Solving equations by balancing and transposing

Try­ing val­ues one by one is slow and can­not find answers like $\displaystyle \frac{9}{4}$ or $-4$. This les­son gives you a method that always works. It starts with the bal­ance idea from the last les­son and ends with a quick short­cut called trans­pos­ing, which is how most peo­ple solve equa­tions every day.

The four bal­ance rules#

An equa­tion is like a level pair of scales. It stays level, and stays true, if you do the same thing to both sides. There are four things you may do:

1. Add the same num­ber to both sides.
2. Sub­tract the same num­ber from both sides.
3. Mul­ti­ply both sides by the same num­ber.
4. Divide both sides by the same num­ber, as long as that num­ber is not zero.

The aim is always the same: get the vari­able alone on one side. What­ever is attached to the vari­able, do the oppo­site to remove it. Addi­tion is undone by sub­trac­tion, and mul­ti­pli­ca­tion is undone by divi­sion.

Exam­ple 1#

Solve $x + 9 = 23$.

The $9$ is added to $x$, so sub­tract $9$ from both sides.

$$\begin{aligned} x + 9 - 9 &= 23 - 9 \\ x &= 14 \end{aligned}$$

Check: $14 + 9 = 23$. Cor­rect.

Exam­ple 2#

Solve $x - 6 = 11$.

The $6$ is taken away from $x$, so add $6$ to both sides.

$$\begin{aligned} x - 6 + 6 &= 11 + 6 \\ x &= 17 \end{aligned}$$

Exam­ple 3#

Solve $4x = 36$.

$4x$ means $4 \times x$, so divide both sides by $4$.

$$\displaystyle \begin{aligned} \frac{4x}{4} &= \frac{36}{4} \\ x &= 9 \end{aligned}$$

Exam­ple 4#

Solve $\displaystyle \frac{x}{5} = 7$.

$x$ is divided by $5$, so mul­ti­ply both sides by $5$.

$$\displaystyle \begin{aligned} \frac{x}{5} \times 5 &= 7 \times 5 \\ x &= 35 \end{aligned}$$

Two steps#

Most equa­tions need more than one step. In $3x + 4 = 25$, the $x$ has been mul­ti­plied by $3$ and then had $4$ added. To undo this, work back­wards: remove the $4$ first, then the $3$. It is like tak­ing off your shoes before your socks: the last thing put on is the first thing taken off.

Exam­ple 5#

Solve $3x + 4 = 25$.

$$\displaystyle \begin{aligned} 3x + 4 - 4 &= 25 - 4 \\ 3x &= 21 \\ \frac{3x}{3} &= \frac{21}{3} \\ x &= 7 \end{aligned}$$

Check: the LHS is $3 \times 7 + 4 = 25$, which is the RHS.

Trans­pos­ing: the short­cut#

Look again at the first step of Exam­ple 5. We sub­tracted $4$ from both sides, and the result was that the $+4$ dis­ap­peared from the left and a $-4$ appeared on the right:

$$3x + 4 = 25 \quad \Longrightarrow \quad 3x = 25 - 4$$

This always hap­pens, so we can skip the mid­dle line. Mov­ing a term from one side of an equa­tion to the other is called trans­pos­ing, and it fol­lows sim­ple rules:

Remem­ber. When a term crosses the equals sign, its oper­a­tion changes to the oppo­site. Trans­pos­ing is not a new rule; it is the bal­ance method writ­ten in fewer lines.

Exam­ple 6#

Solve $5x - 8 = 27$.

Trans­pose $-8$ to the right, where it becomes $+8$. Then the $5$ that mul­ti­plies $x$ crosses as a divi­sion.

$$\displaystyle \begin{aligned} 5x &= 27 + 8 \\ 5x &= 35 \\ x &= \frac{35}{5} \\ x &= 7 \end{aligned}$$

Check: $5 \times 7 - 8 = 35 - 8 = 27$.

Exam­ple 7: a neg­a­tive answer#

Solve $2x + 15 = 7$.

$$\displaystyle \begin{aligned} 2x &= 7 - 15 \\ 2x &= -8 \\ x &= \frac{-8}{2} \\ x &= -4 \end{aligned}$$

Check: $2 \times (-4) + 15 = -8 + 15 = 7$. A neg­a­tive solu­tion is per­fectly nor­mal.

Exam­ple 8: a frac­tion answer#

Solve $4x - 3 = 6$.

$$\displaystyle \begin{aligned} 4x &= 6 + 3 \\ 4x &= 9 \\ x &= \frac{9}{4} \end{aligned}$$

Leave the answer as $\displaystyle \frac{9}{4}$, or write it as $\displaystyle 2\frac{1}{4}$ or $2.25$. Trial would never have found this one.

Exam­ple 9: a neg­a­tive coef­fi­cient#

Solve $10 - 3x = 1$.

Here the $x$ term is neg­a­tive. One clean way is to trans­pose the $x$ term to the side where it will be pos­i­tive, and the num­ber to the other side.

$$\begin{aligned} 10 - 1 &= 3x \\ 9 &= 3x \\ x &= 3 \end{aligned}$$

Check: $10 - 3 \times 3 = 10 - 9 = 1$. You could also write $-3x = 1 - 10 = -9$ and divide by $-3$. Both routes give $x = 3$, because divid­ing a neg­a­tive by a neg­a­tive gives a pos­i­tive.

Exam­ple 10: the vari­able divided#

Solve $\displaystyle \frac{2y}{3} + 1 = 9$.

$$\displaystyle \begin{aligned} \frac{2y}{3} &= 9 - 1 \\ \frac{2y}{3} &= 8 \\ 2y &= 8 \times 3 \\ 2y &= 24 \\ y &= 12 \end{aligned}$$

Check: $\displaystyle \frac{2 \times 12}{3} + 1 = 8 + 1 = 9$.

A habit worth keep­ing#

Always check your answer by putting it back into the orig­i­nal equa­tion, not into one of your work­ing lines. If you made a slip in the work­ing, the work­ing lines will agree with the wrong answer, but the orig­i­nal equa­tion will not. A check takes ten sec­onds and catches most mis­takes.

Prac­tice#

1. Solve $x + 13 = 30$.
2. Solve $y - 8 = 5$.
3. Solve $7m = 63$.
4. Solve $\displaystyle \frac{p}{4} = 9$.
5. Solve $6x + 5 = 41$.
6. Solve $3a - 11 = 13$.
7. Solve $5x + 22 = 7$.
8. Solve $8n - 5 = 7$.
9. Solve $15 - 4t = 3$.
10. Solve $\displaystyle \frac{3k}{5} - 2 = 7$.

Answers#