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A1.1: Algebraic operations

Algebraic operations are critical to many fields of study. In science, they are used to calculate forces to accelerate a car or determine the rate of a chemical reaction. In engineering, they can be used to design safe and efficient systems. They also underpin algorithms and data structures, which are fundamental to software development and solving complex computational problems. This resource introduces you to the basic skills for adding (\(+\)), subtracting (\(-\)), multiplying (\(\times\)) and dividing (\(÷\)) algebraic expressions.

A page in Arabic of one of the earliest books on Algebra
A page from Al-Khwarizmis al-Kitab al-muhtasar fi hisab al-gabr wa-l-muqabala, an early book on algebra. In the public domain, via Wikipedia.

Algebraic terms

Before we dive into algebraic operations, let's recap the components of an algebraic term. An algebraic term consists of a pro-numeral and, often, a coefficient.

  • A pro-numeral is a letter that represents a number or range of numbers.
  • A coefficient is the number written before the pro-numeral. It represents how many lots of the pro-numeral are present.

For example, \(3x\) is an algebraic term with \(3\) as the coefficient and \(x\) as the pro-numeral. What this really means is \(3\times x\) or \(3\) lots of \(x\).

Another example is \(5ab\). This is \(5\times a \times b\). It is the same as \(5ba\). The order of the letters and numbers is not important but we usually write the numbers first and use alphabetical order for the letters. \(5ba\) would be correct, but it is more common to write \(5ab\).

When there is no coefficient written before the pro-numeral, we assume that the coefficient is \(1\).

Video tutorial – algebraic operations

Watch this video to learn about algebraic operations.

M: Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on operations with algebra. First of all make sure you can tell the difference between like terms and unlike terms. Like terms contain exactly the same pro-numerals, in other words letters or variables, so look at the like terms down the left hand column and compare them with the unlike terms in the right hand column, in particular look at the last term, seven E F and eight F E, they are in fact like terms. Only like terms can be added or subtracted. For instance, in this example, seven E plus 10 E, the terms are alike, therefore they can be added to give you 17 E. But look at this one, be careful because here we have two terms, the X squared term and the X term so we have to add them separately, so three X squared minus X squared minus four X squared gives us minus two X squared, and two X, which is different to the X squared term must be left alone so the answer is minus two X squared plus two X.Similarly if you look at this example in red, there's a U V term which is the same as V U and that is different to the U term therefore that gives us minus two U V plus three U, we keep them separate, we can't add minus two U V to three U. And finally six R squared S minus two R S squared, they look similar but they are not the same, these are two unlike terms so we can't simplify this expression.Now before we move on to multiplying and dividing algebraic terms make sure you are familiar with the multiplication and dividing of positive and negative numbers on this slide. So study this carefully before moving onto the next slide.

But let's multiply some pro-numerals. Here's the first example, minus four times three B, multiply the numbers together, minus four and minus three gives you plus 12 and then just tag on the B afterwards so the answer is 12 B. Here's another one, multiply out the numbers first, three times minus 15 gives you minus 15, now multiply the pro-numeral, E and E squared is E cubed, so the answer is minus 15 E cubed.

The next example in red multiply out, as before, the numbers, minus two times minus four is plus eight, then the pro-numerals, there's a U squared term and there are Vs there, V times V is a V squared, so the answer is eight U squared V squared. Notice in these examples that because we're multiplying the actual terms can be different, so make sure you know the difference between adding and subtracting where the pro‑numerals must be exactly the same, multiplying and dividing the pro-numerals can be different.

Finally minus three P Q times minus two Q times P, multiply out the numbers, that gives you plus six, P times P is P squared, Q times Q is Q squared, so the answer is six P squared Q squared. Now in this slide we're dividing pro-numerals, so the first term minus 12 W Y Z divided by the second term three Y Z, express that as a fraction first and look for terms that will cancel top and bottom, in this case the Ys cancel top and bottom and the Zs cancel top and bottom, notice three will divide into 12 four times, so the answer is minus four W.

