Solve {l}{x-2y=6}{x^2+4y^2=42}

$$\left\{ \begin{array} { l } { x - 2 y = 6 } \\ { x ^ { 2 } + 4 y ^ { 2 } = 42 } \end{array} \right.$$

Solve for x, y

$$x=3-2\sqrt{3}\approx -0.464101615\text{, }y=-\sqrt{3}-\frac{3}{2}\approx -3.232050808$$

$$x=2\sqrt{3}+3\approx 6.464101615\text{, }y=\sqrt{3}-\frac{3}{2}\approx 0.232050808$$

Steps Using Substitution and Quadratic Formula

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.

$$x-2y=6,4y^{2}+x^{2}=42$$

Solve x-2y=6 for x by isolating x on the left hand side of the equal sign.

$$x-2y=6$$

Subtract -2y from both sides of the equation.

$$x=2y+6$$

Substitute 2y+6 for x in the other equation, 4y^{2}+x^{2}=42.

$$4y^{2}+\left(2y+6\right)^{2}=42$$

Square 2y+6.

$$4y^{2}+4y^{2}+24y+36=42$$

Add 4y^{2} to 4y^{2}.

$$8y^{2}+24y+36=42$$

Subtract 42 from both sides of the equation.

$$8y^{2}+24y-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4+1\times 2^{2} for a, 1\times 6\times 2\times 2 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-24±\sqrt{24^{2}-4\times 8\left(-6\right)}}{2\times 8}$$

Square 1\times 6\times 2\times 2.

$$y=\frac{-24±\sqrt{576-4\times 8\left(-6\right)}}{2\times 8}$$

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

$$y=\frac{-24±\sqrt{576-32\left(-6\right)}}{2\times 8}$$

Multiply -32 times -6.

$$y=\frac{-24±\sqrt{576+192}}{2\times 8}$$

Add 576 to 192.

$$y=\frac{-24±\sqrt{768}}{2\times 8}$$

Take the square root of 768.

$$y=\frac{-24±16\sqrt{3}}{2\times 8}$$

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

$$y=\frac{-24±16\sqrt{3}}{16}$$

Now solve the equation y=\frac{-24±16\sqrt{3}}{16} when ± is plus. Add -24 to 16\sqrt{3}.

$$y=\frac{16\sqrt{3}-24}{16}$$

Divide -24+16\sqrt{3} by 16.

$$y=\sqrt{3}-\frac{3}{2}$$

Now solve the equation y=\frac{-24±16\sqrt{3}}{16} when ± is minus. Subtract 16\sqrt{3} from -24.

$$y=\frac{-16\sqrt{3}-24}{16}$$

Divide -24-16\sqrt{3} by 16.

$$y=-\sqrt{3}-\frac{3}{2}$$

There are two solutions for y: -\frac{3}{2}+\sqrt{3} and -\frac{3}{2}-\sqrt{3}. Substitute -\frac{3}{2}+\sqrt{3} for y in the equation x=2y+6 to find the corresponding solution for x that satisfies both equations.

$$x=2\left(\sqrt{3}-\frac{3}{2}\right)+6$$

Now substitute -\frac{3}{2}-\sqrt{3} for y in the equation x=2y+6 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\left(-\sqrt{3}-\frac{3}{2}\right)+6$$

The system is now solved.

$$x=2\left(\sqrt{3}-\frac{3}{2}\right)+6,y=\sqrt{3}-\frac{3}{2}\text{ or }x=2\left(-\sqrt{3}-\frac{3}{2}\right)+6,y=-\sqrt{3}-\frac{3}{2}$$