What a linear equation is

An equa­tion is a sen­tence in math­e­mat­ics that says two things are equal. Many every­day puz­zles turn into equa­tions: Ravi thinks of a num­ber, triples it and adds 5, and gets 20. What was the num­ber? In this les­son you will learn what an equa­tion is, what makes an equa­tion lin­ear, what it means to solve one, and how to check an answer. Every later les­son in this chap­ter rests on these ideas.

Expres­sions and equa­tions#

You already know alge­braic expres­sions such as $3x + 5$ or $7 - 2y$. An expres­sion is a col­lec­tion of terms joined by $+$ and $-$ signs. It has no equals sign, so it makes no claim. It sim­ply stands for a num­ber that changes as $x$ changes.

An equa­tion joins two expres­sions with an equals sign:

$$3x + 5 = 20$$

The part to the left of the equals sign is called the left-hand side (LHS), and the part to the right is the right-hand side (RHS). Here the LHS is $3x + 5$ and the RHS is $20$. The equa­tion makes a claim: the two sides have the same value. That claim is true for some val­ues of $x$ and false for oth­ers.

The let­ter $x$ is called the vari­able or the unknown. It stands for a num­ber we have not yet found. Any let­ter will do: $y$, $m$, $p$ and $t$ are all com­mon.

What makes an equa­tion lin­ear#

An equa­tion is lin­ear in one vari­able when two things are true:

1. it has only one vari­able, and
2. the high­est power of that vari­able is $1$.

So $4y - 9 = 11$ is lin­ear: one vari­able, $y$, and it appears only as $y$, never as $y^{2}$ or $y^{3}$. The word "lin­ear" comes from "line": in a later chap­ter you will see that equa­tions of this kind draw straight lines on a graph.

These are lin­ear equa­tions in one vari­able:

$$\displaystyle 2x + 7 = 15 \qquad \frac{m}{3} - 1 = 4 \qquad 5(p - 2) = 3p + 6$$

These are not:

• $x^{2} + 1 = 10$, because $x$ is squared, so the high­est power is $2$.
• $3x + 2y = 7$, because it has two vari­ables. Equa­tions like this are the sub­ject of the next chap­ter.
• $7 + 2m$, because there is no equals sign. It is an expres­sion, not an equa­tion.

A num­ber on its own under a frac­tion line is fine: $\displaystyle \frac{m}{3}$ means $\displaystyle \frac{1}{3} \times m$, and $m$ still has power $1$.

Exam­ple 1#

Which of these are lin­ear equa­tions in one vari­able?

(a) $6a - 4 = 20$ (b) $t^{2} - 5t = 6$ (c) $x + y = 12$ (d) $\displaystyle \frac{k}{2} + 3 = 9$ (e) $8n + 1$

(a) Yes. One vari­able, power $1$.

(b) No. The term $t^{2}$ has power $2$.

(c) No. There are two vari­ables, $x$ and $y$.

(d) Yes. $\displaystyle \frac{k}{2}$ is half of $k$, still power $1$.

(e) No. There is no equals sign, so this is an expres­sion.

The solu­tion of an equa­tion#

The solu­tion of an equa­tion is the value of the vari­able that makes the LHS equal to the RHS. It is also called the root of the equa­tion. A lin­ear equa­tion in one vari­able has exactly one solu­tion.

To solve an equa­tion means to find that value. To check a solu­tion, put the value in place of the vari­able on both sides and see whether the two sides come out equal.

Exam­ple 2#

Is $x = 4$ the solu­tion of $3x - 5 = 7$?

Put $x = 4$ into the LHS:

$$3 \times 4 - 5 = 12 - 5 = 7$$

The RHS is $7$. Both sides are $7$, so $x = 4$ is the solu­tion.

Exam­ple 3#

Is $y = 2$ the solu­tion of $5y + 1 = 3y + 7$? If not, test $y = 3$.

For $y = 2$: the LHS is $5 \times 2 + 1 = 11$ and the RHS is $3 \times 2 + 7 = 13$. The sides are not equal, so $y = 2$ is not the solu­tion.

For $y = 3$: the LHS is $5 \times 3 + 1 = 16$ and the RHS is $3 \times 3 + 7 = 16$. The sides are equal, so $y = 3$ is the solu­tion.

Notice that the vari­able may appear on both sides. You must sub­sti­tute on both sides when you check.

Find­ing a solu­tion by trial#

For small num­bers you can find a solu­tion by try­ing val­ues in turn. Take $2x + 3 = 13$ and work out the LHS for $x = 1, 2, 3, \ldots$

The LHS reaches $13$ when $x = 5$, so the solu­tion is $x = 5$.

Trial works, but it is slow, and it fails when the answer is a frac­tion such as $\displaystyle \frac{9}{4}$ or a neg­a­tive num­ber such as $-6$. The next les­son gives a method that always works.

Writ­ing an equa­tion from words#

Many prob­lems come to you in words. The first job is to turn the words into an equa­tion. Pick a let­ter for the unknown num­ber, then trans­late the sen­tence piece by piece.

Exam­ple 4#

Write an equa­tion for each sen­tence.

(a) Seven more than three times a num­ber is $31$.

(b) When $4$ is taken away from half a num­ber, the result is $6$.

(c) Sita's age now is $x$ years. In $5$ years she will be $18$.

(a) Three times the num­ber is $3x$, seven more is $3x + 7$, so $3x + 7 = 31$.

(b) Half the num­ber is $\displaystyle \frac{x}{2}$, take away $4$, so $\displaystyle \frac{x}{2} - 4 = 6$.

(c) In $5$ years she will be $x + 5$ years old, so $x + 5 = 18$.

Be care­ful with "less than". "4 less than a num­ber" is $x - 4$, not $4 - x$: you start with the num­ber and take $4$ away from it.

An equa­tion is a bal­ance#

Think of an equa­tion as a pair of scales that is level. The LHS sits in one pan and the RHS in the other. What­ever you do to one pan, you must do to the other, or the scales tip. Add $5$ to one side and you must add $5$ to the other. Halve one side and you must halve the other. This one idea is the whole secret of solv­ing equa­tions, and the next les­son puts it to work.

Prac­tice#

1. Which of these are lin­ear equa­tions in one vari­able? (a) $9z - 2 = 25$ (b) $x^{2} = 49$ (c) $4p + q = 10$ (d) $\displaystyle \frac{y}{5} = 3$
2. Write the LHS and the RHS of $6m + 2 = 4m - 8$.
3. Is $x = 6$ the solu­tion of $2x + 9 = 21$?
4. Is $a = 5$ the solu­tion of $4a - 3 = 2a + 9$?
5. Show that $y = 7$ is the solu­tion of $3(y - 2) = 2y + 1$.
6. Use a table of val­ues to find the solu­tion of $4x - 1 = 19$.
7. Write an equa­tion: "Five times a num­ber, decreased by $8$, is $27$."
8. Write an equa­tion: "A third of a num­ber added to $10$ gives $14$."
9. Ravi has $x$ mar­bles. After his friend gives him $12$ more, he has $40$. Write an equa­tion and find $x$ by trial or by think­ing.
10. Which of $m = 2$, $m = 3$ or $m = 4$ is the solu­tion of $7m - 10 = 3m + 6$?

Answers#