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

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

Solve for x, y

$$x=\frac{-3\sqrt{5}-3}{8}\approx -1.213525492\text{, }y=\frac{1-3\sqrt{5}}{4}\approx -1.427050983$$

$$x=\frac{3\sqrt{5}-3}{8}\approx 0.463525492\text{, }y=\frac{3\sqrt{5}+1}{4}\approx 1.927050983$$

Steps Using Substitution and Quadratic Formula

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

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

Consider the second equation. Subtract 2x from both sides.

$$y-2x=1$$

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.

$$y-2x=1,4x^{2}+3y^{2}=12$$

Solve y-2x=1 for y by isolating y on the left hand side of the equal sign.

$$y-2x=1$$

Subtract -2x from both sides of the equation.

$$y=2x+1$$

Substitute 2x+1 for y in the other equation, 4x^{2}+3y^{2}=12.

$$4x^{2}+3\left(2x+1\right)^{2}=12$$

Square 2x+1.

$$4x^{2}+3\left(4x^{2}+4x+1\right)=12$$

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

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

Add 4x^{2} to 12x^{2}.

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

Subtract 12 from both sides of the equation.

$$16x^{2}+12x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4+3\times 2^{2} for a, 3\times 1\times 2\times 2 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 16\left(-9\right)}}{2\times 16}$$

Square 3\times 1\times 2\times 2.

$$x=\frac{-12±\sqrt{144-4\times 16\left(-9\right)}}{2\times 16}$$

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

$$x=\frac{-12±\sqrt{144-64\left(-9\right)}}{2\times 16}$$

Multiply -64 times -9.

$$x=\frac{-12±\sqrt{144+576}}{2\times 16}$$

Add 144 to 576.

$$x=\frac{-12±\sqrt{720}}{2\times 16}$$

Take the square root of 720.

$$x=\frac{-12±12\sqrt{5}}{2\times 16}$$

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

$$x=\frac{-12±12\sqrt{5}}{32}$$

Now solve the equation x=\frac{-12±12\sqrt{5}}{32} when ± is plus. Add -12 to 12\sqrt{5}.

$$x=\frac{12\sqrt{5}-12}{32}$$

Divide -12+12\sqrt{5} by 32.

$$x=\frac{3\sqrt{5}-3}{8}$$

Now solve the equation x=\frac{-12±12\sqrt{5}}{32} when ± is minus. Subtract 12\sqrt{5} from -12.

$$x=\frac{-12\sqrt{5}-12}{32}$$

Divide -12-12\sqrt{5} by 32.

$$x=\frac{-3\sqrt{5}-3}{8}$$

Both solutions for x are the same: \frac{-3+3\sqrt{5}}{8}. Substitute \frac{-3+3\sqrt{5}}{8} for x in the equation y=2x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{3\sqrt{5}-3}{8}+1$$

Now substitute \frac{-3-3\sqrt{5}}{8} for x in the equation y=2x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{-3\sqrt{5}-3}{8}+1$$

The system is now solved.

$$y=2\times \frac{3\sqrt{5}-3}{8}+1,x=\frac{3\sqrt{5}-3}{8}\text{ or }y=2\times \frac{-3\sqrt{5}-3}{8}+1,x=\frac{-3\sqrt{5}-3}{8}$$