3. Algebra
This chapter discusses:
- Algebraic expressions;
- Addition, subtraction, multiplication and division of algebraic expressions;
- Simple rules of algebra for binomial products;
- Algebraic fractions - addition, subtraction, multiplication and division.
Do you need this chapter?
The following questions illustrate the material covered in this chapter. If you can solve all these problems, then you might not need to complete this chapter as you already have the requisite knowledge.
Quiz: Algebra
1. Simplify [latex]x^3+3xy+y^2-5x-xy-5y^2+y^4+5[/latex]
2. Simplify [latex]2 \left( 2x^2-3x +3\right) - 3\left(y-2-y^2\right)-4x[/latex].
3. Simplify [latex]\dfrac{h}{6}+\dfrac{2h}{9}[/latex].
4. Simplify [latex]4yz\times5w^{2}z\div10wy[/latex].
5. Simplify [latex]9b-3b\times2k+2k\times b[/latex].
6. Expand and simplify [latex]9c\left(4-c\right)+2\left(c-7\right)[/latex]
7. Simplify
\begin{align*}
\frac{4m-16}{4m}\div\frac{8m-32}{8m}&
\end{align*}
8. Expand [latex]\left( x+3 \right)^2[/latex].
Answers
1) [latex]y^4 + x^3 -4y^2 +2xy -5x +5[/latex]
2) [latex]4x^2 + 3y^2 -10x -3y +12[/latex]
3) [latex]\dfrac{7h}{18}[/latex]
4) [latex]2wz^2[/latex]
5) [latex]9b-4bk[/latex]
6) [latex]-9c^2 +38c - 14[/latex]
7) [latex]1[/latex]
8) [latex]x^2+6x+9[/latex]
If you have any difficulty with these questions, please go to the relevant section in this chapter.
Algebraic operations
The following sections introduce the basic skills for addition, subtraction, multiplication and division of algebraic expressions. Courses in science, engineering and other fields require you to have these skills.
The following short video is a quick introduction to some essential concepts.
Algebraic operations (8:45 min) by RMIT University Library Videos (YouTube)
Like and unlike terms
A pro-numeral is a symbol that may represent a number. Letters of the alphabet (English and Greek) are commonly used as pronumerals. For example, [latex]x[/latex] is a pro-numeral.
Algebraic expressions are combinations of numbers, brackets, operation signs and pronumerals.
Usually the multiplication sign is not used in algebraic expressions. A simple algebraic expression is [latex]3x[/latex]. When we write [latex]3x[/latex] we really mean [latex]3\times x[/latex]. Another example of an algebraic expression is [latex]5ab[/latex]. This means [latex]5\times a\times b[/latex]. Note that [latex]5ab=5ba[/latex]. The order of the letters and numbers is not important but usually we write numbers first and use alphabetical order for the letters. While [latex]b5a[/latex] is a legitimate expression, it is better to write [latex]5ab[/latex].
Like terms contain exactly the same pronumerals and the pronumerals have the same index. The following table gives some examples of like and unlike terms.
| Like Terms | Unlike Terms |
| [latex]3x,5x[/latex] | [latex]3x,5y[/latex] |
| [latex]2ab,3ab[/latex] | [latex]2a^{2}b,3ab^{2}[/latex] |
| [latex]2a,-3a[/latex] | [latex]2a,-3[/latex] |
| [latex]m^{2},7m^{2}[/latex] | [latex]m^{2},7m[/latex] |
| [latex]3xyz,5xyz[/latex] | [latex]3xyz,3xy[/latex] |
| [latex]3x^{2}yz^5,5z^{5}yx^2[/latex] | [latex]3x^{2}yz^5,5x^{2}yz^4[/latex] |
| [latex]3ef,3fe[/latex] | [latex]3ef,\,3eg[/latex] |
| [latex]3yzx,5xyz[/latex] | [latex]3yzx,5xy[/latex] |
| [latex]6\alpha\beta,\beta\alpha[/latex] | [latex]6\alpha \beta,6\alpha \beta \gamma[/latex] |
Please try the following exercises.
Exercises: Like and unlike terms
Which of the five terms on the right is a like term with the term
on the left?
\begin{array}{llllllll}
\mathbf{1.} \ 3x & | & 3 & 2x & 3x^{2} & 2xy & 4x^{2}a \qquad \qquad \qquad \qquad \qquad\\
\mathbf{2.} \ 2ab & | & 2a & 2b & 3x^{2} & 6abc & 12ab\\
\mathbf{3.} \ 2x^{2} & | & x & 2x & 5x^{2} & 4x & 4x^{3}\\
\mathbf{4.} \ 3xy^{2} & | & 3 & 3xy & 3x^{2}xy^{2} & 3y^{2} & y^{2}x\\
\mathbf{5.} \ 2m^{2}n & | & n & mn^{2} & 8mn & 2m^{2} & 4nm^{2}\\
\mathbf{6.} \ 4ab^{2}c & | & 4 & 2ab^{2}c & 4abc & 8b^{2}c & 4cba.\\
\end{array}
Answers
[latex]\begin{array}{lllllllllllllll} \quad \mathbf{1.}\ 2x & & \mathbf{2.}\ 12ab & & \mathbf{3.}\ 5x^{2} & & \mathbf{4.}\ y^{2}x & & \mathbf{5.}\ 4nm^{2} & & \mathbf{6.}\ 2ab^{2}c\end{array}[/latex]
Addition and subtraction of algebraic expressions
The essential rule is that only like terms may be added or subtracted.
Examples: Addition and subtraction
1. Simplify if possible [latex]7e+10e[/latex].
Solution:
\begin{align*}
7e+10e & =7\times e+10\times e\\
& =\left(7+10\right)e\\
& =17e.
\end{align*}
In practice, we would not put in the first line. It is included to
remind you what [latex]7e[/latex] and [latex]10e[/latex] mean.
2. Simplify [latex]3x^{2}-x^{2}-4x^{2}[/latex].
Solution:
\begin{align*}
3x^{2}-x^{2}-4x^{2} & =\left(3-1-4\right)x^{2}\\
& =-2x^{2}.
