Question

1. \\(\frac{y - 2}{2} = \frac{5}{y - 5}\\)

Explanation:

Step1: Cross - multiply

Cross - multiply the fractions \(\frac{y - 2}{2}=\frac{5}{y - 5}\) to get \((y - 2)(y - 5)=2\times5\).

Step2: Expand the left - hand side

Expand \((y - 2)(y - 5)\) using the FOIL method. \((y - 2)(y - 5)=y^{2}-5y-2y + 10=y^{2}-7y + 10\). And the right - hand side is \(10\). So the equation becomes \(y^{2}-7y + 10 = 10\).

Step3: Simplify the equation

Subtract \(10\) from both sides of the equation \(y^{2}-7y + 10 = 10\) to get \(y^{2}-7y=0\).

Step4: Factor the equation

Factor out \(y\) from \(y^{2}-7y\), we get \(y(y - 7)=0\).

Step5: Solve for \(y\)

Set each factor equal to zero: \(y = 0\) or \(y-7=0\). Solving \(y - 7=0\) gives \(y = 7\). We need to check for extraneous solutions. When \(y = 0\), the original equation \(\frac{0 - 2}{2}=\frac{5}{0 - 5}\), \(\frac{-2}{2}=-1\) and \(\frac{5}{-5}=-1\), so \(y = 0\) is valid. When \(y = 7\), \(\frac{7 - 2}{2}=\frac{5}{2}\) and \(\frac{5}{7 - 5}=\frac{5}{2}\), so \(y = 7\) is also valid.

Answer:

The solutions are \(y = 0\) and \(y = 7\)