Question

question
solve for all values of x.
\\(\frac{x + 1}{x - 6} = \frac{-4}{x}\\)
answer attempt 1 out of 2
additional solution no solution
x =
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Explanation:

Step1: Cross - multiply the equation

To solve the rational equation \(\frac{x + 1}{x-6}=\frac{-4}{x}\), we cross - multiply. Cross - multiplying gives us \(x(x + 1)=-4(x - 6)\).

Step2: Expand both sides

Expand the left - hand side: \(x(x + 1)=x^{2}+x\).
Expand the right - hand side: \(-4(x - 6)=-4x + 24\).
So the equation becomes \(x^{2}+x=-4x + 24\).

Step3: Rearrange into standard quadratic form

Bring all terms to the left - hand side to get a quadratic equation: \(x^{2}+x + 4x-24 = 0\).
Combine like terms: \(x^{2}+5x - 24=0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(-24\) and add up to \(5\). The numbers are \(8\) and \(-3\) since \(8\times(-3)=-24\) and \(8+( - 3)=5\).
So, \(x^{2}+5x - 24=(x + 8)(x - 3)=0\).

Step5: Solve for x

Set each factor equal to zero:

• If \(x + 8=0\), then \(x=-8\).
• If \(x - 3=0\), then \(x = 3\).

Step6: Check for extraneous solutions

We need to check if these solutions make the original denominators zero.
For the original equation \(\frac{x + 1}{x-6}=\frac{-4}{x}\), the denominators are \(x-6\) and \(x\).

• When \(x=-8\): \(x-6=-8 - 6=-14

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

• When \(x = 3\): \(x-6=3 - 6=-3

eq0\) and \(x = 3
eq0\).

Answer:

\(x=-8\) or \(x = 3\)