Given $\dfrac{3x - y}{x + 2y} = \dfrac{1}{3}$, find the ratio x : y. Answer x : y = ... Solve the inequality $\dfrac{2 - 3x}{3} < \dfrac{2x - 1}{6}$. Answer ... Given that $\dfrac{1}{x} + \dfrac{1}{y^2} = \dfrac{1}{w - 3}$, express $y$ in terms of $x$ and $w$. Answer $y =$ ...

1. Clear the fractions by cross-multiplying.

$\frac{3x-y}{x+2y}=\frac{1}{3}$
$3(3x-y)=x+2y$

2. Rearrange to express $x$ and $y$ in a ratio.

$9x-3y=x+2y$
$8x=5y$
$\frac{x}{y}=\frac{5}{8}$
So,
$x:y=5:8$

1. Clear the fractions using the lowest common denominator.

$\frac{2-3x}{3}<\frac{2x-1}{6}$
Multiply both sides by $6$:
$2(2-3x)<2x-1$
$4-6x<2x-1$

2. Collect like terms and isolate $x$.

$4-6x<2x-1$
$5<8x$
$x>\frac{5}{8}$

1. Isolate the term containing $y$.

$\frac{1}{x}+\frac{1}{y^2}=\frac{1}{w-3}$
$\frac{1}{y^2}=\frac{1}{w-3}-\frac{1}{x}$

2. Combine the algebraic fractions into a single fraction.

$\frac{1}{y^2}=\frac{x-(w-3)}{x(w-3)}$
$\frac{1}{y^2}=\frac{x-w+3}{x(w-3)}$

3. Invert and take the square root to obtain $y$.

$y^2=\frac{x(w-3)}{x-w+3}$
$y=\pm\sqrt{\frac{x(w-3)}{x-w+3}}$