Question

question
solve for all values of x.
\\(\frac{x - 5}{x - 2} = \frac{4}{x}\\)
answer attempt 1 out of 2
\\(\odot\\) additional solution \\(\ominus\\) no solution
\\(x = \square\\) submit answer

Explanation:

Step1: Cross - multiply the equation

To solve the rational equation \(\frac{x - 5}{x - 2}=\frac{4}{x}\), we cross - multiply. Cross - multiplying gives us \(x(x - 5)=4(x - 2)\).

Step2: Expand both sides

Expanding the left - hand side: \(x(x - 5)=x^{2}-5x\).
Expanding the right - hand side: \(4(x - 2)=4x-8\).
So our equation becomes \(x^{2}-5x = 4x-8\).

Step3: Rearrange into standard quadratic form

Subtract \(4x\) from both sides and add \(8\) to both sides to get \(x^{2}-5x-4x + 8=0\), which simplifies to \(x^{2}-9x + 8=0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(8\) and add up to \(-9\). The numbers are \(-1\) and \(-8\). So, \(x^{2}-9x + 8=(x - 1)(x - 8)=0\).

Step5: Solve for x

Setting each factor equal to zero:

• If \(x - 1=0\), then \(x = 1\).
• If \(x - 8=0\), then \(x = 8\).

Step6: Check for extraneous solutions

We need to check if these solutions make the original denominators zero.
For the denominator \(x-2\):

• When \(x = 1\), \(x-2=1 - 2=-1

eq0\).

• When \(x = 8\), \(x - 2=8 - 2 = 6

eq0\).
For the denominator \(x\):

• When \(x = 1\), \(x=1

eq0\).

• When \(x = 8\), \(x = 8

eq0\).

So both \(x = 1\) and \(x = 8\) are valid solutions. But since the problem may be expecting one of them (maybe a typo in the problem - setting or a single - answer context), but from the equation, the solutions are \(x = 1\) and \(x = 8\). If we assume we are to find all solutions, but if we consider the initial problem's "Additional Solution" option, and if we made a mistake in the process, but following the steps, the solutions are \(x = 1\) and \(x = 8\). However, let's re - check the cross - multiplication step.

Wait, let's re - do the cross - multiplication: \(\frac{x - 5}{x - 2}=\frac{4}{x}\) implies \(x(x - 5)=4(x - 2)\)

\(x^{2}-5x=4x - 8\)

\(x^{2}-9x + 8 = 0\)

\((x - 1)(x - 8)=0\)

So \(x = 1\) or \(x = 8\).

Answer:

The solutions are \(x = 1\) and \(x = 8\). If we consider the problem's "Additional Solution" and a single - answer context (maybe a mistake in the problem), but the correct solutions from the equation are \(x = 1\) and \(x = 8\). If we have to choose one (maybe the problem was intended to have one solution), but according to the steps, the solutions are \(x = 1\) and \(x = 8\).