Solve the rational equation and check for undefined values

Key takeaway: Rational equations often reward patience more than speed. If a simplification seems too easy, it usually means the domain check is waiting at the end.

Step 1: Find the excluded values

For

$\frac{3}{x}+\frac{12}{x^2-4x}=\frac{-7}{x-4}$

factor the quadratic denominator:

$x^2-4x=x(x-4)$

So the restrictions are

$x\ne 0,\quad x\ne 4$

Step 2: Rewrite with a common denominator

The common denominator is $x(x-4)$.

Rewrite each term:

$\frac{3}{x}=\frac{3(x-4)}{x(x-4)}$

$\frac{12}{x(x-4)}=\frac{12}{x(x-4)}$

$\frac{-7}{x-4}=\frac{-7x}{x(x-4)}$

Step 3: Combine the fractions

$\frac{3(x-4)+12}{x(x-4)}=\frac{-7x}{x(x-4)}$

Simplify the numerator on the left:

$\frac{3x-12+12}{x(x-4)}=\frac{-7x}{x(x-4)}$

So

$\frac{3x}{x(x-4)}=\frac{-7x}{x(x-4)}$

Multiply both sides by $x(x-4)$:

$3x=-7x$

Step 4: Solve

$10x=0$

$x=0$

Step 5: Check restrictions

But $x=0$ is excluded. So there is no valid solution.

Final answer

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

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Pitfalls the pros know 👇 A shortcut can be dangerous here: after canceling an $x$, it may look like the equation became simpler, but cancellation is only legal when the factor is guaranteed nonzero. Since $x=0$ is already excluded, you still have to reject it at the end.

What if the problem changes? If the right-hand side were a different number, the common-denominator method would stay the same. Sometimes you get one valid solution, sometimes none, and sometimes an identity after simplification. The restriction list still comes first.

Tags: common denominator, excluded values, factorization