Solve {l}{-2x=0}{text{Solvefor}ytext{where}}{y={(sqrt{3}-1)}^2+{(2x-2)}{(sqrt{3}-1)}+2}

$$\left. \begin{array} { l } { -2 x = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = {(\sqrt{3} - 1)} ^ {2} + {(2 x - 2)} {(\sqrt{3} - 1)} + 2 } \end{array} \right.$$

Solve for x, y

$$y=8-4\sqrt{3}\approx 1.07179677$$

Solution Steps

Consider the first equation. Divide both sides by -2. Zero divided by any non-zero number gives zero.

$$x=0$$

Consider the second equation. Insert the known values of variables into the equation.

$$y=\left(\sqrt{3}-1\right)^{2}+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

$$y=\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

The square of \sqrt{3} is 3.

$$y=3-2\sqrt{3}+1+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

Add 3 and 1 to get 4.

$$y=4-2\sqrt{3}+\left(2\times 0-2\right)\left(\sqrt{3}-1\right)+2$$

Multiply 2 and 0 to get 0.

$$y=4-2\sqrt{3}+\left(0-2\right)\left(\sqrt{3}-1\right)+2$$

Subtract 2 from 0 to get -2.

$$y=4-2\sqrt{3}-2\left(\sqrt{3}-1\right)+2$$

Use the distributive property to multiply -2 by \sqrt{3}-1.

$$y=4-2\sqrt{3}-2\sqrt{3}+2+2$$

Combine -2\sqrt{3} and -2\sqrt{3} to get -4\sqrt{3}.

$$y=4-4\sqrt{3}+2+2$$

Add 4 and 2 to get 6.

$$y=6-4\sqrt{3}+2$$

Add 6 and 2 to get 8.

$$y=8-4\sqrt{3}$$

The system is now solved.

$$x=0 y=8-4\sqrt{3}$$