Question

solve the rational equation: $\frac{1}{18x - 3}=\frac{-x + 2}{6x - 1}$
$x=-6$ and $x = 1$
$x=-4$
$x=\frac{1}{2}$ and $x=\frac{5}{2}$
$x=\frac{5}{3}$

Explanation:

Step1: Cross - multiply

$(6x - 1)\times1=(18x - 3)(-x + 2)$

Step2: Expand both sides

$6x-1=-18x^{2}+36x + 3x-6$

Step3: Rearrange to standard quadratic form

$18x^{2}-33x + 5 = 0$

Step4: Use quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$

Here $a = 18$, $b=-33$, $c = 5$. First calculate the discriminant $\Delta=b^{2}-4ac=(-33)^{2}-4\times18\times5=1089 - 360 = 729$. Then $x=\frac{33\pm\sqrt{729}}{2\times18}=\frac{33\pm27}{36}$

Step5: Find the two solutions

$x_1=\frac{33 + 27}{36}=\frac{60}{36}=\frac{5}{3}$ and $x_2=\frac{33 - 27}{36}=\frac{6}{36}=\frac{1}{6}$ (we check for extraneous solutions by substituting back into the original equation, and $\frac{1}{6}$ makes the original denominators zero, so we discard it)

Answer:

$x=\frac{5}{3}$