Here's another one slightly more complicated. Now in this slide we're dividing pro‑numerals, so the first term minus 12 W Y Z divided by the second term three Y Z, express that as a fraction first and look for terms that will cancel top and bottom, in this case the Ys cancel top and bottom and the Zs cancel top and bottom, notice three will divide into 12 four times so the answer is minus four W.

Here's another one slightly more complicated. Writing it as a fraction, minus two M squared N over six M N squared, and then in the third term I've expanded the M squared so it's M times N and then I've tacked the N on the end of that and I've done the same thing with the denominator with N squared N times N, so you can clearly see where the Ms and the Ns cancel, giving you an answer of minus M over three N. Notice minus on the top divided by plus on the bottom gives you an answer of minus.

And finally in blue there's a multiplication and a division so you again express that as a fraction, expand the A squared and then cancel top and bottom terms giving you an answer of two A. Now we'll move on to some slightly harder problems with algebra, but notice the rule that we use first of all, in other words if an expression has brackets in it that must be done first, secondly if there are indices you do that next, thirdly you do multiplying and dividing and lastly you do adding and subtraction.

So let's do some examples. Three S T minus three S times four T, notice we have a negative sign and a multiplication sign, so the times is worked out first, so three S times four T is 12 S T, notice now we have subtraction, three S T minus 12 S T, notice the terms are alike therefore we can subtract them, three minus 12 is minus nine S T.

Let's look at this one. We've got two terms in brackets and a division, so the brackets are worked out first, even though there's a plus inside the brackets must be worked out first, which gives us six X Y, then we divide that by two X, express that as a fraction, cancel any terms, Xs cancel here, twos go into six three times so the answer is three Y.

And finally look at this one. We have brackets, multiplication, addition and division. So the brackets is worked out first, that gives as minus 10 A squared B squared in the second line, then I write out the rest, plus two A cubed B cubed divided by A B, what do we do next, well we have division which becomes ... comes before addition, therefore we divide the two A cubed B cubed by A B, writing that as a fraction, simplifying by cancelling like terms and then in the final line we've just got two terms which we can add together. Notice again that the terms are exactly the same, A squared B squared, therefore we can add them, minus 10 plus two gives us minus eight A squared B squared.

Now try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short video.

Like and unlike terms

Like terms contain exactly the same pro-numerals, raised to the same power. For example, \(x\) and \(5x\) are like terms, but \(x\) and \(x^{2}\) are unlike terms. The following table gives some examples of like and unlike terms.

Like terms Unlike terms
\(3x,\,5x\) \(3x,\,5y\)
\(2a,\,3a\) \(3a,\,3\)
\(m^{2},\,7m^{2}\) \(m^{2},\,7m\)
\(2ab,\,3ab\) \(2a^{2}b,\,3ab^{2}\)
\(3xyz,\,5xyz\) \(3xyz,\,3xy\)
\(3ef,\,3fe\) \(3ef,\,3eg\)
\(6\alpha\beta,\,\beta\alpha\) \(6\alpha\beta\gamma,\,6\alpha\beta\)

Your turn – like and unlike terms

For each row in the table, identify the term in the second column that is a like term for the term in the first column.

Term Which is the like term?
\(3x\) \(3,\,2x,\,3x^{2},\,2xy,\,4x^{2}a\)
\(2ab\) \(2a,\,2b,\,3x^{2},\,6abc,\,12ab\)
\(2x^{2}\) \(x,\,2x,\,5x^{2},\,4x,\,4x^{3}\)
\(3xy^{2}\) \(3,\,3xy,\,3x^{2}xy^{2},\,3y^{2},\,y^{2}x\)
\(2m^{2}n\) \(n,\,mn^{2},\,8mn,\,2m^{2},\,4nm^{2}\)
\(4ab^{2}c\) \(4,\,2ab^{2}c,\,4abc,\,8b^{2}c,\,4cba\)

Term Like term
\(3x\) \(2x\)
\(2ab\) \(12ab\)
\(2x^{2}\) \(5x^{2}\)
\(3xy^{2}\) \(y^{2}x\)
\(2m^{2}n\) \(4nm^{2}\)
\(4ab^{2}c\) \(2ab^{2}c\)

Addition and subtraction

The essential rule for adding or subtracting terms is:

Only like terms may be added or subtracted.

