Fraction equations

You order a pizza for 12 euros and split the price evenly. In the end everyone pays 4 euros. How many of you were there? That question is hiding in the equation $\frac{12}{x} = 4$. The only new thing is where the $x$ sits: not in the numerator and not alone on one side, but below the fraction bar. Equations like this are called fraction equations, and three fixed moves solve all of them.

The forbidden numbers come first

You may not divide by 0, that is simply not defined in mathematics. So with a fraction equation you look at the denominator first and ask: which number would make it 0? For $\frac{12}{x}$ that is 0 itself, so $x \neq 0$. For $\frac{6}{x - 2}$ you set $x - 2 = 0$ and get $x = 2$, and that number is forbidden. If the denominator holds a plus, the forbidden number turns negative: $x + 5 = 0$ gives $x = -5$. People say the expression has a definition gap there. Write the forbidden numbers down next to the task before you rearrange anything at all.

Multiplying the fraction away

The balance scale still applies: whatever you do on the left, you also do on the right. The only new move is the one that gets rid of the fraction. You multiply both sides by the denominator, the fraction cancels, and the $x$ is back on top. This step is allowed because the denominator is never 0 for any permitted number. That turns $\frac{12}{x} = 4$ into $12 = 4x$, and dividing by 4 gives $x = 3$. If the denominator holds more than just an $x$, use brackets when you multiply: $\frac{6}{x - 2} = 3$ becomes $6 = 3 \cdot (x - 2)$, so $6 = 3x - 6$, then $3x = 12$ and finally $x = 4$.

The check and the spurious solution

At the end you substitute your solution into the original equation. For $x = 4$ in $\frac{6}{x - 2} = 3$ you calculate $\frac{6}{4 - 2} = \frac{6}{2} = 3$, so both sides agree. With fraction equations a second question comes on top: is the solution allowed at all? In $\frac{x}{x - 3} = \frac{3}{x - 3}$ the number $x = 3$ is forbidden, because both denominators would be 0 there. Yet multiplying by $(x - 3)$ produces exactly the line $x = 3$. In that case $x = 3$ is called a spurious solution, it drops out, and the equation has no solution.