Question

1. \\(\frac{x + 5}{5} = \frac{6}{x - 2}\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{x + 5}{5}=\frac{6}{x - 2}\), we use the cross - multiplication property of proportions. If \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\). So, \((x + 5)(x - 2)=5\times6\).

Step2: Expand the left - hand side

Expand \((x + 5)(x - 2)\) using the FOIL method. \((x+5)(x - 2)=x^{2}-2x+5x - 10=x^{2}+3x - 10\). And the right - hand side is \(30\). So the equation becomes \(x^{2}+3x - 10 = 30\).

Step3: Rearrange into standard quadratic form

Subtract \(30\) from both sides of the equation to get it in standard quadratic form \(ax^{2}+bx + c = 0\). So, \(x^{2}+3x-10 - 30=0\), which simplifies to \(x^{2}+3x - 40 = 0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(- 40\) and add up to \(3\). The numbers are \(8\) and \(-5\) because \(8\times(-5)=-40\) and \(8+( - 5)=3\). So, \(x^{2}+3x - 40=(x + 8)(x - 5)=0\).

Step5: Solve for x

Set each factor equal to zero:

• If \(x + 8=0\), then \(x=-8\).
• If \(x - 5=0\), then \(x = 5\).

We need to check for extraneous solutions by plugging the values back into the original equation.

• For \(x=-8\): Left - hand side \(\frac{-8 + 5}{5}=\frac{-3}{5}\), Right - hand side \(\frac{6}{-8 - 2}=\frac{6}{-10}=-\frac{3}{5}\).
• For \(x = 5\): Left - hand side \(\frac{5+5}{5}=\frac{10}{5}=2\), Right - hand side \(\frac{6}{5 - 2}=\frac{6}{3}=2\).

Both solutions are valid.

Answer:

\(x=-8\) or \(x = 5\)