Worked example – adding and subtracting algebraic terms

\[\begin{align*} 7e+10e & = (7\times e)+(10\times e)\\
& = (7+10)e\\
& = 17e
\end{align*}\]

Here, the terms have been broken up into their coefficients and pro-numerals to show you what they mean. In practice, we do not do this and just go straight to the answer – that is, \(7e+10e=17e\).

\[\begin{align*} 3x^{2}-x^{2}-4x^{2} & = (3-1-4)x^{2}\\
& = -2x^{2}
\end{align*}\]

\[\begin{align*} 3m-4n+6m+n & = (3+6)m+(-4+1)n\\
& = 9m-3n
\end{align*}\]

In this example, the like terms are gathered. All of the terms with \(m\) as the pronumeral are grouped and all the terms with \(n\) as the pronumeral are grouped. In the final answer, they remain separate from each other.

\[\begin{align*} 3a-b-5a+4ab-3b+ab & = (3-5)a+(-1-3)b+(4+1)ab\\
& = -2a-4b+5ab
\end{align*}\]

\[3x-x^{2}\]

There are no like terms, so nothing can be done.

\[\begin{align*} p+2p-3 & = (1+2)p-3\\
& = 3p-3
\end{align*}\]

\[\begin{align*} 8uv+3u-10vu & = (8-10)uv+3u\\
& = -2uv+3u
\end{align*}\]

\[6r^{2}s-2rs^{2}\]

There are no like terms, so nothing can be done.

Your turn – adding and subtracting algebraic terms

  1. \(13x+4-3x-1\)
  2. \(10mn+5m+12mn+6m\)
  3. \(3xy^{2}+2xy+5xy^{2}+3xy\)
  4. \(5xy+6m-2xy-2m\)
  5. \(x+y+2x-y\)
  6. \(3a+5b-a-6b\)
  7. \(x+4y-x-2y\)
  8. \(7x-4m-5x-3m\)
  9. \(4x-5x-3y+5x\)
  10. \(9mn-3m-n+4m^{2}\)

  1. \(10x+3\)
  2. \(22mn|11m\)
  3. \(8xy^{2}+5xy\)
  4. \(3xy+4m\)
  5. \(3x\)
  6. \(2a-b\)
  7. \(2y\)
  8. \(2x-7m\)
  9. \(5x-3y\)
  10. \(9mn-3m-n+4m^{2}\)

Multiplication

Both like and unlike terms can be multiplied.

When multiplying two or more terms:

  1. Determine the sign of the answer.
    • \((+\textrm{ve})\times (+\textrm{ve})=(+\textrm{ve})\)
    • \((+\textrm{ve})\times (-\textrm{ve})=(-\textrm{ve})\)
    • \((-\textrm{ve})\times (-\textrm{ve})=(+\textrm{ve})\)
  2. Calculate the product of the coefficients.
  3. Use index laws to find the powers of the pro-numerals.

Worked example – multiplying algebraic terms

\[(-4) \times (-3b) = 12b\]

\((-\textrm{ve})\times (-\textrm{ve})=(+\textrm{ve})\), so the sign of the answer is \(+\). We take the product of the coefficients: \(4\times 3=12\).

\[-2 \times 6y = -12y\]

\[2e \times (-5e^{2}) = -10e^{3}\]

\[(-2u^{2}v) \times (-4v) = 8u^{2}v^{2}\]

\((-\textrm{ve})\times (-\textrm{ve})=(+\textrm{ve})\), so the sign of the answer is \(+\). We multiply the coefficients: \(2\times 4=8\) and the pro-numerals: \(u^{2}v \times v=u^{2}v^{1+1}=u^{2}v^{2}\). Remember your index laws!

\[-3pq \times (-2q) \times p=6p^{2}q^{2}\]

\((-\textrm{ve})\times (-\textrm{ve}) \times (+\textrm{ve}=(+\textrm{ve})\), so the sign of the answer is \(+\). We multiply the coefficients: \(3\times 2 \times 1=6\) and the pro-numerals: \(pq \times q \times p = p^{1+1}q^{1+1}=p^{2}p^{2}\).

