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

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

Solve for y, x

$$x=\frac{-6\sqrt{6}-8}{19}\approx -1.194575708\text{, }y=\frac{3-12\sqrt{6}}{19}\approx -1.389151416$$

$$x=\frac{6\sqrt{6}-8}{19}\approx 0.352470445\text{, }y=\frac{12\sqrt{6}+3}{19}\approx 1.70494089$$

Steps Using Substitution and Quadratic Formula

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

$$y-2x=1$$

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.

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

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

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

Square 2x+1.

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

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

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

Add 3x^{2} to 16x^{2}.

$$19x^{2}+16x+4=12$$

Subtract 12 from both sides of the equation.

$$19x^{2}+16x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+4\times 2^{2} for a, 4\times 1\times 2\times 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 19\left(-8\right)}}{2\times 19}$$

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

$$x=\frac{-16±\sqrt{256-4\times 19\left(-8\right)}}{2\times 19}$$

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

$$x=\frac{-16±\sqrt{256-76\left(-8\right)}}{2\times 19}$$

Multiply -76 times -8.

$$x=\frac{-16±\sqrt{256+608}}{2\times 19}$$

Add 256 to 608.

$$x=\frac{-16±\sqrt{864}}{2\times 19}$$

Take the square root of 864.

$$x=\frac{-16±12\sqrt{6}}{2\times 19}$$

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

$$x=\frac{-16±12\sqrt{6}}{38}$$

Now solve the equation x=\frac{-16±12\sqrt{6}}{38} when ± is plus. Add -16 to 12\sqrt{6}.

$$x=\frac{12\sqrt{6}-16}{38}$$

Divide -16+12\sqrt{6} by 38.

$$x=\frac{6\sqrt{6}-8}{19}$$

Now solve the equation x=\frac{-16±12\sqrt{6}}{38} when ± is minus. Subtract 12\sqrt{6} from -16.

$$x=\frac{-12\sqrt{6}-16}{38}$$

Divide -16-12\sqrt{6} by 38.

$$x=\frac{-6\sqrt{6}-8}{19}$$

Both solutions for x are the same: \frac{6\sqrt{6}-8}{19}. Substitute \frac{6\sqrt{6}-8}{19} for x in the equation y=2x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{6\sqrt{6}-8}{19}+1$$

Now substitute \frac{-8-6\sqrt{6}}{19} for x in the equation y=2x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{-6\sqrt{6}-8}{19}+1$$

The system is now solved.

$$y=2\times \frac{6\sqrt{6}-8}{19}+1,x=\frac{6\sqrt{6}-8}{19}\text{ or }y=2\times \frac{-6\sqrt{6}-8}{19}+1,x=\frac{-6\sqrt{6}-8}{19}$$