\end{align*}
3. Simplify [latex]3m-4n+6m+n[/latex].
Solution:
\begin{align*}
3m-4n+6m+n & =\left(3+6\right)m+\left(-4+1\right)n\\
& =9m-3n.
\end{align*}
This example shows how we can gather different like terms. In this
case [latex]m[/latex] and [latex]n[/latex].
4. Simplify [latex]3a-b-5a+4ab-3b+ab[/latex].
Solution:
\begin{align*}
3a-b-5a+4ab-3b+ab & =\left(3-5\right)a+\left(-1-3\right)b+\left(4+1\right)ab\\
& =-2a-4b+5ab.
\end{align*}
5. Simplify [latex]3x-x^{2}[/latex].
Solution:
\begin{align*}
3x-x^{2} & \quad\mathrm{there\textrm{ are no like terms so nothing can be done.}}
\end{align*}
6. Simplify [latex]p+2p-3[/latex].
Solution:
\begin{align*}
p+2p-3 & =3p-3.
\end{align*}
7. Simplify [latex]8uv+3u-10vu[/latex].
Solution:
\begin{align*}
8uv+3u-10vu&=-2uv+3u.
\end{align*}
8. Simplify [latex]6r^{2}s-2rs^{2}[/latex].
Solution:
\begin{align*}
6r^{2}s-2rs^{2}&\quad\mathrm{there\textrm{ are no like terms so nothing can be done.}}
\end{align*}
9. Simplify [latex]x^3+3xy+y^2-5x-xy-5y^2+y^4+5[/latex].
Solution:
\begin{align*}
x^3+3xy+y^2-5x-xy-5y^2+y^4+5&=x^3+\left(3-1 \right)xy+ \left(1-5) \right)y^2 +y^4-5x+5\\
&=x^3+2xy-4y^2+y^4-5x+5.
\end{align*}
Please try the following exercises.
Exercises: Addition and subtraction
Simplify each of the following:
[latex]\begin{array}{lllllll} \quad & \mathbf{1.}\ 3x+4-3x-1 & & & & \mathbf{2.}\;10mn+5m+12mn+6m\\ \quad & \mathbf{3.}\ 3xy^{2}+2xy+5xy^{2}+3xy & & & & \mathbf{4.}\ 5xy+6m-2xy-2m\\ \quad & \mathbf{5.}\ x+y+2x-y & & & & \mathbf{6.}\ 3a+5b-a-6b\\ \quad & \mathbf{7.}\ x+4y-x-2y & & & & \mathbf{8.}\ 7x-4m-5x-3m\\ \quad & \mathbf{9.}\ 4x-5x-3y+5x & & & & \mathbf{10.}\ 9mn-3m-n+4m^{2} \end{array}[/latex]
Answers
[latex]\begin{array}{llllllllll} \quad & \mathbf{1.}\;3 & \mathbf{2.}\;22mn+11m & \mathbf{3.}\;8xy^{2}+5xy & & \mathbf{4.}\;3xy+4m\\ \quad & \mathbf{5.}\;3x & \mathbf{6.}\;2a-b & \mathbf{7.}\;2y & & \mathbf{8.}\;2x-7m \\ \quad & \mathbf{9.}\;4x-3y & \mathbf{10.}\;9mn-3m-n+4m^{2} \end{array}[/latex]
Multiplication of algebraic expressions
Both like and unlike terms may be multiplied. When multiplying two or more terms consider:
- The sign of the answer. Use the same rules for determining the sign as when multiplying numbers:
\begin{align*}
\left(+ve\right)\times\left(+ve\right) & =\left(+ve\right)\\
\left(+ve\right)\times\left(-ve\right) & =\left(-ve\right)\\
\left(-ve\right)\times\left(+ve\right) & =\left(-ve\right)\\
\left(-ve\right)\times\left(-ve\right) & =\left(+ve\right).
\end{align*} - The product of the terms.
Examples: Multiplication
[latex]\begin{array}{l} \quad \mathbf{1.} \ \left(-4\right)\times\left(-3b\right)=12b\\ \quad \mathbf{2.}\ -2\times6y=-12y\\ \quad \mathbf{3.}\ 2e\times\left(-5e^{2}\right)=-10e^{3}\\ \quad \mathbf{4.}\ \left(-2u^{2}v\right)\times\left(-4v\right)=8u^{2}v^{2}\\ \quad \mathbf{5.} -3pq\times\left(-2q\right)\times p=6p^{2}q^{2}. \end{array}[/latex]
Please try the following exercises.
Exercises: Multiplication
Simplify the following algebraic expressions:
[latex]\begin{array}{lllllll} \quad \mathbf{1.} \ 5\times2k & & & & \quad \mathbf{2.} \ 4a\times3ab\\ \quad \mathbf{3.} \ y\times3y & & & & \quad \mathbf{4.} \ 4m\times\left(-3mn\right)\\ \quad \mathbf{5.} \ m\times3p\times5 & & & & \quad \mathbf{6.} \ 2ab\times3bc\times\left(-4\right)\\ \quad \mathbf{7.} \ 2ab^{2}\times3ac & & & & \quad \mathbf{8.} \ -4m \ \times\left(-5kmp\right)\\ \end{array}[/latex]
Answers
[latex]\begin{array}{llllllll} \quad \mathbf{1.} \ 10k & & \quad \mathbf{2.} \ 12a^{2}b & & \quad \mathbf{3.} \ 3y^{2} & & \quad \mathbf{4.} \ -12m^{2}n\\ \quad \mathbf{5.} \ 15mp & & \quad \mathbf{6.} \ -24ab^{2}c & &\quad \mathbf{7.} \ 6a^{2}b^{2}c & &\quad \mathbf{8.} \ 20km^{2}p\\ \end{array}[/latex]
Division of algebraic expressions
Consider the fraction [latex]\dfrac{2}{5}[/latex]. The top term [latex]2[/latex], is called the numerator, the bottom term [latex]5[/latex], is called the denominator and the horizontal line between them is called the fraction bar, division bar or vinculum.
