Question

solve for all values of x.
\\(\frac{x - 5}{x - 8} = \frac{1}{x}\\)
answer attempt 1 out of 2
\\(\oplus\\) additional solution \\(\ominus\\) no solution
\\(x = \square\\) submit an

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{x - 5}{x - 8}=\frac{1}{x}\), we cross - multiply. Cross - multiplying gives us \(x(x - 5)=(x - 8)\times1\).

Step2: Expand both sides

Expanding the left - hand side: \(x(x - 5)=x^{2}-5x\), and the right - hand side is \(x - 8\). So our equation becomes \(x^{2}-5x=x - 8\).

Step3: Rearrange into standard quadratic form

Subtract \(x\) and add \(8\) to both sides to get \(x^{2}-5x - x + 8 = 0\), which simplifies to \(x^{2}-6x + 8 = 0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(8\) and add up to \(-6\). The numbers are \(-2\) and \(-4\). So, \(x^{2}-6x + 8=(x - 2)(x - 4)=0\).

Step5: Solve for x

Using the zero - product property, if \((x - 2)(x - 4)=0\), then either \(x - 2 = 0\) or \(x - 4 = 0\). Solving \(x - 2 = 0\) gives \(x = 2\), and solving \(x - 4 = 0\) gives \(x = 4\). We also need to check for extraneous solutions. The original equation has denominators \(x-8\) and \(x\), so \(x
eq8\) and \(x
eq0\). Both \(2\) and \(4\) are not equal to \(8\) or \(0\), so they are valid solutions.

Answer:

\(x = 2\) or \(x = 4\)