12. $$ \begin{array}{l}(2-\sqrt{3}) x-y=\sqrt{3} \ 3 x+(2+\sqrt{3}) y=1-2 \sqrt{3} \ x= ext { ? } \ \begin{array}{lll} ext { A) } \sqrt{3} & ext { B) } 1+\sqrt{3} & ext { C) } 2+\sqrt{3}\end{array} \ \begin{array}{lll} ext { D) } 1 & ext { E) } 2 & \end{array}\end{array} $$

Question

12. $$ \begin{array}{l}(2-\sqrt{3}) x-y=\sqrt{3} \ 3 x+(2+\sqrt{3}) y=1-2 \sqrt{3} \ x=	ext { ? } \ \begin{array}{lll}	ext { A) } \sqrt{3} & 	ext { B) } 1+\sqrt{3} & 	ext { C) } 2+\sqrt{3}\end{array} \ \begin{array}{lll}	ext { D) } 1 & 	ext { E) } 2 & \end{array}\end{array} $$

UpStudy AI · Accepted Answer

**Step 1.** Write the given system of equations:
$$
(2-\sqrt{3})x - y = \sqrt{3}
$$
$$
3x + (2+\sqrt{3})y = 1-2\sqrt{3}
$$

**Step 2.** Solve the first equation for $$ y $$:
$$
(2-\sqrt{3})x - y = \sqrt{3} \quad \Longrightarrow \quad y = (2-\sqrt{3})x - \sqrt{3}
$$

**Step 3.** Substitute the expression for $$ y $$ into the second equation:
$$
3x + (2+\sqrt{3})\left[(2-\sqrt{3})x - \sqrt{3}\right] = 1-2\sqrt{3}
$$

**Step 4.** Simplify the product $$(2+\sqrt{3})(2-\sqrt{3})$$:
$$
(2+\sqrt{3})(2-\sqrt{3}) = 4 - (\sqrt{3})^2 = 4 - 3 = 1
$$

**Step 5.** Substitute the simplified product into the equation:
$$
3x + \left[1 \cdot x - (2+\sqrt{3})\sqrt{3}\right] = 1-2\sqrt{3}
$$
$$
3x + x - (2\sqrt{3} + (\sqrt{3})^2) = 1-2\sqrt{3}
$$
$$
3x + x - (2\sqrt{3} + 3) = 1-2\sqrt{3}
$$

**Step 6.** Combine like terms:
$$
4x - 2\sqrt{3} - 3 = 1-2\sqrt{3}
$$

**Step 7.** Add $$ 2\sqrt{3} + 3 $$ to both sides:
$$
4x = 1 - 2\sqrt{3} + 2\sqrt{3} + 3 = 4
$$

**Step 8.** Divide by 4 to solve for $$ x $$:
$$
x = \frac{4}{4} = 1
$$

Thus, the value of $$ x $$ is $$ 1 $$, which corresponds to option **D**.
$$ x = 1 $$, choice D

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