Your turn – multiplying algebraic terms

  1. \(5\times 2k\)
  2. \(4a \times 3ab\)
  3. \(y \times 3y\)
  4. \(4m \times (-3mn)\)
  5. \(m \times 3p \times 5\)
  6. \(2ab \times 3bc \times (-4)\)
  7. \(2ab^{2} \times 3ac\)
  8. \(4m \times (-5kmp)\)

  1. \(10k\)
  2. \(12a^{2}b\)
  3. \(3y^{2}\)
  4. \(-12m^{2}n\)
  5. \(15mp\)
  6. \(-24ab^{2}c\)
  7. \(6a^{2}b^{2}c\)
  8. \(-20km^{2}p\)

Division

When we divide numbers, we can rewrite them as fractions where the first number is the numerator and the second number that it is divided by is the denominator. For example:

\[2\div5=\frac{2}{5}\]

We can do the same for algebraic terms:

\[2x^{2}yz \div 3xy^{2}= \frac{2x^{2}yz}{2xy^{2}}\]

Like and unlike terms can be divided.

When dividing algebraic terms:

  1. Determine the sign of the answer.
    • \((+\textrm{ve})\div (+\textrm{ve})=(+\textrm{ve})\)
    • \((+\textrm{ve})\div (-\textrm{ve})=(-\textrm{ve})\)
    • \((-\textrm{ve})\div (+\textrm{ve})=(-\textrm{ve})\)
    • \((-\textrm{ve})\div (-\textrm{ve})=(+\textrm{ve})\)
  2. Calculate the quotient of the coefficients or cancel any common factors.
  3. Use index laws to find the powers of the pro-numerals.

Worked example – dividing algebraic terms

Divide \(-12wyz\) by \(3yz\).

\[\begin{align*} -12wyz \div 3yz & = \frac{-12wyz}{3yz}\\
& = - \frac{12wyz}{3yz}\\
& = -4w
\end{align*}\]

After rewriting the question as a fraction, the sign was determined: \(-\textrm{ve} \div +\textrm{ve} = -\textrm{ve}\).

Then, the quotient of the coefficients was found: \(12 \div 3=4\). This is the same as identifying the common factor of \(3\) and dividing both the numerator and denominator by \(3\). \(12 \div 3=4\) so the numerator becomes \(4\) and \(3 \div 3=1\) so the denominator becomes \(1\).

Looking at the pro-numerals, the numerator has \(wyz\) and the denominator has \(yz\). The \(y\) and \(z\) can be cancelled out to leave \(w\) in the numerator.

Divide \(2m^{2}n\) by \(-6mn^{2}\).

\[\begin{align*} 2m^{2}n \div -6mn^{2} & = \frac{2m^{2}n}{-6mn^{2}} \\
& = -\frac{2m^{2}n}{6mn^{2}}\\
& = -\frac{m}{3n}
\end{align*}\]

Simplify \(6a^{2} \times 4ab \div 12ab\).

\[\begin{align*} 6a^{2} \times 4ab \div 12ab & = \frac{6a^{2} \times 4ab}{12ab} \\
& = \frac{24a^{3}}{12ab}\\
& = 2a^{2}
\end{align*}\]

After rewriting as a fraction, the terms in the numerator can be multiplied. The product of the coefficients is \(6\times 4=12\) and the product of the pro-numerals is \(a^{2} \times ab=a^{2+1}b=a^{3}b\). Putting them together gives \(24a^{3}b\) for the numerator.

We can then divide the coefficients in the fraction and cancel the common factors. \(24 \div 12=2\) and \(a^{3}b \div ab=a^{3-1}b^{1-1}=a^{2}\).

Divide \(m+2\) by \(4m\).

\[(m+2) \div 4m= \frac{m+2}{4m}\]

There are no common factors, so it is not possible to simplify further.

