Question

solve: \\(\frac{x}{3} = \frac{1}{x - 2}\\)\
\\(\bigcirc\\) x=3, x=-1\
\\(\bigcirc\\) x=2\
\\(\bigcirc\\) x=-3, x=1\
\\(\bigcirc\\) x=0

Explanation:

Step1: Cross - multiply the equation

Given the equation \(\frac{x}{3}=\frac{1}{x - 2}\), cross - multiplying (which is valid when \(x
eq2\) to avoid division by zero) gives us \(x(x - 2)=3\times1\).

Step2: Expand and simplify the equation

Expanding the left - hand side, we get \(x^{2}-2x = 3\). Rearranging this equation to the standard quadratic form \(ax^{2}+bx + c = 0\), we have \(x^{2}-2x-3 = 0\).

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(- 3\) and add up to \(-2\). The numbers are \(-3\) and \(1\). So, the quadratic equation \(x^{2}-2x - 3=0\) can be factored as \((x - 3)(x+1)=0\).

Step4: Solve for x

Using the zero - product property, if \((x - 3)(x + 1)=0\), then either \(x-3 = 0\) (which gives \(x = 3\)) or \(x + 1=0\) (which gives \(x=-1\)). We also need to check that these values do not make the original denominator zero. For \(x = 3\), the denominator \(x - 2=3 - 2 = 1
eq0\). For \(x=-1\), the denominator \(x - 2=-1 - 2=-3
eq0\).

Answer:

\(x = 3,x=-1\) (corresponding to the option "x = 3, x=-1")