Question

solve for j.
\\(\frac{-1}{j - 2} = \frac{-2}{j + 3}\\)
there may be 1 or 2 solutions.
j = \boxed{} or j = \boxed{}

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{-1}{j - 2}=\frac{-2}{j + 3}\), we cross - multiply. Cross - multiplying gives us \(-1\times(j + 3)=-2\times(j - 2)\).

Step2: Expand both sides

Expanding the left - hand side: \(-1\times j+(-1)\times3=-j - 3\).
Expanding the right - hand side: \(-2\times j+(-2)\times(-2)=-2j + 4\).
So the equation becomes \(-j - 3=-2j + 4\).

Step3: Add \(2j\) to both sides

Adding \(2j\) to both sides of the equation \(-j - 3=-2j + 4\), we get \(-j+2j - 3=-2j + 2j+4\).
Simplifying, we have \(j - 3 = 4\).

Step4: Add 3 to both sides

Adding 3 to both sides of the equation \(j - 3 = 4\), we get \(j-3 + 3=4 + 3\).
Simplifying, we find that \(j = 7\).
We also need to check for extraneous solutions. The original equation has denominators \(j - 2\) and \(j + 3\). When \(j = 7\), \(j-2=7 - 2 = 5
eq0\) and \(j + 3=7+3 = 10
eq0\), so \(j = 7\) is a valid solution. Wait, let's re - check the cross - multiplication and expansion.
Wait, let's start over. Cross - multiply: \(-1(j + 3)=-2(j - 2)\)
\(-j-3=-2j + 4\)
Add \(2j\) to both sides: \(-j + 2j-3=-2j+2j + 4\)
\(j-3 = 4\)
Add 3 to both sides: \(j=4 + 3=7\). Wait, but let's check if there was a mistake in the sign. Wait, the original equation is \(\frac{-1}{j - 2}=\frac{-2}{j + 3}\). Let's substitute \(j = 7\):
Left - hand side: \(\frac{-1}{7-2}=\frac{-1}{5}\)
Right - hand side: \(\frac{-2}{7 + 3}=\frac{-2}{10}=\frac{-1}{5}\). So it works. Wait, but the problem said there may be 1 or 2 solutions. Did we miss a solution?
Wait, let's re - examine the cross - multiplication step. The equation is \(\frac{-1}{j - 2}=\frac{-2}{j + 3}\). Cross - multiplying: \(-1\times(j + 3)=-2\times(j - 2)\)
\(-j-3=-2j + 4\)
\(-j + 2j=4 + 3\)
\(j=7\). So there is only one solution? Wait, maybe I made a mistake in the initial assumption. Let's check the algebra again.
Starting with \(\frac{-1}{j - 2}=\frac{-2}{j + 3}\)
Cross multiply: \(-(j + 3)=-2(j - 2)\)
\(-j-3=-2j + 4\)
Add \(2j\) to both sides: \(j-3 = 4\)
Add 3 to both sides: \(j = 7\). So the only solution is \(j = 7\). Wait, but the problem said "there may be 1 or 2 solutions". Maybe I made a mistake in the cross - multiplication. Let's check with \(j = 1\) (a random value). No, let's do the algebra again.
\(\frac{-1}{j - 2}=\frac{-2}{j + 3}\)
Multiply both sides by \((j - 2)(j + 3)\) (the least common denominator, noting that \(j
eq2\) and \(j
eq - 3\)):
\(-1\times(j + 3)=-2\times(j - 2)\)
\(-j-3=-2j + 4\)
\(-j + 2j=4 + 3\)
\(j = 7\). So the solution is \(j = 7\).

Answer:

\(j=\boxed{7}\) (Note: There is only one solution in this case)