Putting numbers back in

For­mu­las are every­where: the perime­ter of a rec­tan­gle, the cost of sev­eral note­books, the dis­tance trav­elled in a given time. Each for­mula is writ­ten with let­ters so that it works for every case. The moment you have an actual rec­tan­gle or an actual jour­ney, you put its num­bers in place of the let­ters and work out the answer. That step is called sub­sti­tu­tion, and it is one of the most used skills in all of alge­bra.

A let­ter in alge­bra stands for a num­ber, and sub­sti­tu­tion sim­ply means putting that num­ber back in.

The chap­ter says it in older words: replac­ing numer­i­cal val­ues in the place of lit­eral num­bers. (Lit­eral num­bers is just an old name for the let­ters.)

A let­ter used this way is a vari­able, which you met in the Early Years course "Intro­duc­tion to Alge­bra".

Remem­ber. To sub­sti­tute is to swap each let­ter for the num­ber it stands for.

Why this mat­ters#

An expres­sion is let­ters and num­bers joined by signs like $+$, $-$ and $\times$, with no equals sign in it.

An expres­sion like $5x$ is not an answer; it is a recipe for find­ing one. It says: take $x$ and mul­ti­ply it by $5$.

You can't fol­low that recipe while $x$ is still a let­ter. Give $x$ a value, though, and the rule can run, turn­ing the expres­sion into one plain num­ber.

That is the beauty of it. Write $5x$ once and it works for $x = 3$, for $x = 50$, and for any value you choose.

One let­ter, one step#

Find the value of $5x$ when $x = 3$

$$\begin{aligned}&5x \\ &= 5 \times 3 \\ &= 15\end{aligned}$$

You wrote $3$ where $x$ used to be, and then you did the mul­ti­ply­ing.

Notice that the mul­ti­pli­ca­tion sign is hid­ing in $5x$. Writ­ing $5 \times 3$ brings it back into view. Miss it and you might write $53$ by mis­take, which hap­pens more often than you'd think.

One let­ter, two steps#

Find the value of $3a + 2$ when $a = 4$

$$\begin{aligned}&3a + 2 \\ &= 3 \times 4 + 2 \\ &= 12 + 2 \\ &= 14\end{aligned}$$

Here $a$ became $4$, then $3 \times 4$ gave $12$, and finally $12 + 2$ gave $14$.

Mul­ti­ply­ing is always done before adding. That order is fixed, so that every­one read­ing your work­ing gets the same answer.

Sup­pose you added first: $4 + 2$ is $6$, and $3 \times 6$ is $18$. That answer is wrong.

The trap: $x^{2}$ is not $2x$#

This one catches a lot of stu­dents, so read slowly. $x^{2}$ means $x \times x$. It does not mean $2 \times x$. The small $2$ counts how many $x$ val­ues are mul­ti­plied together.

$x^{2}$ is called a power, and it's the small raised num­ber that makes it one.

Find the value of $x^{2}$ when $x = 5$

$$\begin{aligned}&x^{2} \\ &= 5 \times 5 \\ &= 25\end{aligned}$$

Find the value of $2x$ when $x = 5$

$$\begin{aligned}&2x \\ &= 2 \times 5 \\ &= 10\end{aligned}$$

Same let­ter, same value, and two very dif­fer­ent answers.

The chart shows why the two must never be con­fused. At $x = 2$ they hap­pen to agree, because $2 \times 2$ and $2 + 2$ are both $4$. For every larger value, $x^{2}$ is big­ger, and the gap keeps grow­ing.

Remem­ber. A small raised num­ber tells you how many of the let­ter are mul­ti­plied together. A num­ber in front tells you what to mul­ti­ply the let­ter by.

When a power sits inside a longer expres­sion, you work it out first.

Find the value of $x^{2} + 1$ when $x = 3$

$$\begin{aligned}&x^{2} + 1 \\ &= 3 \times 3 + 1 \\ &= 9 + 1 \\ &= 10\end{aligned}$$

The power came first: $3 \times 3$ gave $9$, and only then did the $1$ go on.

Find the value of $2x^{2}$ when $x = 3$

$$\begin{aligned}&2x^{2} \\ &= 2 \times 3 \times 3 \\ &= 2 \times 9 \\ &= 18\end{aligned}$$

Only the $x$ is squared, not the $2$. If the whole of $2x$ were meant to be squared, it would be writ­ten $(2x)^{2}$, and that is $6 \times 6 = 36$.

More than one let­ter#

Find the value of $2a + 3b$ when $a = 1$ and $b = 2$

$$\begin{aligned}&2a + 3b \\ &= 2 \times 1 + 3 \times 2 \\ &= 2 + 6 \\ &= 8\end{aligned}$$

Both mul­ti­pli­ca­tions came first, and then the two results were added.

An expres­sion can hold two let­ters or more, each with its own value. Swap them all in, then sim­plify, which means work­ing it out until one num­ber is left.

