Solve a rational equation with two different denominators

Expert intro: Once the denominators are different, the equation stops being a quick linear cleanup. You have to clear both denominators carefully, then watch for invalid roots at the end.

Detailed walkthrough

Step 1: Identify the restrictions

From

$4-\frac{3x}{x-9}=\frac{5x-72}{x+1}$

we need

$x\ne 9,\quad x\ne -1$

Step 2: Move the constant

Subtract 4 from both sides:

$-\frac{3x}{x-9}=\frac{5x-72}{x+1}-4$

Write $4$ with denominator $x+1$:

$-\frac{3x}{x-9}=\frac{5x-72-4(x+1)}{x+1}$

Simplify the numerator:

$-\frac{3x}{x-9}=\frac{x-76}{x+1}$

Step 3: Cross-multiply

$-3x(x+1)=(x-76)(x-9)$

Expand both sides:

$-3x^2-3x=x^2-85x+684$

Step 4: Bring everything to one side

$-4x^2+82x-684=0$

Divide by $-2$ to simplify:

$2x^2-41x+342=0$

Step 5: Solve

Use the quadratic formula:

$x=\frac{41\pm\sqrt{41^2-4(2)(342)}}{4}$

The discriminant is

$41^2-2736=1681-2736=-1055$

Since it is negative, there are no real solutions.

Final answer

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

💡 Pitfall guide

A common slip is treating $(x-9)$ and $(x+1)$ as if they cancel somewhere in the algebra. They do not. Also, if the quadratic discriminant is negative, that means no real solution even if the intermediate steps looked neat.

🔄 Real-world variant

If the right-hand denominator changed from $x+1$ to another linear factor, the setup would be the same: list restrictions first, clear denominators, then solve the resulting quadratic. A different constant could easily change the discriminant from negative to zero or positive.

🔍 Related terms

cross-multiplication, quadratic equation, restricted values