Question

question
solve for all values of x.
\\(\frac{x + 2}{x - 4} = \frac{-1}{x}\\)

Explanation:

Step1: Cross - multiply to eliminate fractions

To solve the equation \(\frac{x + 2}{x-4}=\frac{-1}{x}\), we cross - multiply. Cross - multiplying gives us \(x(x + 2)=-1(x - 4)\).

Step2: Expand both sides

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

Step3: Rearrange into standard quadratic form

Add \(x\) to both sides and subtract \(4\) from both sides to get the standard quadratic equation \(ax^{2}+bx + c = 0\).
\(x^{2}+2x+x - 4=0\), which simplifies to \(x^{2}+3x - 4=0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(-4\) and add up to \(3\). The numbers are \(4\) and \(-1\).
So, \(x^{2}+3x - 4=(x + 4)(x - 1)=0\).

Step5: Solve for \(x\) using the zero - product property

If \((x + 4)(x - 1)=0\), then either \(x+4 = 0\) or \(x - 1=0\).

• For \(x+4 = 0\), we subtract \(4\) from both sides to get \(x=-4\).
• For \(x - 1=0\), we add \(1\) to both sides to get \(x = 1\).

We also need to check for extraneous solutions. The original equation has denominators \(x-4\) and \(x\), so \(x
eq4\) and \(x
eq0\). Both \(-4\) and \(1\) are not equal to \(4\) or \(0\), so they are valid solutions.

Answer:

The solutions are \(x=-4\) and \(x = 1\)