Fractions involve division. For example,
\[
\frac{2}{5}=2\div5.
\]
Similarly for algebraic terms:
\[
\frac{2x^{2}yz}{3xy^{2}}=2x^{2}yz\div3xy^{2}.
\]
When dividing algebraic terms:
- Rewrite as a fraction if necessary;
- Multiply numbers and expand powers of pronumerals using index laws;
- Establish the sign of the answer. Use the same rules for determining the sign as when dividing numbers:
\begin{align*}
\left(+ve\right)\div\left(+ve\right) & =\left(+ve\right)\\
\left(+ve\right)\div\left(-ve\right) & =\left(-ve\right)\\
\left(-ve\right)\div\left(+ve\right) & =\left(-ve\right)\\
\left(-ve\right)\div\left(-ve\right) & =\left(+ve\right).
\end{align*} - Cancel any common factors of numbers and pronumerals using index laws.
Examples: Division
1. Divide [latex]-12wyz \text{ by } 3yz[/latex].
Solution:
\begin{align*}
-12wyz\div3yz & =\frac{-12wyz}{3yz}\quad\textrm{(write as a fraction)}\\
& =-\frac{12wyz}{3yz}\quad\textrm{(determine sign)}\\
& =-\frac{^4\cancel{12}w\cancel{y}\cancel{z}}{\cancel{3}_1\cancel{y}\cancel{z}}\quad\textrm{(cancel common factors)}\\
& =-4w.
\end{align*}
2. Divide [latex]2m^{2}n \text{ by } -6mn^{2}[/latex].
Solution:
\begin{align*}
2m^{2}n\div 6mn^{2} & =\frac{2m^{2}n}{-6mn^{2}} \quad \textrm{(write as a fraction)}\\
& =-\frac{2m^{2}n}{6mn^{2}} \quad\textrm{(determine sign)}\\
& =-\frac{^1\cancel{2}m^{2-1}\cancel{n}}{\cancel{6}_3\cancel{m}n^{2-1}} \quad\textrm{(cancel common factors)}\\
& =-\frac{m}{3n} .
\end{align*}
3. Simplify [latex]6a^{2}\times4ab\div12ab[/latex].
Solution:
\begin{align*}
6a^{2}\times4ab\div12ab & =\frac{6a^{2}\times4ab}{12ab}\quad\textrm{write as a fraction}\\
& =\frac{24a^{3}b}{12ab}\quad\textrm{establish sign, multiply numbers and expand powers}\\
& =\frac{^2\cancel{24}a^{3-1}\cancel{b}}{\cancel{12}_1\cancel{a}\cancel{b}}\quad\textrm{cancel common factors}\\
& =2a^{2}.
\end{align*}
4. Divide [latex]m+2 \text{ by } 4m[/latex].
Solution:
\[
\left(m+2\right)\div 4m=\frac{m+2}{4m}.
\]
There are no common factors and so it is not possible to simplify
further.
Please try the following exercises.
Exercises: Division
Simplify the following algebraic expressions:
[latex]\begin{array}{lllllll} & & \mathbf{1.}\ 18ef\div6f & & & & \mathbf{2.}\ -100uvw\div100w\\ & & \mathbf{3.}\ 24gh^{2}\div8gh & & & & \mathbf{4.}\ 3m^{2}n\div12mn^{2}\\ & & \mathbf{5.}\ s\times2st\div2s & & & & \mathbf{6.}\ 3jk\times12km\div9jkm\\ & & \mathbf{7.}\ 10p\times3qp\div16pq & & & & \mathbf{8.}\ 4yz\times5w^{2}z\div10wy \end{array}[/latex]
Answers
[latex]\begin{array}{llllllll} & & \mathbf{1.}\ 3e & & \mathbf{2.}\ -uv & & \mathbf{3.}\ 3h & & \mathbf{4.}\ m/4n\\ & & \mathbf{5.}\ st & & \mathbf{6.}\ 4k & & \mathbf{7.}\ 15p/8 & &\mathbf{8.}\ 2wz^{2} \end{array}[/latex]
Order of operations
The basic operations of arithmetic are multiplication, division, addition and subtraction. Usually the order in which we perform the operations is important.
For example, what is the answer to the following arithmetic problem: [latex]4+3\times2+5[/latex]?
At first sight there are several possible answers.
Working from left to right we get:
\begin{align*}
4+3\times2+5 & =7\times2+5\\
& =14+5\\
& =19.
\end{align*}
Working from right to left we get:
\begin{align*}
4+3\times2+5 & =4+3\times7\\
& =4+21\\
& =25.
\end{align*}
If we do the multiplication first we get another answer:
\begin{align*}
4+3\times2+5 & =4+6+5\\
& =15.
\end{align*}
Yet another answer is obtained by doing the additions first:
\begin{align*}
4+3\times2+5 & =7\times7\\
& =49.
\end{align*}
Mathematicians don't like imprecision. In order to allow only one
answer to the problem we perform operations in the following order:
- Brackets
- Indices or Orders
- Multiplication and Division (from left to right)
- Addition and Subtraction
Using the first letters (bold) we get the abbreviation BIMDAS. You may know this as BODMAS.
Applying this rule the answer to our problem is:
\begin{align*}
4+3\times2+5 & =4+6+5\quad\textrm{multiplication first}\\
& =15\quad\textrm{then addition}.
\end{align*}
Here are some numerical examples of using BIMDAS.
Examples: BIMDAS (BODMAS)
1. [latex]3\times2+4=6+4=10.[/latex]
2. [latex]3+2\times4=3+8=11.[/latex]
3. [latex]\left(3+2\right)\times4=5\times4=20.[/latex]
4. [latex]3-2^{2}=3-2\times2=3-4=-1.[/latex]
5. [latex]\left(3-2\right)^{2}=\left(3-2\right)\times\left(3-2\right)=1\times1=1.[/latex]
6. [latex]3^{2}-2^{2}=3\times3-2\times2=9-4=5.[/latex]
BIMDAS or BODMAS also applies to algebraic expressions as shown in the following examples.
