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

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

Solve for x, y

$$x=\frac{-6\sqrt{2}-4}{7}\approx -1.783611625\text{, }y=\frac{3-6\sqrt{2}}{7}\approx -0.783611625$$

$$x=\frac{6\sqrt{2}-4}{7}\approx 0.640754482\text{, }y=\frac{6\sqrt{2}+3}{7}\approx 1.640754482$$

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 x from both sides.

$$y-x=1$$

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.

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

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

$$y-x=1$$

Subtract -x from both sides of the equation.

$$y=x+1$$

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

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

Square x+1.

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

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

$$3x^{2}+4x^{2}+8x+4=12$$

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

$$7x^{2}+8x+4=12$$

Subtract 12 from both sides of the equation.

$$7x^{2}+8x-8=0$$

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

$$x=\frac{-8±\sqrt{8^{2}-4\times 7\left(-8\right)}}{2\times 7}$$

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

$$x=\frac{-8±\sqrt{64-4\times 7\left(-8\right)}}{2\times 7}$$

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

$$x=\frac{-8±\sqrt{64-28\left(-8\right)}}{2\times 7}$$

Multiply -28 times -8.

$$x=\frac{-8±\sqrt{64+224}}{2\times 7}$$

Add 64 to 224.

$$x=\frac{-8±\sqrt{288}}{2\times 7}$$

Take the square root of 288.

$$x=\frac{-8±12\sqrt{2}}{2\times 7}$$

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

$$x=\frac{-8±12\sqrt{2}}{14}$$

Now solve the equation x=\frac{-8±12\sqrt{2}}{14} when ± is plus. Add -8 to 12\sqrt{2}.

$$x=\frac{12\sqrt{2}-8}{14}$$

Divide -8+12\sqrt{2} by 14.

$$x=\frac{6\sqrt{2}-4}{7}$$

Now solve the equation x=\frac{-8±12\sqrt{2}}{14} when ± is minus. Subtract 12\sqrt{2} from -8.

$$x=\frac{-12\sqrt{2}-8}{14}$$

Divide -8-12\sqrt{2} by 14.

$$x=\frac{-6\sqrt{2}-4}{7}$$

There are two solutions for x: \frac{-4+6\sqrt{2}}{7} and \frac{-4-6\sqrt{2}}{7}. Substitute \frac{-4+6\sqrt{2}}{7} for x in the equation y=x+1 to find the corresponding solution for y that satisfies both equations.

$$y=\frac{6\sqrt{2}-4}{7}+1$$

Now substitute \frac{-4-6\sqrt{2}}{7} for x in the equation y=x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{-6\sqrt{2}-4}{7}+1$$

The system is now solved.

$$y=\frac{6\sqrt{2}-4}{7}+1,x=\frac{6\sqrt{2}-4}{7}\text{ or }y=\frac{-6\sqrt{2}-4}{7}+1,x=\frac{-6\sqrt{2}-4}{7}$$