Question

solve for j.
\\(\frac{6}{j + 6} = \frac{-1}{j - 1}\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{6}{j + 6}=\frac{-1}{j - 1}\), we use the cross - multiplication property of fractions. If \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\). So, we have \(6\times(j - 1)=-1\times(j + 6)\).

Step2: Expand both sides of the equation

Expand the left - hand side: \(6\times(j - 1)=6j-6\).
Expand the right - hand side: \(-1\times(j + 6)=-j - 6\).
So the equation becomes \(6j-6=-j - 6\).

Step3: Add \(j\) to both sides

Add \(j\) to each side of the equation \(6j-6=-j - 6\) to get \(6j + j-6=-j + j-6\), which simplifies to \(7j-6=-6\).

Step4: Add 6 to both sides

Add 6 to each side of the equation \(7j-6=-6\). We have \(7j-6 + 6=-6 + 6\), which simplifies to \(7j = 0\).

Step5: Solve for \(j\)

Divide both sides of the equation \(7j = 0\) by 7. So, \(j=\frac{0}{7}=0\).
We need to check if this solution makes the original denominators zero. For \(j = 0\), \(j + 6=0 + 6 = 6
eq0\) and \(j-1=0 - 1=-1
eq0\). So \(j = 0\) is a valid solution.

Answer:

\(j = 0\)