Examples: BIMDAS (BODMAS) in algebra
1. Simplify [latex]3x \times 4 -3x[/latex].
Solution:
\begin{align*}
3x \times 4 -3x & = 12x - 3x\\
&= 9x.
\end{align*}
2. Simplify [latex]5\left( 3y-1 \right) -2-10y[/latex].
Solution:
\begin{align*}
5\left( 3y-1 \right) -2-10y & = 15y-5-2-10y\\
& =5y-7 .
\end{align*}
3. Simplify [latex]2 \left( 2x^2-3x +3\right) - 3\left(y-2-y^2\right)-4x[/latex].
Solution:
\begin{align*}
2 \left( 2x^2-3x +3\right) - 3\left(y+2-y^2\right)-4x &= 4x^2-6x+6-3y-6-3y^2 -4x\\
&=4x^2-10x-3y-3y^2.\\
\end{align*}
Please try the following exercises.
Exercises: BIMDAS (BODMAS) in algebra
Simplify the following expressions:
[latex]\begin{array}{lllll} \qquad \mathbf{1.}\ 18+\left(3\times\left(-5\right)\right) & & & &\mathbf{2.}\ 3\times\left(-4\right)+\left(8\times2\right)\\ \qquad \mathbf{3.}\ 10-5^{2}+3 & & & & \mathbf{4.}\ \left(10-5\right)^{2}+3\\ \qquad \mathbf{5.}\ 10^{2}-5^{2} & & & & \mathbf{6.}\ \left(10-5\right)^{2}\\ \qquad \mathbf{7.}\ 3m+2\times3m & & & & \mathbf{8.}\ 6ab-3a\times4b\\ \qquad \mathbf{9.}\ 16gh-4gh\times4 & & & & \mathbf{10.}\ 9b-3b\times2k+2k\times b \end{array}[/latex]
Answers
[latex]\begin{array}{lllllllll} \qquad \mathbf{1.}\ 3 & & \mathbf{2.}\ 4 & & \mathbf{3.}\ -12 & & \mathbf{4.}\ 28 & & \mathbf{5.}\ 75\\ \qquad \mathbf{6.}\ 25 & & \mathbf{7.}\ 9m & & \mathbf{8.}\ -6ab & & \mathbf{9.}\ 0 & & \mathbf{10.}\ 9b-4bk \end{array}[/latex]
Some basic rules of algebra
Manipulation of algebraic expressions is subject to some formal rules. These are explained in this section.
Brackets - The distributive law
Brackets often occur in algebraic expressions. Frequently they have to be removed to rearrange expressions. The important thing to remember is that whatever is outside the bracket affects all the terms within the bracket. This is known as the distributive law which says:
[latex]a \left( b + c \right) = ac + bc[/latex].
In words, the pronumeral [latex]a[/latex] multiplies both the pronumerals [latex]b[/latex] and [latex]c[/latex]. This law works with numerals, pronumerals, and combinations of both. Using the distributive law to remove brackets is called expansion.
Examples: Distributive law
1. Expand and simplify [latex]2\left(3+5\right)[/latex].
Solution:
We have
\begin{align*}
2\left(3+5\right) & =2\times3+2\times5\\
&=6+10\\
&=16.
\end{align*}
2. Expand [latex]3\left(x+2\right)[/latex].
Solution:
We have
\begin{align*}
3\left(x+2\right) & =3\times x+3\times2\\
& =3x+6.
\end{align*}
3. Expand [latex]-5\left(m+4\right)[/latex].
Solution:
We have
\begin{align*}
-5\left(m+4\right) & =\left(-5\right)\times m+\left(-5\right)\times4\\& =-5m-20.
\end{align*}Usually we omit the first step shown above and expand directly as shown in the following examples.
4. Expand [latex]-2\left(p-7\right)[/latex].
Solution:
\begin{align*}
-2\left(p-7\right) & =-2p+14.
\end{align*}
5. Expand [latex]e\left(e+2\right)[/latex].
Solution:
\begin{align*}e\left(e+2\right) & =e^{2}+2e.
\end{align*}
6. Expand [latex]-\left(4p-3\right)[/latex].
Solution:
\begin{align*}
-\left(4p-3\right) & =-4p+3.
\end{align*}
7. Expand and simplify [latex]4\left(x-3\right)+2\left(x+1\right)[/latex].
Solution:
\begin{align*}
4\left(x-3\right)+2\left(x+1\right) & =4x-12+2x+2\\
& =6x-10.
\end{align*}
8. Expand and simplify [latex]6k\left(2k+3\right)-3k\left(k-3\right)[/latex].
Solution:
\begin{align*}
6k\left(2k+3\right)-3k\left(k-3\right) & =12k^{2}+18k-3k^{2}+9k\\
& =9k^{2}+27k.
\end{align*}
Please try the following exercises.
Exercises: Distributive law
1. Remove the brackets:
[latex]\begin{array}{lll}a)\;2\left(x-5\right) & b)\;3\left(6-b\right) & c)\;\left(b+5\right)c\\d)\;2x\left(2a+3b\right) & e)\;-6\left(p-2q+4\right) & f)\;3y\left(yz-2y+1\right)\end{array}[/latex]
2. Expand and simplify:
[latex]\begin{array}{lll}a)\;3\left(x+1\right)+2\left(x+2\right) & b)\;2\left(p-1\right)-\left(p-3\right)\\ c)\;2\left(4m-3\right)-5\left(9-2m\right) &d)\;2w\left(3w+1\right)+3w\left(2w-5\right) \\ e)\;2a\left(a-1\right)-a\left(a-3\right) & f)\;5j\left(3j+2\right)-2j\left(4j-1\right)\\ g)\;3q\left(5q+4\right)+6\left(8-3q\right) & h)\;2r\left(3r-1\right)-3r\left(3+2r\right). \end{array}[/latex]
