,4.

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

Consider the second equation. Add \frac{1}{2}x to both sides.

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

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.

$$y+\frac{1}{2}x=2,x^{2}+4y^{2}=16$$

Solve y+\frac{1}{2}x=2 for y by isolating y on the left hand side of the equal sign.

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

Subtract \frac{1}{2}x from both sides of the equation.

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

Substitute -\frac{1}{2}x+2 for y in the other equation, x^{2}+4y^{2}=16.

$$x^{2}+4\left(-\frac{1}{2}x+2\right)^{2}=16$$

Square -\frac{1}{2}x+2.

$$x^{2}+4\left(\frac{1}{4}x^{2}-2x+4\right)=16$$

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

$$x^{2}+x^{2}-8x+16=16$$

Add x^{2} to x^{2}.

$$2x^{2}-8x+16=16$$

Subtract 16 from both sides of the equation.

$$2x^{2}-8x=0$$

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

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

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 2}$$

The opposite of 4\times 2\left(-\frac{1}{2}\right)\times 2 is 8.

$$x=\frac{8±8}{2\times 2}$$

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

$$x=\frac{8±8}{4}$$

Now solve the equation x=\frac{8±8}{4} when ± is plus. Add 8 to 8.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{8±8}{4} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

There are two solutions for x: 4 and 0. Substitute 4 for x in the equation y=-\frac{1}{2}x+2 to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{1}{2}\times 4+2$$

Multiply -\frac{1}{2} times 4.

$$y=-2+2$$

Add -\frac{1}{2}\times 4 to 2.

$$y=0$$

Now substitute 0 for x in the equation y=-\frac{1}{2}x+2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2$$

The system is now solved.

$$y=0,x=4\text{ or }y=2,x=0$$