Question

method: cross multiplication
① \\(\frac{x + 5}{5} = \frac{6}{x - 2}\\)
\\((x + 5)(x - 2) = 30\\)
\\(x^2 - 2x + 5x - 10 = 30\\)
\\(x^2 + 3x - 10 = 30\\)
\\(\underline{\quad - 30 \quad - 30}\\)
\\(x^2 + 3x - 40 = 0\\)
\\(\

$$\begin{array}{|c|c|} \\hline -40 & +3 \\\\ \\hline -1 + 40 & \\\\ \\hline -2 + 20 & \\\\ \\hline -4 + 10 & \\\\ \\hline -5 + 8 & -5 + 8 = +3 \\\\ \\hline \\end{array}$$

\\)
\\((x - 5)(x + 8) = 0\\)
\\(x - 5 = 0\\)
\\(\underline{\quad +5 \quad +5}\\)
\\(\boxed{x = 5}\\)
\\(x + 8 = 0\\)
\\(\underline{\quad -8 \quad -8}\\)
\\(\boxed{x = -8}\\)
② \\(\frac{x + 4}{2} = \frac{4x}{x - 6}\\)
\\((x + 4)(x - 6) = 8x\\)
\\(x^2 - 6x + 4x - 24 = 8x\\)
\\(x^2 - 2x - 24 = 8x\\)
\\(\underline{\quad +8x \quad -8x}\\)
\\(x^2 - 2x - 32x = 0\\)
\\(\

$$\begin{array}{|c|c|} \\hline -32 & -2 \\\\ \\hline -1 + 32 & \\\\ \\hline -2 + 16 & \\\\ \\hline -8 + 4 & \\\\ \\hline \\end{array}$$

\\)

Explanation:

Step1: Solve the first equation \(\frac{x + 5}{5}=\frac{6}{x - 2}\)

Cross - multiply: \((x + 5)(x - 2)=5\times6 = 30\)
Expand the left - hand side: \(x^{2}-2x + 5x-10 = 30\)
Simplify: \(x^{2}+3x-10 - 30=0\), so \(x^{2}+3x - 40 = 0\)
Factor the quadratic equation: We need two numbers that multiply to \(- 40\) and add up to \(3\). The numbers are \(-5\) and \(8\). So \((x - 5)(x + 8)=0\)
Set each factor equal to zero:

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

Step2: Solve the second equation \(\frac{x + 4}{2}=\frac{4x}{x - 6}\)

Cross - multiply: \((x + 4)(x - 6)=2\times4x=8x\)
Expand the left - hand side: \(x^{2}-6x + 4x-24 = 8x\)
Simplify: \(x^{2}-2x-24 - 8x=0\), so \(x^{2}-10x - 24 = 0\) (Note: There was a mistake in the original calculation where \(-2x-32x\) should be \(-2x - 8x=-10x\) instead of \(-2x-32x\))
Factor the quadratic equation: We need two numbers that multiply to \(-24\) and add up to \(-10\). The numbers are \(-12\) and \(2\). So \((x - 12)(x + 2)=0\)
Set each factor equal to zero:

• If \(x - 12=0\), then \(x = 12\)
• If \(x + 2=0\), then \(x=-2\) (We also need to check for extraneous solutions. For the original equation \(\frac{x + 4}{2}=\frac{4x}{x - 6}\), \(x

eq6\). Both \(x = 12\) and \(x=-2\) are valid as they do not make the denominator zero)

Answer:

For the equation \(\frac{x + 5}{5}=\frac{6}{x - 2}\), the solutions are \(x = 5\) and \(x=-8\); for the equation \(\frac{x + 4}{2}=\frac{4x}{x - 6}\) (after correcting the calculation error), the solutions are \(x = 12\) and \(x=-2\)