Question

solve for u. \\(\frac{2}{u - 8} = \frac{-5}{u - 1}\\) there may be 1 or 2 solutions. \\(u = \square\\) or \\(u = \square\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{2}{u - 8}=\frac{-5}{u - 1}\), we cross - multiply. Cross - multiplying gives us \(2(u - 1)=-5(u - 8)\). The reason for cross - multiplying is that if \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d = c\times b\).

Step2: Expand both sides

Expand the left - hand side: \(2(u - 1)=2u-2\). Expand the right - hand side: \(-5(u - 8)=-5u + 40\). So our equation becomes \(2u-2=-5u + 40\).

Step3: Add \(5u\) to both sides

Adding \(5u\) to both sides of the equation \(2u-2=-5u + 40\) gives \(2u+5u-2=-5u + 5u+40\), which simplifies to \(7u-2 = 40\).

Step4: Add 2 to both sides

Adding 2 to both sides of the equation \(7u-2 = 40\) gives \(7u-2 + 2=40 + 2\), so \(7u=42\).

Step5: Solve for \(u\)

Divide both sides of the equation \(7u = 42\) by 7. \(\frac{7u}{7}=\frac{42}{7}\), which gives \(u = 6\). We also need to check for extraneous solutions. The original equation has denominators \(u - 8\) and \(u - 1\). When \(u = 6\), \(u-8=6 - 8=-2
eq0\) and \(u - 1=6 - 1 = 5
eq0\). Since there is only one solution (we made a mistake earlier in thinking there could be two, but let's re - check the cross - multiplication and solving process). Wait, let's re - do the cross - multiplication:

Starting over: \(\frac{2}{u - 8}=\frac{-5}{u - 1}\)

Cross - multiply: \(2(u - 1)=-5(u - 8)\)

\(2u-2=-5u + 40\)

\(2u + 5u=40 + 2\)

\(7u=42\)

\(u = 6\)

Wait, maybe the initial thought of 1 or 2 solutions was a bit misleading. Let's check the domain: \(u
eq8\) and \(u
eq1\). When we solved, we got \(u = 6\), which is in the domain. So there is only one solution? Wait, maybe I made a mistake in cross - multiplying. Let's check again.

Wait, the equation is \(\frac{2}{u - 8}=\frac{-5}{u - 1}\)

Cross - multiply: \(2(u - 1)=-5(u - 8)\)

\(2u-2=-5u + 40\)

\(2u+5u=40 + 2\)

\(7u=42\)

\(u = 6\)

Yes, so the solution is \(u = 6\). Wait, maybe the problem had a typo? Or maybe I misread the equation. Let me check the original equation again. The original equation is \(\frac{2}{u - 8}=\frac{-5}{u - 1}\). So cross - multiplying gives \(2(u - 1)=-5(u - 8)\), which leads to \(u = 6\). So there is only one solution.

Wait, perhaps I made a mistake in the number of solutions. Let's consider the general case of a rational equation \(\frac{a}{x - b}=\frac{c}{x - d}\). Cross - multiplying gives \(a(x - d)=c(x - b)\), which is a linear equation, so it will have at most one solution, provided that the solution is not in the excluded values (\(x
eq b,x
eq d\)). In this case, the solution \(u = 6\) is not equal to 8 or 1, so it is a valid solution. So there is only one solution, \(u = 6\).

Answer:

\(u=\boxed{6}\) (and there is no other solution, so maybe the problem's "1 or 2 solutions" was a bit of a red herring, or perhaps a mistake in the problem statement)