3. Expand the following expressions and simplify where possible remembering the correct order of operations.
[latex]\begin{array}{lll}a)\;4\left(2m-3\right)+8 & b)\;9-3\left(4b+3\right) & c)\;1-4\left(x-1\right)\\d)\;4-\left(5-2x\right) & e)\;8m-3\left(1-2m\right)+6 & f)\;9c\left(4-c\right)+2\left(c-7\right). \end{array}[/latex]
Answers
[latex]\begin{array}{lllllll} \mathbf{1.}\ &a)\;2x-10 & b)\;18-3b & c)\;bc+5c \\ \ \qquad &d)\;4ax+6bx & e)\;-6p+12q-24 & f)\;3y^{2}z-6y^{2}+3y\end{array}[/latex]
[latex]\begin{array}{lllll} \mathbf{2.}\ & a)\;5x+7 & b)\;p+1 & c)\;18m-51 & d)\;12w^{2}-13w\\ \ \qquad & e)\;a^{2}+a & f)\;7j^{2}+12j & g)15q^{2}-6q+48 & h)\;-11r. \end{array}[/latex]
[latex]\begin{array}{llllllc} \mathbf{3.}\ & a)\;8m-4 & b)\;-12b & c)\;5-4x \\ \ \qquad & d)\;-1+2x & e)\;14m+3 & f)\;38c-9c^{2}-14.\end{array}[/latex]
Binomial products
A binomial is an algebraic expression comprising two terms. Examples are [latex]a+3,\, 1+c, \,a+b[/latex] and [latex]3xy+2y[/latex]. When multiplying two binomials they are each put into a bracket. For instance to multiply [latex]a+3[/latex] by [latex]1+c[/latex], we would write
\begin{align*}
\left( a+3 \right)\left(c + 1 \right).&
\end{align*}
To carry out the multiplication, each term in the first bracket multiplies each term in the second. That is
\begin{align*}
\left( a+3 \right)\left(c + 1 \right)&=a\left(c + 1 \right)+3\left(c + 1 \right)\\
&=ac+a+3c+3.
\end{align*}
Often, after multiplying out the brackets, you can simplify the expression. For example:
\begin{align*}
\left(m+3 \right)\left(2m + 1 \right)&=m\left(2m + 1 \right)+3\left(2m + 1 \right)\\
&=2m^2+m+2m+3\\
&=2m^2+3m+3
\end{align*}
The cases when the two binomials are the same lead to two important formulas.
\begin{align*}
\left( a+b \right)\left(a +b \right)&=a\left(a + b \right)+b\left(a + b \right)\\
&=a^2+ab+ba+b^2\\
&=a^2+2ab+b^2
\end{align*}
and
\begin{align*}
\left( a-b \right)\left(a -b \right)&=a\left(a - b \right)+b\left(a -b \right)\\
&=a^2-ab-ba+b^2\\
&=a^2-2ab+b^2.
\end{align*}
That is
\begin{align*}
\left( a+b \right)^2&=a^2+2ab+b^2\\
\left( a-b \right)^2&=a^2-2ab+b^2.
\end{align*}
These two formulas are used frequently and you should memorize them.
Examples: Binomial products
1. Expand [latex]\left(2f+3\right)\left(g+5h\right)[/latex].
Solution:
\begin{align*}
\left(2f+3\right)\left(g+5h\right) & =2fg+\left(2f\right)\left(5h\right)+3g+15h\\
& =2fg+10fh+3g+15h.
\end{align*}
In this case no further simplification is possible.
2. Expand [latex]\left(p-2\right)\left(p-7\right)[/latex].
Solution:
\begin{align*}
\left(p-2\right)\left(p-7\right) & =p^{2}-7p-2p+14\\
& =p^{2}-9p+14.
\end{align*}
3. Expand [latex]\left(2w+3\right)\left(w-4\right)[/latex].
Solution:
\begin{align*}
\left(2w+3\right)\left(w-4\right) & =2w^{2}-8w+3w-12\\
& =2w^{2}-5w-12.
\end{align*}
4. Expand [latex]\left(4r-3s\right)^{2}[/latex].
Solution:
\begin{align*}
\left(4r-3s\right)^{2} & =\left(4r-3s\right)\left(4r-3s\right)\\
& =16r^{2}-12rs-12sr+9s^{2}\\
& =16r^{2}-24rs+9s^{2}.
\end{align*}
This can be done faster if you remember the formula above:
\begin{align*}
\left(4r-3s\right)^{2} & =\left(4r\right)^{2}-2\left(4r\right)\left(3s\right)+\left(-3s\right)^{2}\\
& =16r^{2}-24rs+9s^{2}.
\end{align*}
Please try the following exercises.
Exercises: Binomial products
1. Expand the following binomial products:
[latex]\quad \begin{array}{lll}a)\;\left(x+5\right)\left(x+3\right) & b)\;\left(u-5\right)\left(u+3\right) & c)\;\left(k+6\right)\left(j+1\right)\\d)\;\left(y+2\right)\left(y-2\right) & e)\;\left(e-7\right)\left(e-8\right) & f)\;\left(t-3\right)\left(5-t\right)\\g)\;\left(3q+2\right)\left(q+1\right) & h)\;\left(8a+5\right)\left(a-3\right) & i)\;\left(2c-d\right)\left(3c+2d\right)\\j)\;\left(3a-2b\right)\left(3a+2b\right).\end{array}[/latex]
2. Remove the brackets:
[latex]\quad \begin{array}{lll}a)\;\left(d+3\right)^{2} & b)\;\left(x-2\right)^{2} & c)\;\left(3v+2\right)^{2}\\d)\;\left(y-3z\right)^{2} & e)\;\left(3-2q\right)^{2} & f)\;\left(4f-3gh\right)^{2}.\end{array}[/latex]
Answers
[latex]\begin{array}{llllll} \mathbf{1.}\ & a)\;x^{2}+8x+15 & b)\;u^{2}-2u-15 & c)\;kj+6j+k+6 \\ \quad & d)\;y^{2}-4 & e)\;e^{2}-15e+56 & f)\;-t^{2}-15+8t \\ \quad & g)\;3q^{2}+5q+2 & h)\;8a^{2}-19a-15 & i)\;6c^{2}+cd-2d^{2} \\ \quad & j)\;9a^{2}-4b^{2}. \end{array}[/latex]
[latex]\begin{array}{llllll} \mathbf{2.}\ & a)\;d^{2}+6d+9 & b)\;x^{2}-4x+4 & c)\;9v^{2}+12v+4 \\ \ \quad & d)\;y^{2}-6yz+9z^{2} & e)\;9-12q+4q^{2}& f)\;16f^{2}-24fgh+9g^{2}h^{2}. \end{array}[/latex]
Algebraic fractions - addition and subtraction
Algebraic fractions are fractions that involve pro-numerals. For example:
\[
\frac{a}{2},\,\frac{2b}{3},\,\frac{1}{2c},\,\frac{e+1}{2},\,\frac{3}{a-2},\,\frac{a}{b},\,\frac{a}{b+c}
\]
are algebraic fractions.
