Question

solve for x.
\\(\frac{-16}{x + 3} = \frac{-8}{x}\\)
\\(x = \square\\)

Explanation:

Step1: Cross - multiply the fractions

Given the equation \(\frac{-16}{x + 3}=\frac{-8}{x}\), cross - multiplying (which is valid when \(x
eq0\) and \(x
eq - 3\)) gives \(-16x=-8(x + 3)\).

Step2: Expand the right - hand side

Expand \(-8(x + 3)\) using the distributive property \(a(b + c)=ab+ac\). Here, \(a=-8\), \(b = x\), and \(c = 3\), so \(-8(x + 3)=-8x-24\). The equation becomes \(-16x=-8x - 24\).

Step3: Add \(16x\) to both sides

Adding \(16x\) to both sides of the equation \(-16x=-8x - 24\) gives \(0 = 8x-24\).

Step4: Add 24 to both sides

Adding 24 to both sides of \(0 = 8x-24\) gives \(24 = 8x\).

Step5: Solve for \(x\)

Divide both sides of \(24 = 8x\) by 8. \(\frac{24}{8}=x\), so \(x = 3\). We need to check if this solution is valid. Substitute \(x = 3\) into the original equation: the left - hand side is \(\frac{-16}{3 + 3}=\frac{-16}{6}=-\frac{8}{3}\), and the right - hand side is \(\frac{-8}{3}\). So \(x = 3\) is a valid solution.

Answer:

\(3\)