Solve a rational equation and identify extraneous solutions

Key concept: This kind of equation looks straightforward until the denominator is cleared. The real work is not only solving for $x$, but also checking whether the answer makes any denominator zero.

Step by step

Step 1: Note the restriction

Because the denominator is $x-9$, we must have

$x \ne 9$

Step 2: Clear the denominator

Start with

$4-\frac{3x}{x-9}=\frac{5x-72}{x-9}$

Multiply every term by $(x-9)$:

$4(x-9)-3x=5x-72$

Step 3: Expand and simplify

$4x-36-3x=5x-72$

Combine like terms:

$x-36=5x-72$

Step 4: Solve for $x$

Add 72 to both sides:

$36=4x$

Divide by 4:

$x=9$

Step 5: Check the result

The value $x=9$ is not allowed because it makes $x-9=0$. So even though it solves the algebraic steps, it is not a valid solution.

Final answer

$\boxed{\text{No solution}}$

Pitfall alert

The most common mistake is stopping as soon as you get a number. For rational equations, that number still has to survive the domain check. Here, $x=9$ cancels the denominator, so it must be rejected as extraneous.

Try different conditions

If the constant on the left changed, the algebra could still lead to a single candidate answer, but the final step would stay the same: check whether that candidate makes any denominator zero. If the denominator were something like $x+1$ instead of $x-9$, then $x=9$ might become valid, because the restriction would change to $x\ne -1$.

Further reading

extraneous solution, rational equation, domain restriction