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Algebra
Algebra involves the representation of numbers using letters, such as \(a, b, c, x, y, z\) etc. We use letters to represent unknown numbers.
In this article, we are going to look at what we need to know for algebra:
- Using letters to represent unknown numbers
- Algebraic rules for multiplying and dividing
- Simplifying simple algebraic expressions
- Evaluating simple algebraic expressions
1. Using Letters To Represent Unknown Numbers
Practice Problems
Example 1a:
Jason is \(3\) years younger than Kenny.
(a) If Jason is \(7\) years old, how old is Kenny?
Solution:
Jason’s age \(= 7\) years old
Kenny’s age
\(= 7+3\)
\( = 10\) years old
Answer:
\(10\) years old
Example 1b:
(b) If Jason is \(x\) years old, how old is Kenny?
Solution:
Jason’s age \(= x\) years old
Kenny’s age \(= (x+3)\) years old
Answer:
\((x+3)\) years old
Note: Remember to put a pair of brackets for algebraic expressions with units as the units belong to the whole expression.
Example 2:
Peter bought \(y\) apples. He ate \(4\) apples. How many apples had he left in terms of \(y\)?
Solution:
Number of apples left \(= (y-4)\) apples
Answer:
\((y-4)\) apples
Note: Did you observe that algebraic terms and numbers cannot be simplified further? This is because they are called unlike terms. Only algebraic terms of the same kind can be grouped together. You will learn more about this below.
2. Algebraic Rules For Multiplying And Dividing
- For algebraic expressions involving multiplication, do not write the multiplication sign. For example:
\(\begin{align} 4 \times y &= 4y \\[2ex] 7 \times x &= 7x \\[2ex] b \times 6 &= 6b \end{align}\)
- For algebraic terms, numbers are always in front of the letters. For example:
\(\begin{align} 9 \times y &= 9y, \;\;\text{ not }\;\; y9\\[2ex] a \times 8 &= 8a, \;\;\text{ not }\;\; a8 \end{align}\)
- For the product of \(1\) and any letter, the number \(1\) is not written. For example:
\(\begin{align} 1 \times y &= y \\[2ex] a \times 1 &= a \end{align}\)
- For algebraic expressions involving division, do not write the division sign. For example:
\(\begin{align} 8y \div 5 &= \frac {8y}{5} \\[2ex] 9a \div 2 &= \frac {9a}{2} \\[2ex] 7 \div 3b &= \frac {7}{3b} \end{align}\)
The first number in the division statement is the numerator while the second number is the denominator.
Note: Take note of the last example, \(3b\) is written as the denominator. The expression \(\displaystyle{\frac {7}{3b}}\) is not the same as \(\displaystyle{\frac {7}{3}b}\) or \(\displaystyle{\frac {7b}{3}}\). Think about it, why?
Practice Problems
Example 1:
There are \(4\) bags. If there are \(m\) books in each bag, how many books are there altogether?
Solution:
\(\begin{align} \text{Total number of books} &= 4\times m\\[2ex] &=4m \end{align} \)
Answer:
\(4m\) books
Example 2:
What is the product of \(3\) and \(b\)?
Solution:
\(\begin{align} \text{Product of 3 and }b &= 3\times b\\[2ex] &=3b \end{align}\)
Answer:
\(3b\)
Example 3:
Gary bought some chocolate bars for \(7\) students. He gave the same number of chocolate bars to each student. If he bought \(w\) chocolate bars, how many chocolate bars would each student receive?
Solution:
\(\begin{align} \text{No. of chocolate bars each student receive} &= w\div 7\\[2ex] &=\frac {w}{7} \end{align}\)
Answer:
\(\displaystyle{\frac {w}{7}}\) chocolate bars
3. Simplifying Algebraic Expressions
We can only add or subtract algebra of the same letters. For example:
\(\begin{align} 5a - 3a &= 2a \\[2ex] 5b - 8 &= (5b-8) \end{align}\)
Note: Terms that do not have the same letters cannot be simplified. Only algebraic terms of the same kind can be simplified by addition or subtraction. We call them like terms.
To simplify algebraic expressions:
Step 1:
Identify all the terms with the same letters first
Step 2:
Rearrange them (remember the sign belongs to the term behind it)
Step 3:
Add or subtract algebra of the same letters or like terms
Practice Problems
Example 1:
Simplify \(3b+2-b+5\)
Solution:
\(\begin{align} &3b+2-b+5\\[2ex] &= 3b-b+2+5 \\[2ex] &=2b+7 \end{align}\)
Answer:
\(2b+7\)
Example 2:
Simplify \(6a+10-2a-3\)
Solution:
\(\begin{align} &6a+10-2a-3\\[2ex] &= 6a-2a+10-3 \\[2ex] &=4a+7 \end{align}\)
Answer:
\(4a+7\)
Example 3:
Simplify \(10w-3-5w+13+3w\)
Solution:
\(\begin{align} &10w-3-5w+13+3w\\[2ex] &=10w-5w+3w-3+13 \\[2ex] &=8w+10 \end{align}\)
Answer:
\(8w+10\)
4. Evaluating Simple Algebraic Expressions
To evaluate algebraic expressions,
Step 1:
Substitute the letter by a given value.
Step 2:
Work out the final answer.
Practice Problems
Example 1:
Evaluate the following equation given that \(a=3\) .
\(5a+13\)
Solution:
\(\begin{align*} &5a + 13\\[2ex] &= 5 \times 3 + 13 \\[2ex] &= 15 + 13 \\[2ex] &= 28 \end{align*}\)
Answer:
\(28\)
Summary
In this topic, we learnt how to use letters to represent unknown numbers.
- These letters can represent any numerical value.
- For algebraic expressions involving multiplication, do not write the multiplication sign.
- For algebraic terms, numbers are always in front of the letters.
- For the product of 1 and any letter, the number 1 is not written.
- For algebraic expressions involving division, do not write the division sign.
Test Yourself
Simplify (7m + 5 − 2m) ÷ 3
Ans: (3) (5m + 5) / 3
A watermelon cost \($3\). Mrs Boh bought two such watermelons and had \($\text{J}\) left after buying the watermelons. How much money did she have before buying the watermelons in terms of j?
Ans: (4) $ ( j + 6 )
Adam had thrice as much money as Bell. Bell had twice as much money as Chris. If Adam, Bell and Chris had $36y altogether, how much money did Bell have?
Ans: (4) $ 8 y
Christian had m packets of game cards and Ahmad had thrice as many packets of game cards as Christian. The two boys had a total of 28 packets of game cards. If Ahmad gave away 2 packets of game cards, how many packets of game cards had he left?
Ans: (3) 19 packets
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