Find the value of $4p - q$ when $p = 3$ and $q = 5$

$$\begin{aligned}&4p - q \\ &= 4 \times 3 - 5 \\ &= 12 - 5 \\ &= 7\end{aligned}$$

A for­mula: the perime­ter of a rec­tan­gle

The perime­ter of a rec­tan­gle is $P = 2(l + b)$, where $l$ is the length and $b$ is the breadth. Find $P$ when $l = 7$ cm and $b = 4$ cm.

$$\begin{aligned}P &= 2(l + b) \\ &= 2 \times (7 + 4) \\ &= 2 \times 11 \\ &= 22\end{aligned}$$

So the perime­ter is 22 cm. Notice the bracket was worked out first, then the mul­ti­ply­ing.

When the value is neg­a­tive#

When the value is neg­a­tive, the minus sign is very easy to lose. Brack­ets keep it where it belongs.

Find the value of $5 - x$ when $x = -2$

$$\begin{aligned}&5 - x \\ &= 5 - (-2) \\ &= 5 + 2 \\ &= 7\end{aligned}$$

Write $(-2)$ in the place of $x$, brack­ets and all. The expres­sion takes $x$ away from $5$.

Tak­ing away means mov­ing down the num­ber line, and adding means mov­ing up. Now $-2$ sits two steps below zero, and tak­ing away some­thing below zero pushes you back up. That is why $5 - (-2)$ is $7$ and not $3$.

With­out brack­ets you'd be writ­ing two minus signs side by side, which is easy to mis­read. The brack­ets hold the minus sign firmly onto its own num­ber.

Find the value of $3(a + b)$ when $a = 2$ and $b = -5$

$$\begin{aligned}&3(a + b) \\ &= 3 \times (2 + (-5)) \\ &= 3 \times (-3) \\ &= -9\end{aligned}$$

Find the value of $a^{2} - b$ when $a = -3$ and $b = 4$

$$\begin{aligned}&a^{2} - b \\ &= (-3) \times (-3) - 4 \\ &= 9 - 4 \\ &= 5\end{aligned}$$

A neg­a­tive num­ber times a neg­a­tive num­ber is pos­i­tive, so $(-3)^{2} = 9$. With­out the brack­ets, $-3^{2}$ would be read as $-(3 \times 3) = -9$, which is a dif­fer­ent num­ber.

Remem­ber. Always put brack­ets round a neg­a­tive value when you sub­sti­tute it.

The safe order#

Sub­sti­tute first, sim­plify sec­ond. Try­ing to do both at once is exactly how the hid­den mul­ti­pli­ca­tion sign gets lost, and $2 \times 3$ turns into $23$.

1. Copy the expres­sion out.
2. Replace each let­ter with its value. Use brack­ets round any neg­a­tive value.
3. Put back the mul­ti­pli­ca­tion signs that were hid­den.
4. Sim­plify: brack­ets first, then pow­ers, then mul­ti­ply­ing, then adding and tak­ing away.
5. Check that no let­ter is left. A let­ter left behind means you missed one.

Step five is well worth the few sec­onds it takes, because a let­ter is easy to miss in a long expres­sion.

Prac­tice#

1. Find the value of $4x$ when $x = 2$.
2. Find the value of $2a + 5$ when $a = 3$.
3. Find the value of $x^{2}$ when $x = 4$.
4. Find the value of $3m + 2n$ when $m = 2$ and $n = 5$.
5. Find the value of $10 - y$ when $y = -3$.
6. Find the value of $x^{2} + 5$ when $x = 6$.

Com­mon mis­takes#

• Writ­ing the dig­its side by side. $5x$ with $x = 3$ is $5 \times 3 = 15$, not $53$.
• Read­ing $x^{2}$ as $2x$. With $x = 5$, $x^{2} = 25$ but $2x = 10$.
• Adding before mul­ti­ply­ing. $3a + 2$ with $a = 4$ is $12 + 2 = 14$, not $3 \times 6 = 18$.
• Drop­ping the brack­ets round a neg­a­tive value. $10 - y$ with $y = -3$ is $10 - (-3) = 13$, not $7$.
• Squar­ing the num­ber in front. $2x^{2}$ with $x = 3$ is $18$, not $36$.
• Leav­ing a let­ter behind. In $3m + 2n$ both let­ters need their val­ues before you sim­plify.

Key terms#

Vari­able

A let­ter that stands for a num­ber, such as $x$.

Lit­eral num­ber

An older name for a let­ter used in place of a num­ber.

Expres­sion

Let­ters and num­bers joined by signs such as $+$ and $\times$, with no equals sign.

Sub­sti­tute

Swap each let­ter for the num­ber it stands for.

Sim­plify

Work the expres­sion out until one num­ber is left.

Power

A num­ber or let­ter with a small raised num­ber, such as $x^{2}$, mean­ing $x \times x$.

For­mula

A rule writ­ten with let­ters, such as $P = 2(l + b)$, that works for every value.

Answers#

Show answers

1. $4x = 4 \times 2 = 8$
2. $2a + 5 = 2 \times 3 + 5 = 6 + 5 = 11$
3. $x^{2} = 4 \times 4 = 16$
4. $3m + 2n = 3 \times 2 + 2 \times 5 = 6 + 10 = 16$
5. $10 - y = 10 - (-3) = 10 + 3 = 13$
6. $x^{2} + 5 = 6 \times 6 + 5 = 36 + 5 = 41$