Question

solve for t. \\(\frac{t - 3}{t - 7} = \frac{t + 6}{t - 10}\\) there may be 1 or 2 solutions. \\(t = \square\\) or \\(t = \square\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{t - 3}{t - 7}=\frac{t + 6}{t - 10}\), 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, \((t - 3)(t - 10)=(t + 6)(t - 7)\)

Step2: Expand both sides

Expand the left - hand side: \((t - 3)(t - 10)=t\times t-10t-3t + 30=t^{2}-13t + 30\)
Expand the right - hand side: \((t + 6)(t - 7)=t\times t-7t+6t - 42=t^{2}-t - 42\)

Step3: Simplify the equation

Now we have the equation \(t^{2}-13t + 30=t^{2}-t - 42\)
Subtract \(t^{2}\) from both sides of the equation:
\(t^{2}-13t + 30-t^{2}=t^{2}-t - 42-t^{2}\)
This simplifies to \(-13t + 30=-t - 42\)

Step4: Solve for t

Add \(13t\) to both sides:
\(-13t+30 + 13t=-t - 42+13t\)
\(30 = 12t-42\)
Add 42 to both sides:
\(30 + 42=12t-42 + 42\)
\(72 = 12t\)
Divide both sides by 12:
\(t=\frac{72}{12}=6\)

Wait, let's check for extraneous solutions. We need to ensure that \(t eq7\) and \(t eq10\) (because the denominators \(t - 7\) and \(t - 10\) cannot be zero). When \(t = 6\), \(t-7=6 - 7=-1 eq0\) and \(t - 10=6-10=-4 eq0\). Wait, did we make a mistake in the expansion?

Let's re - expand:

Left - hand side: \((t - 3)(t - 10)=t^{2}-10t-3t + 30=t^{2}-13t + 30\) (correct)

Right - hand side: \((t + 6)(t - 7)=t^{2}-7t+6t - 42=t^{2}-t - 42\) (correct)

Then \(t^{2}-13t + 30=t^{2}-t - 42\)

Subtract \(t^{2}\): \(-13t + 30=-t - 42\)

Add \(13t\): \(30 = 12t-42\)

Add 42: \(72 = 12t\), \(t = 6\). Wait, but let's check the original equation:

Left - hand side: \(\frac{6 - 3}{6 - 7}=\frac{3}{-1}=-3\)

Right - hand side: \(\frac{6+6}{6 - 10}=\frac{12}{-4}=-3\). So \(t = 6\) is a solution. But the problem says 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 \(\frac{t - 3}{t - 7}=\frac{t + 6}{t - 10}\), \(t eq7\), \(t eq10\)

Cross - multiply: \((t - 3)(t - 10)=(t + 6)(t - 7)\)

\(t^{2}-10t-3t + 30=t^{2}-7t+6t - 42\)

\(t^{2}-13t + 30=t^{2}-t - 42\)

Subtract \(t^{2}\) from both sides: \(-13t+30=-t - 42\)

\(-13t + t=-42 - 30\)

\(-12t=-72\)

\(t = 6\). Wait, that's the same as before. So why does the problem say there may be 1 or 2 solutions? Maybe I made a mistake in the sign when moving terms.

Wait, let's do the algebra again:

Starting from \(-13t + 30=-t - 42\)

Add \(t\) to both sides: \(-12t+30=-42\)

Subtract 30: \(-12t=-72\)

Divide by \(-12\): \(t = 6\). So there is only one solution? But the problem has two boxes. Wait, maybe I made a mistake in the expansion.

Wait, let's re - expand \((t + 6)(t - 7)\):

\(t\times t=t^{2}\), \(t\times(-7)=-7t\), \(6\times t = 6t\), \(6\times(-7)=-42\). So \(t^{2}-7t + 6t-42=t^{2}-t - 42\) (correct)

\((t - 3)(t - 10)=t^{2}-10t-3t + 30=t^{2}-13t + 30\) (correct)

So \(t^{2}-13t + 30=t^{2}-t - 42\)

\(-13t + 30=-t - 42\)

\(-13t+t=-42 - 30\)

\(-12t=-72\)

\(t = 6\). So the solution is \(t = 6\). But the problem has two boxes. Maybe there was a typo in the problem, or maybe I made a mistake.

Wait, let's check the original equation again. If we consider the equation \(\frac{t - 3}{t - 7}=\frac{t + 6}{t - 10}\), cross - multiply:

\((t - 3)(t - 10)-(t + 6)(t - 7)=0\)

\(t^{2}-13t + 30-(t^{2}-t - 42)=0\)

\(t^{2}-13t + 30 - t^{2}+t + 42=0\)

\(-12t+72 = 0\)

\(-12t=-72\)

\(t = 6\). So there is only one solution. Maybe the problem's "may be 1 or 2" is just a statement, and in this case, there is one solution. But let's check again.

Wait, if we made a mistake in the sign when expanding \((t + 6)(t - 7)\). Let's do it again: \((t + 6)(t - 7)=t\times t+t\times(-7)+6\times t+6\times(-7)=t^{2}-7t + 6t-42=t^{2}-t - 42\) (correct).

So the only solution is \(t = 6\).

Answer:

\(t = \boxed{6}\) (and since there is only one solution, the second box can be considered as having no additional solution, but based on the calculation, the solution is \(t = 6\))