Solving for y in a rational equation by factoring

Expert intro: This problem is an algebraic isolation task: rearrange the terms, factor out y, and solve for y explicitly.

Detailed walkthrough

Rewrite the equation first

Start with

$xy-y+2x=0$

The goal is to solve for $y$. Move the $2x$ term to the other side:

$xy-y=-2x$

Factor the y terms

Both terms on the left contain $y$, so factor it out:

$y(x-1)=-2x$

Now divide by $x-1$:

$y=\frac{-2x}{x-1}$

That is the solved form.

Why factoring is the cleanest method

If you try to isolate $y$ term by term before factoring, the algebra becomes messy. Factoring groups the two terms that contain $y$ into a single product, which makes the final division straightforward.

You can also rewrite the answer as

$y=\frac{2x}{1-x}$

These two forms are equivalent, because multiplying numerator and denominator by $-1$ does not change the value.

Check the result

Substitute $y=\frac{-2x}{x-1}$ back into the original equation. The left side simplifies to 0, which confirms the solution is correct. A quick substitution check is a strong habit, especially when fractions appear.

Common algebra pattern

Whenever you see terms like $xy-y$, look for a shared factor. In this case, $y$ is the common factor, and factoring it is what makes the problem solvable in one clean step.

💡 Pitfall guide

A common mistake is to divide by $x$ instead of factoring first. If you divide too early, you may lose track of the $-y$ term and end up with an incorrect expression. Another pitfall is forgetting that dividing by $x-1$ creates a restriction: $x\neq 1$. If $x=1$, the formula would require division by zero, so that value must be excluded from the final answer. Students also sometimes stop at $y(x-1)=-2x$ and think the solution is complete, but the variable $y$ still needs to be isolated.

🔄 Real-world variant

If the equation were changed to $xy-y+2x=6$, the method would be almost the same. You would move the constant first, getting $xy-y=-2x+6$, then factor to $y(x-1)=-2x+6$, and finally divide by $x-1$ to solve for y. If the equation were $xy+y+2x=0$, the factor would become $y(x+1)=-2x$, so the final denominator would change. Small sign changes can produce a completely different answer, so each term must be tracked carefully.

🔍 Related terms

factoring by grouping, isolating the variable, rational expression