Solve {l}{y=x+2}{x^2/4+y^2/3=1}

$$\left\{ \begin{array} { l } { y = x + 2 } \\ { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1 } \end{array} \right.$$

Solve for y, x

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

$$x=-\frac{2}{7}\approx -0.285714286\text{, }y=\frac{12}{7}\approx 1.714285714$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=2$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x^{2}+4y^{2}=12$$

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.

$$y-x=2,3x^{2}+4y^{2}=12$$

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

$$y-x=2$$

Subtract -x from both sides of the equation.

$$y=x+2$$

Substitute x+2 for y in the other equation, 3x^{2}+4y^{2}=12.

$$3x^{2}+4\left(x+2\right)^{2}=12$$

Square x+2.

$$3x^{2}+4\left(x^{2}+4x+4\right)=12$$

Multiply 4 times x^{2}+4x+4.

$$3x^{2}+4x^{2}+16x+16=12$$

Add 3x^{2} to 4x^{2}.

$$7x^{2}+16x+16=12$$

Subtract 12 from both sides of the equation.

$$7x^{2}+16x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+4\times 1^{2} for a, 4\times 2\times 1\times 2 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 7\times 4}}{2\times 7}$$

Square 4\times 2\times 1\times 2.

$$x=\frac{-16±\sqrt{256-4\times 7\times 4}}{2\times 7}$$

Multiply -4 times 3+4\times 1^{2}.

$$x=\frac{-16±\sqrt{256-28\times 4}}{2\times 7}$$

Multiply -28 times 4.

$$x=\frac{-16±\sqrt{256-112}}{2\times 7}$$

Add 256 to -112.

$$x=\frac{-16±\sqrt{144}}{2\times 7}$$

Take the square root of 144.

$$x=\frac{-16±12}{2\times 7}$$

Multiply 2 times 3+4\times 1^{2}.

$$x=\frac{-16±12}{14}$$

Now solve the equation x=\frac{-16±12}{14} when ± is plus. Add -16 to 12.

$$x=-\frac{4}{14}$$

Reduce the fraction \frac{-4}{14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{2}{7}$$

Now solve the equation x=\frac{-16±12}{14} when ± is minus. Subtract 12 from -16.

$$x=-\frac{28}{14}$$

Divide -28 by 14.

$$x=-2$$

There are two solutions for x: -\frac{2}{7} and -2. Substitute -\frac{2}{7} for x in the equation y=x+2 to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{2}{7}+2$$

Add -\frac{2}{7} to 2.

$$y=\frac{12}{7}$$

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

$$y=-2+2$$

Add -2 to 2.

$$y=0$$

The system is now solved.

$$y=\frac{12}{7},x=-\frac{2}{7}\text{ or }y=0,x=-2$$