Question

1. \\(\frac{x + 5}{x^2} - \frac{x - 1}{x^2} = \frac{1}{3x}\\)

Explanation:

Step1: Simplify the left - hand side

Since the two fractions on the left - hand side have the same denominator \(x^{2}\), we can subtract the numerators:
\(\frac{(x + 5)-(x - 1)}{x^{2}}=\frac{x+5 - x + 1}{x^{2}}=\frac{6}{x^{2}}\)
So the equation becomes \(\frac{6}{x^{2}}=\frac{1}{3x}\)

Step2: Cross - multiply to eliminate the fractions

Cross - multiplying gives us \(6\times(3x)=1\times x^{2}\)
Which simplifies to \(18x=x^{2}\)

Step3: Rearrange the equation into standard quadratic form

Subtract \(18x\) from both sides to get \(x^{2}-18x = 0\)

Step4: Factor the quadratic equation

Factor out an \(x\) from the left - hand side: \(x(x - 18)=0\)

Step5: Solve for \(x\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
So \(x=0\) or \(x - 18=0\), which gives \(x = 18\). But we need to check for extraneous solutions because in the original equation, \(x
eq0\) (since \(x\) is in the denominator of \(\frac{1}{3x}\) and \(\frac{x + 5}{x^{2}}\), \(\frac{x-1}{x^{2}}\)). So we discard \(x = 0\).

Answer:

\(x = 18\)