Clearing fractions and cross multiplication

Equa­tions full of frac­tions, such as $\displaystyle \frac{x + 1}{3} - \frac{x - 1}{4} = 1$, look scary. But one move clears every frac­tion at once: mul­ti­ply both sides by the LCM of the bot­toms. In this les­son you will learn that move, and a sec­ond short cut called cross mul­ti­pli­ca­tion for equa­tions with one frac­tion on each side.

Clear­ing the frac­tions#

The LCM (low­est com­mon mul­ti­ple) of some num­bers is the small­est num­ber that all of them divide into exactly. The LCM of $2$ and $3$ is $6$. The LCM of $4$ and $6$ is $12$.

If you mul­ti­ply every term on both sides by the LCM of the bot­toms, every frac­tion becomes a whole num­ber. The two sides stay equal, because you did the same thing to both.

Exam­ple 1#

Solve $\displaystyle \frac{x}{2} + \frac{x}{3} = 10$.

The LCM of $2$ and $3$ is $6$. Mul­ti­ply every term by $6$.

$$\displaystyle \begin{aligned} \frac{x}{2} + \frac{x}{3} &= 10 \\ \Rightarrow 6 \times \frac{x}{2} + 6 \times \frac{x}{3} &= 6 \times 10 \\ \Rightarrow 3x + 2x &= 60 \\ \Rightarrow 5x &= 60 \\ \Rightarrow x &= 12 \end{aligned}$$

Check: $\displaystyle \frac{12}{2} + \frac{12}{3} = 6 + 4 = 10$. It works.

Exam­ple 2#

Solve $\displaystyle \frac{x}{4} - \frac{x}{6} = \frac{1}{3}$.

The LCM of $4$, $6$ and $3$ is $12$.

$$\displaystyle \begin{aligned} \frac{x}{4} - \frac{x}{6} &= \frac{1}{3} \\ \Rightarrow 3x - 2x &= 4 \\ \Rightarrow x &= 4 \end{aligned}$$

Each term was mul­ti­plied by $12$: $\displaystyle 12 \times \frac{x}{4} = 3x$, $\displaystyle 12 \times \frac{x}{6} = 2x$ and $\displaystyle 12 \times \frac{1}{3} = 4$.

Exam­ple 3#

Solve $\displaystyle \frac{x + 1}{3} - \frac{x - 1}{4} = 1$.

The LCM of $3$ and $4$ is $12$. Keep each top in a bracket when you mul­ti­ply. This pro­tects the signs.

$$\displaystyle \begin{aligned} \frac{x + 1}{3} - \frac{x - 1}{4} &= 1 \\ \Rightarrow 4(x + 1) - 3(x - 1) &= 12 \\ \Rightarrow 4x + 4 - 3x + 3 &= 12 \\ \Rightarrow x + 7 &= 12 \\ \Rightarrow x &= 5 \end{aligned}$$

Notice $-3(x - 1) = -3x + 3$. The minus sign changes both terms.

Exam­ple 4: dec­i­mals#

Solve $0.25x + 0.5 = 0.75$.

Dec­i­mals can be cleared the same way. The most dig­its after any point is two, so mul­ti­ply every term by $100$.

$$\begin{aligned} 0.25x + 0.5 &= 0.75 \\ \Rightarrow 25x + 50 &= 75 \\ \Rightarrow 25x &= 25 \\ \Rightarrow x &= 1 \end{aligned}$$

Remem­ber. Mul­ti­ply every term on both sides by the LCM of the bot­toms. Put each top in a bracket first.

Cross mul­ti­pli­ca­tion#

When an equa­tion is one frac­tion equal to another frac­tion, there is a quicker way.

$$\displaystyle \frac{a}{b} = \frac{c}{d} \quad \Rightarrow \quad a \times d = b \times c$$

Mul­ti­ply the top of each side by the bot­tom of the other side. This is called cross mul­ti­pli­ca­tion. It is really mul­ti­ply­ing both sides by $b \times d$.

For exam­ple, $\displaystyle \frac{2}{3} = \frac{4}{6}$, and indeed $2 \times 6 = 3 \times 4 = 12$.

Exam­ple 5#

Solve $\displaystyle \frac{x + 1}{3} = \frac{x - 1}{2}$.

$$\displaystyle \begin{aligned} \frac{x + 1}{3} &= \frac{x - 1}{2} \\ \Rightarrow 2(x + 1) &= 3(x - 1) \\ \Rightarrow 2x + 2 &= 3x - 3 \\ \Rightarrow 2x - 3x &= -3 - 2 \\ \Rightarrow -x &= -5 \\ \Rightarrow x &= 5 \end{aligned}$$

Check: $\displaystyle \frac{5 + 1}{3} = 2$ and $\displaystyle \frac{5 - 1}{2} = 2$. It works.

Exam­ple 6: x on the bot­tom#

Solve $\displaystyle \frac{2x - 1}{x + 2} = \frac{3}{4}$.

$$\displaystyle \begin{aligned} \frac{2x - 1}{x + 2} &= \frac{3}{4} \\ \Rightarrow 4(2x - 1) &= 3(x + 2) \\ \Rightarrow 8x - 4 &= 3x + 6 \\ \Rightarrow 8x - 3x &= 6 + 4 \\ \Rightarrow 5x &= 10 \\ \Rightarrow x &= 2 \end{aligned}$$

Check: $\displaystyle \frac{2 \times 2 - 1}{2 + 2} = \frac{3}{4}$. It works.

When $x$ is on the bot­tom, always check that your answer does not make the bot­tom $0$. Here the bot­tom is $2 + 2 = 4$, so all is well.

Com­mon mis­takes#

• Mul­ti­ply­ing only some of the terms by the LCM. Every term on both sides must be mul­ti­plied.
• For­get­ting the bracket around a top, so that $-3(x - 1)$ becomes $-3x - 1$. It should be $-3x + 3$.
• Cross mul­ti­ply­ing when there is more than one term on a side, such as $\displaystyle \frac{x}{2} + 1 = \frac{x}{3}$. Clear the frac­tions with the LCM instead.
• Not check­ing an answer that sits under a frac­tion line.

Prac­tice#

1. Solve $\displaystyle \frac{x}{2} + \frac{x}{4} = 9$.
2. Solve $\displaystyle \frac{x}{3} - \frac{x}{5} = 2$.
3. Solve $\displaystyle \frac{x}{2} - \frac{1}{3} = \frac{x}{3} + \frac{1}{2}$.
4. Solve $\displaystyle \frac{x}{5} + \frac{x + 1}{2} = 4$.
5. Solve $\displaystyle \frac{2x}{3} - \frac{x - 1}{2} = 1$.
6. Solve $\displaystyle \frac{x + 3}{5} - \frac{x - 1}{3} = 0$.
7. Solve $0.2x + 0.15 = 0.05x + 0.6$.
8. Solve $\displaystyle \frac{x - 2}{3} = \frac{x + 1}{4}$.
9. Solve $\displaystyle \frac{3x + 1}{2x - 1} = 2$.
10. Solve $\displaystyle \frac{x + 4}{x - 2} = \frac{5}{2}$.

Answers#