Algebraic fractions can be added and subtracted in a similar way to ordinary fractions. The same concepts apply. To add or subtract two algebraic fractions, their denominators must be the same.
The following video provides a quick introduction to addition and subtraction of algebraic fractions.
Algebraic fractions: Addition and subtraction (4:31 min)
Algebraic fractions: Addition and subtraction (4:31 min) by RMIT University Library Videos (YouTube)
Here are some examples that further explain these concepts.
Examples: Algebraic fractions - addition and subtraction
1. Simplify [latex]\dfrac{h}{6}+\dfrac{2h}{9}[/latex].
Strategy: The lowest common denominator is [latex]18[/latex] so multiply the first fraction by [latex]\dfrac{3}{3}[/latex] to get [latex]18[/latex] as the denominator. Multiply the second fraction by [latex]\dfrac{2}{2}[/latex] to get [latex]18[/latex] as the denominator. Now add the numerators.
Solution:
\begin{align*}
\frac{h}{6}+\frac{2h}{9} & =\frac{h}{6}\times\frac{3}{3}+\frac{2h}{9}\times\frac{2}{2} \quad\text{common denominator is }18\\
& =\frac{3h}{18}+\frac{4h}{18}\\
& =\frac{3h+4h}{18}\\
& =\frac{7h}{18}.
\end{align*}Alternative solution:
Use cross multiplication to get a common denominator of [latex]6\times 9=54[/latex]. That is,
\begin{align*}
\frac{h}{6}+\frac{2h}{9} & =\frac{h\times9+2h\times 6}{54} \quad\text{common denominator is }54\\
& =\frac{9h+12h}{54}\\
& =\frac{21h}{54}\\
& =\frac{^7\cancel{21}h}{^{18}\cancel{54}}\quad \text{dividing top and bottom by }3\\
& =\frac{7h}{18}.
\end{align*}
2. Simplify [latex]\dfrac{e+1}{2}+\dfrac{e}{5}[/latex].
Solution:
\begin{align*}
\frac{e+1}{2}+\frac{e}{5} & =\frac{e+1}{2}\times\frac{5}{5}+\frac{e}{5}\times\frac{2}{2}\quad \text{ common denominator is }10\\
& =\frac{5\left(e+1\right)}{10}+\frac{2e}{10}\\
& =\frac{5\left(e+1\right)+2e}{10}\\
& =\frac{5e+5+2e}{10}\\
& =\frac{7e+5}{10}.
\end{align*}
Alternative solution:
Using cross multiplication,
\begin{align*}
\frac{e+1}{2}+\frac{e}{5} & =\frac{5 \left( e+2 \right) + 2e}\\
& =\frac{5e+5+2e}{10}\\
& =\frac{7e+5}{10}.
\end{align*}For the following examples, we will use the cross-multiplication method.
3. Simplify [latex]\dfrac{5}{2a}-\dfrac{3}{4}[/latex].
Solution:
\begin{align*}
\frac{5}{2a}-\frac{3}{4} & =\frac{4\times 5 - 3 \times 2a}{8a}\\
& =\frac{10}{4a}-\frac{3a}{4a}\\
& =\frac{20-6a}{8a}\\
&=\frac{^{10}\cancel{20}-^{3}\cancel{6}a}{^{4}\cancel{8}a}\\
&=\frac{10-3a}{4a}
\end{align*}
4. Simplify [latex]\dfrac{5}{2a}-\dfrac{3}{4b}[/latex].
Solution:
\begin{align*}
\frac{5}{2a}-\frac{3}{4b} & =\frac{5\times 4b - 3\times 2a}{8ab}\\
& =\frac{10b}{4ab}-\frac{3a}{4ab}\\
& =\frac{20b-6a}{8ab}\\
&=\frac{^10\cancel{20}b-^{3}\cancel{6}^{3}a}{^{4}\cancel{8}ab}\\
&=\frac{10b-3a}{4ab}.
\end{align*}
Please try the following exercises.
Exercises: Algebraic fractions - addition and subtraction
Simplify the following:
[latex]\begin{align*} \quad \mathbf{1.}\ & \frac{4}{5}+\frac{3}{4} & \mathbf{2.}\ & \frac{x}{3}-\frac{x}{5}\;\; & \mathbf{3.}\ & \frac{2p}{7}-\frac{p}{4} \\ \\ \mathbf{4.}\ & \frac{2g}{3}+\frac{g+1}{4} & \mathbf{5.}\ & \frac{d+3}{2}+\frac{1-d}{4} & \mathbf{6.}\ & \frac{5}{9}-\frac{3}{b} \\ \\ \mathbf{7.}\ & \frac{3x+2}{5}-\frac{x-3}{10}\;\; & \mathbf{8.}\ & \frac{3}{v}+\frac{2}{v+1} \end{align*}[/latex]
Answers
[latex]\begin{align*} \quad \mathbf{1.} & \frac{31}{20} & \mathbf{2.} & \frac{2x}{15} & \mathbf{3.}& \frac{p}{28} & \mathbf{4.} & \frac{11g+3}{12} \\ \\ \mathbf{5.} & \frac{d+7}{4} & \mathbf{6.}\ & \frac{5b-27}{9b} & \mathbf{7.} & \frac{5x+7}{10} & \mathbf{8.} & \frac{5v+3}{v\left(v+1\right)} \end{align*}[/latex]
Simplifying algebraic fractions
Just as numerical fractions may be simplified so too can algebraic fractions. For instance, [latex]\dfrac{2}{6}[/latex] may be simplified to [latex]\dfrac{1}{3}[/latex] by dividing the top and bottom numbers (called the numerator and denominator, respectively) by [latex]2[/latex]. Simplifying fractions by dividing the same number into both the numerator and denominator is called cancelling.
