Two-step equations

Many equa­tions need two moves, not one. In $\displaystyle 2x + \frac{1}{3} = \frac{5}{3}$, the $x$ is first mul­ti­plied by $2$ and then has $\displaystyle \frac{1}{3}$ added. To free the $x$, you undo those jobs in the reverse order. In this les­son you will solve such two-step equa­tions with frac­tions and dec­i­mals, and also equa­tions whose answers are roots.

The order of the moves#

Think of putting on socks and then shoes. To take them off, you remove the shoes first, then the socks. An equa­tion is the same.

1. First move the num­ber that is added or taken away.
2. Then move the num­ber that mul­ti­plies or divides $x$.

Remem­ber. Undo adding and tak­ing away first. Undo mul­ti­ply­ing and divid­ing last.

Exam­ple 1#

Solve $\displaystyle 2x + \frac{1}{3} = \frac{5}{3}$.

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

Check: $\displaystyle 2 \times \frac{2}{3} + \frac{1}{3} = \frac{4}{3} + \frac{1}{3} = \frac{5}{3}$. It works.

Exam­ple 2#

Solve $\displaystyle 3x - \frac{1}{2} = \frac{7}{4}$.

$$\displaystyle \begin{aligned} 3x - \frac{1}{2} &= \frac{7}{4} \\ \Rightarrow 3x &= \frac{7}{4} + \frac{1}{2} \\ \Rightarrow 3x &= \frac{7}{4} + \frac{2}{4} = \frac{9}{4} \\ \Rightarrow x &= \frac{9}{4} \div 3 \\ \Rightarrow x &= \frac{3}{4} \end{aligned}$$

Exam­ple 3#

Solve $0.4x - 1.2 = 2$.

$$\displaystyle \begin{aligned} 0.4x - 1.2 &= 2 \\ \Rightarrow 0.4x &= 2 + 1.2 \\ \Rightarrow 0.4x &= 3.2 \\ \Rightarrow x &= \frac{3.2}{0.4} = \frac{32}{4} \\ \Rightarrow x &= 8 \end{aligned}$$

Check: $0.4 \times 8 - 1.2 = 3.2 - 1.2 = 2$. It works.

Exam­ple 4#

Solve $\displaystyle \frac{x}{5} + \frac{1}{2} = \frac{3}{10}$.

$$\displaystyle \begin{aligned} \frac{x}{5} + \frac{1}{2} &= \frac{3}{10} \\ \Rightarrow \frac{x}{5} &= \frac{3}{10} - \frac{5}{10} \\ \Rightarrow \frac{x}{5} &= -\frac{2}{10} = -\frac{1}{5} \\ \Rightarrow x &= -\frac{1}{5} \times 5 \\ \Rightarrow x &= -1 \end{aligned}$$

Exam­ple 5#

Solve $\displaystyle \frac{2x}{3} + 1.5 = 4.5$.

Here $x$ is mul­ti­plied by $2$ and divided by $3$. Undo the adding first, then the divid­ing, then the mul­ti­ply­ing.

$$\displaystyle \begin{aligned} \frac{2x}{3} + 1.5 &= 4.5 \\ \Rightarrow \frac{2x}{3} &= 4.5 - 1.5 = 3 \\ \Rightarrow 2x &= 3 \times 3 = 9 \\ \Rightarrow x &= \frac{9}{2} = 4.5 \end{aligned}$$

Exam­ple 6: x with a minus sign#

Solve $\displaystyle 5 - \frac{3x}{4} = \frac{1}{2}$.

$$\displaystyle \begin{aligned} 5 - \frac{3x}{4} &= \frac{1}{2} \\ \Rightarrow -\frac{3x}{4} &= \frac{1}{2} - 5 = -\frac{9}{2} \\ \Rightarrow \frac{3x}{4} &= \frac{9}{2} \\ \Rightarrow 3x &= \frac{9}{2} \times 4 = 18 \\ \Rightarrow x &= 6 \end{aligned}$$

When both sides have a minus sign, you may remove it from both. That is the same as mul­ti­ply­ing both sides by $-1$.

Equa­tions with roots#

The real num­bers include roots like $\sqrt{3}$. The rules do not change. Like roots are added and taken away just as in the chap­ter on real num­bers.

Exam­ple 7#

Solve $x + \sqrt{3} = 4\sqrt{3}$.

$$\begin{aligned} x + \sqrt{3} &= 4\sqrt{3} \\ \Rightarrow x &= 4\sqrt{3} - \sqrt{3} \\ \Rightarrow x &= 3\sqrt{3} \end{aligned}$$

Exam­ple 8#

Solve $2x - \sqrt{5} = 3\sqrt{5}$.

$$\displaystyle \begin{aligned} 2x - \sqrt{5} &= 3\sqrt{5} \\ \Rightarrow 2x &= 3\sqrt{5} + \sqrt{5} = 4\sqrt{5} \\ \Rightarrow x &= \frac{4\sqrt{5}}{2} = 2\sqrt{5} \end{aligned}$$

Exam­ple 9#

Solve $\sqrt{2}\,x = 6$.

$$\displaystyle \begin{aligned} \sqrt{2}\,x &= 6 \\ \Rightarrow x &= \frac{6}{\sqrt{2}} \\ \Rightarrow x &= \frac{6 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{6\sqrt{2}}{2} \\ \Rightarrow x &= 3\sqrt{2} \end{aligned}$$

We moved the root off the bot­tom by mul­ti­ply­ing the top and the bot­tom by $\sqrt{2}$.

Check: $\sqrt{2} \times 3\sqrt{2} = 3 \times 2 = 6$. It works.

Com­mon mis­takes#

• Divid­ing first. In $\displaystyle 2x + \frac{1}{3} = \frac{5}{3}$, move the $\displaystyle \frac{1}{3}$ before divid­ing by $2$.
• Divid­ing only part of a side. When you divide by $2$, the whole of the other side is divided.
• Los­ing a minus sign, as in $\displaystyle \frac{1}{2} - 5$, which is $\displaystyle -\frac{9}{2}$.
• Adding a plain num­ber to a root. In $2x - \sqrt{5} = 3\sqrt{5}$, the $\sqrt{5}$ joins $3\sqrt{5}$ to make $4\sqrt{5}$.

Prac­tice#

1. Solve $\displaystyle 2x + \frac{1}{4} = \frac{9}{4}$.
2. Solve $\displaystyle 3x - \frac{2}{5} = 1$.
3. Solve $5x + 0.5 = 3$.
4. Solve $0.3x - 0.9 = 1.5$.
5. Solve $\displaystyle \frac{x}{4} - \frac{1}{2} = \frac{3}{4}$.
6. Solve $\displaystyle \frac{3x}{5} + 2 = -4$.
7. Solve $\displaystyle 7 - \frac{2x}{3} = 3$.
8. Solve $x + 2\sqrt{7} = 5\sqrt{7}$.
9. Solve $3x - \sqrt{2} = 5\sqrt{2}$.
10. Solve $\sqrt{3}\,x = 12$.

Answers#