Question

Fixing a rational-equation simplification error

Original question: Problem 5: 4 \frac{3x}{x-9}=\frac{5x-72}{x-9} -\frac{3x}{x-9}=\frac{5x-72}{x-9}-4 -\frac{3x}{x-9}-\frac{5x-72}{x-9}=-4 \frac{-8x-72}{x-9} -\frac{8(x-9)}{x-9}=-4 -8=4

Expert Verified Solution

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Key takeaway: This is the kind of work where one sign error can send everything sideways. The cleanest way to debug it is to go back to the original equation and combine terms with care.

Key idea

The expression

4βˆ’3xxβˆ’9=5xβˆ’72xβˆ’94-\frac{3x}{x-9}=\frac{5x-72}{x-9}

has the same denominator on both fractions, so the safest move is to multiply through by (xβˆ’9)(x-9).

Step 1: Multiply both sides by (xβˆ’9)(x-9)

4(xβˆ’9)βˆ’3x=5xβˆ’724(x-9)-3x=5x-72

Step 2: Expand

4xβˆ’36βˆ’3x=5xβˆ’724x-36-3x=5x-72

Step 3: Combine like terms

xβˆ’36=5xβˆ’72x-36=5x-72

Step 4: Solve

Add 72 to both sides:

36=4x36=4x

So

x=9x=9

Step 5: Check the domain

But the original equation is undefined at x=9x=9 because the denominator becomes zero:

xβˆ’9=0x-9=0

So x=9x=9 must be rejected.

Final answer

NoΒ solution\boxed{\text{No solution}}


Pitfalls the pros know πŸ‘‡ Don’t trust a candidate answer just because the arithmetic works. In rational equations, the last line of defense is always substitution into the original expression, not the simplified one.

What if the problem changes? If the denominator were xβˆ’8x-8 instead of xβˆ’9x-9, the same algebra would likely produce a different candidate and a different validity check. The method stays the same; only the forbidden values change.

Tags: domain restriction, extraneous root, rational equation

FAQ

Why does x = 9 appear during the algebra?

It appears because the linear equation after clearing denominators leads to x = 9, but that value is not allowed in the original equation.

How do I know the solution is extraneous?

If substituting the answer into the original equation makes any denominator zero, the solution is extraneous.

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