Similarly, an algebraic fraction like [latex]\dfrac{4x^{2}y}{6y^{2}}[/latex] can be simplified as follows:
\begin{align*}
\frac{4x^{2}y}{6y^{2}} & =\frac{2x^{2}y}{3y^{2}}\quad\text{ dividing top and bottom by }2\\
& =\frac{2x^{2}}{3y}\quad\text{ dividing top and bottom by }y.
\end{align*}
More complex fractions can also be simplified. For instance,
\begin{align*}
\frac{3p^{2}q+6p}{3p}&=\frac{p^{2}q+2p}{p} \quad \text{ dividing top and bottom by }3\\
&=\frac{pq+2}{1} \quad \text{ dividing top and bottom by }p\\
&=pq+2.
\end{align*}
Examples: Simplifying algebraic fractions
1. Simplify [latex]\dfrac{a(b+2c)}{2ab}[/latex].
Solution:
\begin{align*}
\frac{a(b+2c)}{2ab}&=\frac{b+2c}{2b} \quad \text{ dividing top and bottom by }a
\end{align*}
2. Simplify [latex]\dfrac{m-n}{\left(m-n\right)^{2}}[/latex].
Strategy:
Since [latex]m[/latex] and [latex]n[/latex] are just numbers, [latex]m-n[/latex] is also a number and so we can divide top and bottom by [latex]m-n[/latex] provided that [latex]m-n\neq0[/latex].
Solution:
\begin{align*}
\frac{m-n}{\left(m-n\right)^{2}} & =\frac{m-n}{\left(m-n\right)\left(m-n\right)}\quad\left(m\neq n\right)\\
& =\frac{1}{m-n}\quad\mathrm{\mathrm{dividing}\ top\ and\ bottom\ by}\:m-n.
\end{align*}
3. Simplify [latex]\dfrac{3x^{2}y}{6x+9y}[/latex].
Strategy:
Notice that [latex]3[/latex] divides into the numerator and both terms in the denominator. So we divide top and bottom by [latex]3[/latex].
Solution:
\begin{align*}
\frac{3x^{2}y}{6x+9y} & =\frac{x^{2}y}{2x+3y}\quad \text{ dividing top and bottom by }3
\end{align*}
Please try the following exercises.
Exercises: Simplifying algebraic fractions
Simplify the following algebraic fractions.
[latex]\begin{array}{lll} \quad \mathbf{1.}\ \dfrac{12ab^{2}}{8bc}& \mathbf{2.}\ \dfrac{5x-20}{5}& \mathbf{3.}\ \dfrac{b}{b^{2}+7b} \end{array}[/latex]
Answers
[latex]\begin{array}{lll} \quad \mathbf{1.}& \dfrac{3ab}{2c} & & & \mathbf{2.}& x-4 & & & \mathbf{3.}\ & \dfrac{1}{b+7} \end{array}[/latex]
Algebraic fractions - multiplication and division
The following video gives an introduction to multiplication and division of algebraic fractions.
Multiplication
When we multiply two fractions, we multiply the numerators and the
denominators of each fraction. For example,
\[
\frac{3}{4}\times\frac{7}{5}=\frac{21}{20}.
\]
It is best to simplify each fraction, if possible, before you do the
multiplication. For example,
\begin{align*}
\frac{15}{8}\times\frac{24}{35} & =\frac{15}{1}\times\frac{3}{35}\quad\mathrm{dividing\:the\;24\:in\:the\:top\:line\:and\:the\:8\:in\:the\:bottom\:line\:by\:8}\\
& =\frac{3}{1}\times\frac{3}{7}\quad\mathrm{dividing\:the\;15\:in\:the\:top\:line\:and\:the\:35\:in\:the\:bottom\:line\:by\:5}\\
& =\frac{9}{7}.
\end{align*}
You can also use these ideas with algebraic fractions. For example,
\begin{align*}
\frac{5a}{7}\times\frac{14}{a} & =\frac{5}{7}\times\frac{14}{1}\quad\mathrm{dividing\:the\;5\mathit{a}\:in\:the\:top\:line\:and\:the\:\mathit{a}\:in\:the\:bottom\:line\:by\:\mathit{a}}\\
& =\frac{5}{1}\times\frac{2}{1}\quad\:\mathrm{dividing\:the\;14\:in\:the\:top\:line\:and\:the\:\mathit{7}\:in\:the\:bottom\:line\:by\:\mathit{7}}\\
& =\frac{10}{1}\\
& =10.
\end{align*}
Here are some more examples. Note that we are showing every step in the examples below and so the solutions may appear long and complicated. You don't have to do this. You can take as many steps as you like. As you get more familiar with algebra you will naturally use fewer steps to get a result.
Examples: Algebraic fractions - multiplication
1. Simplify
\begin{align*}&\frac{x}{6\left(x-2\right)}\times\frac{3\left(x-2\right)}{x^{2}}.\end{align*}
Solution:
\begin{align*}
\frac{x}{6\left(x-2\right)}\times\frac{3\left(x-2\right)}{x^{2}} & =\frac{x}{6}\times\frac{3}{x^{2}}\ \ \text{dividing the top and bottom lines by }x-2\\
& =\frac{x}{2}\times\frac{1}{x^{2}}\quad\text{dividing the top and bottom lines by }3\\
& =\frac{1}{2}\times\frac{1}{x}\quad\text{dividing the top and bottom lines by }x\\
& =\frac{1}{2x}.
