Question

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

Explanation:

Step1: Cross - multiply the fractions

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

Step2: Expand both sides

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

Step3: Move all terms to one side

Add \(2x\) to both sides and subtract \(8\) from both sides to get a quadratic equation: \(x^{2}+5x + 2x-8 = 0\), which simplifies to \(x^{2}+7x - 8=0\).

Step4: Factor the quadratic equation

We factor \(x^{2}+7x - 8\). We need two numbers that multiply to \(-8\) and add up to \(7\). The numbers are \(8\) and \(-1\). So, \(x^{2}+7x - 8=(x + 8)(x - 1)=0\).

Step5: Solve for x

Set each factor equal to zero:

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

Step6: Check for extraneous solutions

We need to check if these solutions make the original denominators zero.

• For \(x=-8\): The denominator \(x-4=-8 - 4=-12

eq0\) and \(x=-8
eq0\).

• For \(x = 1\): The denominator \(x-4=1 - 4=-3

eq0\) and \(x = 1
eq0\). Both solutions are valid. But let's check in the original equation:
For \(x = 1\): Left - hand side \(\frac{1 + 5}{1-4}=\frac{6}{-3}=-2\), Right - hand side \(\frac{-2}{1}=-2\).
For \(x=-8\): Left - hand side \(\frac{-8 + 5}{-8-4}=\frac{-3}{-12}=\frac{1}{4}\), Right - hand side \(\frac{-2}{-8}=\frac{1}{4}\).

Answer:

\(x = 1\) or \(x=-8\) (But since the box in the original problem seems to expect one answer, and if we consider the initial attempt with \(x = 1\), but the full solution has two solutions. However, based on the solving steps, the solutions are \(x = 1\) and \(x=-8\). If we take the one in the box's context, \(x = 1\) is a valid solution. But the complete set of solutions is \(x = 1\) and \(x=-8\))