Question

solve for m. \\(\frac{m - 4}{m - 9} = \frac{m - 8}{m - 3}\\) there may be 1 or 2 solutions. \\(m = \square\\) or \\(m = \square\\)

Explanation:

Step1: Cross - multiply the equation

To solve the equation \(\frac{m - 4}{m - 9}=\frac{m - 8}{m - 3}\), we cross - multiply. Cross - multiplying gives \((m - 4)(m - 3)=(m - 8)(m - 9)\).

Step2: Expand both sides

Expand the left - hand side: \((m - 4)(m - 3)=m^{2}-3m-4m + 12=m^{2}-7m + 12\).
Expand the right - hand side: \((m - 8)(m - 9)=m^{2}-9m-8m + 72=m^{2}-17m + 72\).
So our equation becomes \(m^{2}-7m + 12=m^{2}-17m + 72\).

Step3: Simplify the equation

Subtract \(m^{2}\) from both sides of the equation:
\(m^{2}-7m + 12-m^{2}=m^{2}-17m + 72-m^{2}\)
This simplifies to \(-7m+12=-17m + 72\).

Step4: Solve for m

Add \(17m\) to both sides: \(-7m + 17m+12=-17m + 17m + 72\)
\(10m+12 = 72\).
Subtract 12 from both sides: \(10m+12 - 12=72 - 12\)
\(10m=60\).
Divide both sides by 10: \(m=\frac{60}{10}=6\).
We need to check for extraneous solutions. The original equation has denominators \(m - 9\) and \(m - 3\). When \(m = 6\), \(m-9=6 - 9=-3
eq0\) and \(m - 3=6 - 3 = 3
eq0\). So \(m = 6\) is a valid solution. Wait, let's check our expansion again. Wait, maybe we made a mistake in the expansion. Let's re - expand:

\((m - 4)(m - 3)=m\times m-m\times3-4\times m + 4\times3=m^{2}-3m-4m + 12=m^{2}-7m + 12\)

\((m - 8)(m - 9)=m\times m-m\times9-8\times m+8\times9=m^{2}-9m-8m + 72=m^{2}-17m + 72\)

Then \(m^{2}-7m + 12=m^{2}-17m + 72\)

Subtract \(m^{2}\) from both sides: \(-7m + 12=-17m+72\)

Add \(17m\) to both sides: \(10m + 12=72\)

Subtract 12: \(10m=60\), \(m = 6\). Wait, but let's plug \(m = 6\) into the original equation:

Left - hand side: \(\frac{6 - 4}{6 - 9}=\frac{2}{-3}=-\frac{2}{3}\)

Right - hand side: \(\frac{6 - 8}{6 - 3}=\frac{-2}{3}=-\frac{2}{3}\). So it works. 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 \(\frac{m - 4}{m - 9}=\frac{m - 8}{m - 3}\), \(m
eq9\), \(m
eq3\).

Cross - multiplying: \((m - 4)(m - 3)=(m - 8)(m - 9)\)

\(m^{2}-3m-4m + 12=m^{2}-9m-8m + 72\)

\(m^{2}-7m + 12=m^{2}-17m + 72\)

Subtract \(m^{2}\) from both sides: \(-7m+12=-17m + 72\)

Add \(17m\) to both sides: \(10m+12 = 72\)

Subtract 12: \(10m=60\)

\(m = 6\). So there is only one solution? But the problem said there may be 1 or 2 solutions. Wait, maybe we made a mistake in the cross - multiplication. Wait, no, cross - multiplication is valid when \(m
eq9\) and \(m
eq3\). So the solution is \(m = 6\).

Wait, let's check again. Let's solve the equation \(\frac{m - 4}{m - 9}=\frac{m - 8}{m - 3}\)

Cross - multiply: \((m - 4)(m - 3)=(m - 8)(m - 9)\)

\(m^{2}-7m + 12=m^{2}-17m + 72\)

\(m^{2}-m^{2}-7m + 17m=72 - 12\)

\(10m=60\)

\(m = 6\). So the solution is \(m = 6\).

Answer:

\(m = 6\) (Wait, but the problem had two boxes. Maybe we made a mistake. Let's re - examine the problem. Wait, maybe the original equation was written incorrectly? Or maybe we made a mistake in the sign. Wait, let's check the original equation again. The equation is \(\frac{m - 4}{m - 9}=\frac{m - 8}{m - 3}\). Let's try to solve it again.

Wait, another way: \(\frac{m - 4}{m - 9}-\frac{m - 8}{m - 3}=0\)

Find a common denominator \((m - 9)(m - 3)\)

\(\frac{(m - 4)(m - 3)-(m - 8)(m - 9)}{(m - 9)(m - 3)}=0\)

The numerator must be zero (and the denominator non - zero).

\((m - 4)(m - 3)-(m - 8)(m - 9)=0\)

Expand:

\((m^{2}-7m + 12)-(m^{2}-17m + 72)=0\)

\(m^{2}-7m + 12 - m^{2}+17m - 72=0\)

\(10m-60 = 0\)

\(10m=60\)

\(m = 6\). So there is only one solution. So the two boxes may be a bit misleading, but the solution is \(m = 6\). Wait, maybe I made a mistake in the initial assumption. Let's check with \(m = 12\) (just to test). If \(m = 12\), left - hand side: \(\frac{12 - 4}{12 - 9}=\frac{8}{3}\), right - hand side: \(\frac{12 - 8}{12 - 3}=\frac{4}{9}\), not equal. If \(m = 0\), left - hand side: \(\frac{-4}{-9}=\frac{4}{9}\), right - hand side: \(\frac{-8}{-3}=\frac{8}{3}\), not equal. So the only solution is \(m = 6\).