\end{align*}
2. Simplify
\begin{align*}&\frac{6 \left( p+q \right)^2}{pq} \times \frac{pq^2}{3}.\end{align*}
Solution:
\begin{align*}
\frac{6 \left( p+q \right)^2}{pq} \times \frac{pq^2}{3}& =\frac{2 \left( p+q \right)^2}{pq} \times \frac{pq^2}{1}\quad\text{dividing the top and bottom lines by }3\\
& =\frac{2 \left( p+q \right)^2}{1} \times \frac{q}{1}\quad\text{dividing the top and bottom lines by }pq\\
& =2q\left( p+q \right)^2.\\
\end{align*}
Please try the following exercises.
Exercises: Algebraic fractions - multiplication
Simplify the following algebraic fractions.
[latex]\begin{array}{lll} \quad \mathbf{1.}\ \dfrac{4}{5}\times \dfrac{15}{16}&&& \mathbf{2.}\ \dfrac{4a}{3}\times \dfrac{9}{a}\\ \quad \mathbf{3.}\ \dfrac{32h^2}{9j}\times \dfrac{27j}{48h}&&& \mathbf{4.}\ \dfrac{3d-2}{3} \times \dfrac{4}{3d-2} \end{array}[/latex]
Answers
[latex]\begin{array}{lll} \quad \mathbf{1.}\ \dfrac{3}{4} & & & \mathbf{2.}\ 12 \\ \quad \mathbf{3.}\ 2h &&& \mathbf{4.}\ \dfrac{4}{3} \end{array}[/latex]
Algebraic fractions - division
The reciprocal of a fraction is just the fraction turned upside down. So the reciprocal of [latex]\dfrac{3}{4}[/latex] is [latex]\dfrac{4}{3}[/latex], the reciprocal of [latex]\dfrac{1}{7}[/latex] is [latex]\dfrac{7}{1}=7[/latex] and the reciprocal of [latex]5[/latex] is [latex]\dfrac{1}{5}[/latex].
Dividing by a fraction is the same as multiplying by the reciprocal. When dividing fractions change the divide sign to times and turn the last fraction upside down. For example:
\begin{alignat*}{1}
\frac{5}{4}\div\frac{19}{8} & =\frac{5}{4}\times\frac{8}{19}\\
& =\frac{5}{1}\times\frac{2}{19}\\
& =\frac{10}{19}.
\end{alignat*}
The same technique for dividing numerical fractions is used for dividing algebraic fractions. So
\begin{align*}
\frac{6a^{2}bc}{7} \div \frac{b^2c^3}{21}&=\frac{6a^{2}bc}{7} \times\frac{21}{b^2c^3}.\\
\end{align*}
Now we just carry on as for multiplication:
\begin{align*}
\frac{6a^{2}bc}{7} \div \frac{b^2c^3}{21}&=\frac{6a^{2}bc}{7} \times\frac{21}{b^2c^3}\\
&=\frac{6a^{2}bc}{1} \times\frac{3}{b^2c^3}\\
&=\frac{18a^{2}bc}{b^2c^3} \\
&=\frac{18a^{2}}{bc^2}. \\
\end{align*}
Here are some examples.
Examples: Algebraic fractions - division
1. Simplify [latex]\dfrac{7p}{12}\div\dfrac{3}{8}[/latex].
Solution:
\begin{align*}
\frac{7p}{12}\div\frac{3}{8} & =\frac{7p}{12}\times\frac{8}{3}\quad\mathrm{changing\:sign\:and\:inverting\:the\:last\:fraction}\\
& =\frac{7p}{3}\times\frac{2}{3}\quad\mathrm{dividing\:top\:and\:bottom\;by\:4}\\
& =\frac{14p}{9}.
\end{align*}
2. Simplify [latex]\dfrac{m^{2}}{n}\div6m[/latex].
Solution:
\begin{align*}
\frac{m^{2}}{n}\div6m & =\frac{m^{2}}{n}\div\frac{6m}{1}\\
& =\frac{m^{2}}{n}\times\frac{1}{6m}\quad\mathrm{changing\:sign\:and\:inverting\:the\:last\:fraction}\\
& =\frac{m}{n}\times\frac{1}{6}\quad\mathrm{dividing\:top\:and\:bottom\;by\:}m\\
& =\frac{m}{6n}.
\end{align*}
3. Simplify [latex]\dfrac{4\left(x+3\right)}{9}\div\dfrac{24}{5x}[/latex].
Solution:
\begin{align*}
\frac{4\left(x+3\right)}{9}\div\frac{24}{5x} & =\frac{4\left(x+3\right)}{9}\times\frac{5x}{24}\quad\mathrm{changing\:sign\:and\:inverting\:the\:last\:fraction}\\
& =\frac{\left(x+3\right)}{9}\times\frac{5x}{6}\quad\mathrm{dividing\:top\:and\:bottom\;by\:4}\\
& =\frac{5x\left(x+3\right)}{54}.
\end{align*}
Please try the following exercises.
Exercises: Algebraic fractions - division
Simplify the following.
[latex]\begin{align*} \quad \mathbf{1.}\ \frac {6pq^3}{5} \div \frac{\left(2pq^3-pq \right)}{4}. \end{align*}[/latex]
[latex]\begin{align*} \quad \mathbf{2.}\ \frac{4m-16}{4m}\div\frac{8m-32}{8m}. \end{align*}[/latex]
Answers
[latex]\begin{array}{lll} \quad \ \mathbf{1.}\ \dfrac{24q^2}{5\left( 2q^2-1 \right)} &&& \mathbf{2.}\ 1 \end{array}[/latex]
Key takeaways
1. A pro-numeral is a symbol that may represent a number. Letters of the alphabet (especially English and Greek) are commonly used as pronumerals.
2. Algebraic expressions are combinations of numbers, brackets, operational symbols and pronumerals.
3. Only like terms may be added and subtracted.
4. Order of operations rules (BIMDAS or BODMAS) also apply to algebraic expressions.
5. The distributive law:
\begin{align*}a\left(b+c \right)=ab+ac.\end{align*}
6. Square of a binomial:
\begin{align*}\left(a+b \right)^2&=a^2 + 2ab + b^2\\
\left(a-b \right)^2&=a^2 - 2ab + b^2.\end{align*}