Your turn – dividing algebraic terms

  1. \(18ef \div 6f\)
  2. \(-100uvw \div 100w\)
  3. \(24gh^{2} \div 8gh\)
  4. \(3m^{2}n \div 12mn^{2}\)
  5. \(rs \times 2st \div 2s\)
  6. \(3jk \times 12km \div 9jkm\)
  7. \(10p \times 3qp \div 16pq\)
  8. \(4yz \times 5w^{2}z \div 10wy\)

  1. \(3e\)
  2. \(-uv\)
  3. \(3h\)
  4. \(\frac{m}{4n}\)
  5. \(rst\)
  6. \(4k\)
  7. \(\frac{15q}{4}\)
  8. \(2wz^{2}\)

Order of operations

The basic operations of arithmetic are multiplication, division, addition and subtraction. The order in which we perform the operations is important.

Let's consider the following arithmetic problem:

Head with thinking bubble that reads 4 plus 3 times 2 plus 5 equals question mark

At first sight, there are several possible answers.

  • Working from left to right:
    \[\begin{align*} 4+3 \times 2+5 & = 7 \times 2+5\\
    & = 14+5\\
    & = 19
    \end{align*}\]
  • Working from right to left:
    \[\begin{align*} 4+3 \times 2+5 & = 4+3 \times 7\\
    & = 4+21\\
    & = 25
    \end{align*}\]
  • If we do the multiplication first:
    \[\begin{align*} 4+3 \times 2+5 & = 4+6+5\\
    & = 15
    \end{align*}\]
  • If we do the additions first:
    \[\begin{align*} 4+3 \times 2+5 & = 7 \times 7\\
    & = 49
    \end{align*}\]

Mathematicians don’t like imprecision. To allow only one answer to the problem, we perform mathematical operations in the following order:

  1. Brackets
  2. Indices
  3. Multiplication and Division from left to right
  4. Addition and Subtraction

Using the bolded first letters, we get the abbreviation BIMDAS.

When we apply BIMDAS to the previous problem, we get:

\[\begin{align*} 4+3 \times 2+5 & = 4+6+5\\
& =15
\end{align*}\]

Worked examples – applying BIMDAS

\[\begin{align*}3 \times 2+4 & = 6+4\\
& = 10
\end{align*}\]

The multiplication is completed first, so \(3 \times2=6\). Then, the addition \(6+4=10\).

\[\begin{align*} 3+2 \times 4 & = 3+8\\
& = 11
\end{align*}\]

\[\begin{align*} (3+2) \times 4 & = 5 \times 4\\
& = 20
\end{align*}\]

The operation in the brackets is completed first, so \(3+2=5\). Then, we complete the multiplication, \(5\times 4=20\).

\[\begin{align*} 3-2^{2} & = 3-2 \times 2\\
& = 3-4\\
& = -1
\end{align*}\]

\[\begin{align*} (3-2)^{2} & = (3-2) \times (3-2)\\
& = 1 \times 1\\
& = 1
\end{align*}\]

\[\begin{align*} 3^{2}-2^{2} & = 3 \times 3-2 \times 2\\
& = 9-4\\
& = 5
\end{align*}\]

\[\begin{align*} 3st-3s \times 4t & = 3st-12st\\
& = -9st
\end{align*}\]

According to BIMDAS, the multiplication is completed first: \(3s \times 4t=12st\). Then, the subtraction: \(3st-12st=-9st\).

Your turn – applying BIMDAS

Simplify the following expressions:

  1. \((18+(3\times (-5))\)
  2. \(3 \times (-4) + (8\times 2)\)
  3. \(10-5^{2}+3\)
  4. \(10-5)^{2}+3\)
  5. \(10^{2}-5^{2}\)
  6. \((10-5)^{2}\)
  7. \(3m+2\times 3m\)
  8. \(6ab-3a\times 4b\)
  9. \(16gh-4gh\times 4\)
  10. \(9b-3b\times 2k+2k\times b\)

  1. \(3\)
  2. \(4\)
  3. \(-12\)
  4. \(28\)
  5. \(75\)
  6. \(25\)
  7. \(9m\)
  8. \(-6ab\)
  9. \(0\)
  10. \(9b-4bk\)

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