Question

solve for j.
\\(\frac{j + 5}{j + 2} = \frac{j - 5}{j - 6}\\)
there may be 1 or 2 solutions.
j = \\(\square\\) or j = \\(\square\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{j + 5}{j + 2}=\frac{j-5}{j - 6}\), we use the cross - multiplication property of fractions. If \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d=c\times b\).
So, \((j + 5)(j - 6)=(j - 5)(j + 2)\)

Step2: Expand both sides

Expand the left - hand side: \((j + 5)(j - 6)=j\times j-6\times j+5\times j - 5\times6=j^{2}-6j + 5j-30=j^{2}-j - 30\)
Expand the right - hand side: \((j - 5)(j + 2)=j\times j+2\times j-5\times j - 5\times2=j^{2}+2j-5j - 10=j^{2}-3j - 10\)

Step3: Simplify the equation

Set the expanded forms equal to each other: \(j^{2}-j - 30=j^{2}-3j - 10\)
Subtract \(j^{2}\) from both sides of the equation:
\(j^{2}-j - 30-j^{2}=j^{2}-3j - 10-j^{2}\)
This simplifies to \(-j - 30=-3j - 10\)

Step4: Solve for \(j\)

Add \(3j\) to both sides: \(-j+3j - 30=-3j + 3j-10\)
\(2j-30=-10\)
Add 30 to both sides: \(2j-30 + 30=-10 + 30\)
\(2j=20\)
Divide both sides by 2: \(j = 10\)

We also need to check for extraneous solutions. The original equation has restrictions \(j
eq - 2\) and \(j
eq6\) (because the denominators \(j + 2\) and \(j - 6\) cannot be zero). When \(j = 10\), \(j+2=12
eq0\) and \(j - 6 = 4
eq0\), so \(j = 10\) is a valid solution. Wait, let's check our expansion again. Wait, maybe we made a mistake in the expansion. Let's re - expand:

Left - hand side: \((j + 5)(j - 6)=j^{2}-6j+5j - 30=j^{2}-j - 30\)
Right - hand side: \((j - 5)(j + 2)=j^{2}+2j-5j - 10=j^{2}-3j - 10\)
Then \(j^{2}-j - 30=j^{2}-3j - 10\)
Subtract \(j^{2}\) from both sides: \(-j-30=-3j - 10\)
Add \(3j\) to both sides: \(2j-30=-10\)
Add 30 to both sides: \(2j=20\), \(j = 10\). Wait, but let's check the original equation:

Left - hand side: \(\frac{10 + 5}{10+2}=\frac{15}{12}=\frac{5}{4}\)
Right - hand side: \(\frac{10 - 5}{10-6}=\frac{5}{4}\)
So \(j = 10\) is a solution. Wait, but the problem said there may be 1 or 2 solutions. Did we miss a solution?

Wait, let's go back to the cross - multiplication step. The equation is \(\frac{j + 5}{j + 2}=\frac{j-5}{j - 6}\), \(j
eq - 2,j
eq6\)

Cross - multiply: \((j + 5)(j - 6)=(j - 5)(j + 2)\)

\(j^{2}-6j+5j - 30=j^{2}+2j-5j - 10\)

\(j^{2}-j - 30=j^{2}-3j - 10\)

Subtract \(j^{2}\) from both sides: \(-j-30=-3j - 10\)

Add \(3j\) to both sides: \(2j-30=-10\)

Add 30 to both sides: \(2j = 20\)

\(j = 10\)

Wait, maybe there is only one solution. Let's check again.

Answer:

\(j=\boxed{10}\) (Wait, but maybe we made a mistake. Let's re - do the cross - multiplication:

\((j + 5)(j - 6)=(j - 5)(j + 2)\)

\(j^{2}-6j + 5j-30=j^{2}+2j-5j - 10\)

\(j^{2}-j - 30=j^{2}-3j - 10\)

Subtract \(j^{2}\) from both sides: \(-j-30=-3j - 10\)

\(2j=20\)

\(j = 10\)

Yes, so the solution is \(j = 10\))