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

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

Solve for x, y

$$x=\frac{-2\sqrt{174}-48}{67}\approx -1.110176297\text{, }y=\frac{9-8\sqrt{174}}{67}\approx -1.440705189$$

$$x=\frac{2\sqrt{174}-48}{67}\approx -0.322659524\text{, }y=\frac{8\sqrt{174}+9}{67}\approx 1.709361905$$

Steps Using Substitution and Quadratic Formula

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

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

Consider the second equation. Subtract 4x from both sides.

$$y-4x=3$$

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.

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

Solve y-4x=3 for y by isolating y on the left hand side of the equal sign.

$$y-4x=3$$

Subtract -4x from both sides of the equation.

$$y=4x+3$$

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

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

Square 4x+3.

$$3x^{2}+4\left(16x^{2}+24x+9\right)=12$$

Multiply 4 times 16x^{2}+24x+9.

$$3x^{2}+64x^{2}+96x+36=12$$

Add 3x^{2} to 64x^{2}.

$$67x^{2}+96x+36=12$$

Subtract 12 from both sides of the equation.

$$67x^{2}+96x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+4\times 4^{2} for a, 4\times 3\times 2\times 4 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-96±\sqrt{96^{2}-4\times 67\times 24}}{2\times 67}$$

Square 4\times 3\times 2\times 4.

$$x=\frac{-96±\sqrt{9216-4\times 67\times 24}}{2\times 67}$$

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

$$x=\frac{-96±\sqrt{9216-268\times 24}}{2\times 67}$$

Multiply -268 times 24.

$$x=\frac{-96±\sqrt{9216-6432}}{2\times 67}$$

Add 9216 to -6432.

$$x=\frac{-96±\sqrt{2784}}{2\times 67}$$

Take the square root of 2784.

$$x=\frac{-96±4\sqrt{174}}{2\times 67}$$

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

$$x=\frac{-96±4\sqrt{174}}{134}$$

Now solve the equation x=\frac{-96±4\sqrt{174}}{134} when ± is plus. Add -96 to 4\sqrt{174}.

$$x=\frac{4\sqrt{174}-96}{134}$$

Divide -96+4\sqrt{174} by 134.

$$x=\frac{2\sqrt{174}-48}{67}$$

Now solve the equation x=\frac{-96±4\sqrt{174}}{134} when ± is minus. Subtract 4\sqrt{174} from -96.

$$x=\frac{-4\sqrt{174}-96}{134}$$

Divide -96-4\sqrt{174} by 134.

$$x=\frac{-2\sqrt{174}-48}{67}$$

There are two solutions for x: \frac{-48+2\sqrt{174}}{67} and \frac{-48-2\sqrt{174}}{67}. Substitute \frac{-48+2\sqrt{174}}{67} for x in the equation y=4x+3 to find the corresponding solution for y that satisfies both equations.

$$y=4\times \frac{2\sqrt{174}-48}{67}+3$$

Now substitute \frac{-48-2\sqrt{174}}{67} for x in the equation y=4x+3 and solve to find the corresponding solution for y that satisfies both equations.

$$y=4\times \frac{-2\sqrt{174}-48}{67}+3$$

The system is now solved.

$$y=4\times \frac{2\sqrt{174}-48}{67}+3,x=\frac{2\sqrt{174}-48}{67}\text{ or }y=4\times \frac{-2\sqrt{174}-48}{67}+3,x=\frac{-2\sqrt{174}-48}{67}$$