Study S1 Mathematics Maths - Basic Algebra - Geniebook

Basic Algebra And Algebraic Manipulation 1

In this chapter, we will be discussing the below-mentioned topics in detail:

  • Using letters to represent numbers
  • Interpreting algebraic notations
  • Evaluation of algebraic expressions
  • Addition and subtraction of linear expressions

 

Algebraic Notations

Example 1:

Arithmetic Algebra
It is essentially just numbers.  It is using letters to represent numbers
The sum of \(2\) and \(8\) is \(10\). the sum of \(p\) and \(q\) is \(r\)
\(2 + 8 = 10\) \(p + q = r\)

 

Example 2:

Arithmetic Algebra
The difference between \(9\) and \(3\) is \(6\). The difference between \(a\) and \(b\) is \(c\), where \(a\) is greater than \(b\)
\(9 - 3 = 6\) \(a - b = c\)

 

So, this is an example of algebraic notation where we read a particular sentence and then translate it into a mathematical statement that has algebra in it.

 

Example 3:

Arithmetic

Algebra

The product of \(4 \) and \(5 \) is \(20 \).

The product of \(m \) and \(n \) is \(k \).

\(4 \times 5 = 20 \)

Similarly, \(5 \times 4 = 20\)

It does not matter what order we put the numbers or algebra in while multiplication.

\(m \times n = k \) or \(mn = k \)

Similarly, \(n \times m = k \) or \(nm = k \)

 

While doing algebra, sometimes we can avoid using multiplication signs i.e. “\(\times \)”. 

\(mn \) is actually referring to \(m \times n \). It is actually very common to remove multiplication symbols while doing algebra.

 

Example 4:

Arithmetic

Algebra

The division of \(16 \) by \(2 \) gives a quotient of \(8 \).

The division of \(a \) by \(b \) gives a quotient of \(c \).

\(16 \div 2 = 8 \)

\(a \div b = c \)

 

Example 5:

Arithmetic

Algebra

The square of \(5 \) is \(25 \).

The square of \(c \) is \(c^2 \).

\(\begin{align*} 5^2 &= 5 \times 5 \\ &= 25 \end{align*} \)

\(c^2 = c \times c \)

 

Example 6:

Arithmetic

Algebra

The cube of \(2 \) is \(8 \).

The cube of \(a \) is \(a^3 \).

\(2^3 = 2 \times 2 \times 2 \) 
 \(= 8 \)

\(a^3 = a \times a \times a \)

 

 

Algebraic Terms

Term

Coefficient

Variable (s)

\(2a \)

\(2\)

\(a \)

\(-5b \)

\(-5 \)

\(b \)

\(c \)

\(1 \)

\(c \)

\(17abc \)

\(17 \)

\(a,b,c \)

 

A coefficient is an integer which is written along with a variable or multiplied by the variable. It is the numerical factor of a term containing constant and variables.

For example, In the term \(2x \), \(2 \) is the coefficient.

A variable is a term that represents an unknown number or unknown value or unknown quantity. These are used in the case of algebraic expression or algebra. 

For example, \(x+9=4 \) is a linear equation where \(x \) is a variable, where \(9 \) and \(4 \) are constants.

Hence, an algebraic term refers to the whole thing i.e. it consists of the coefficient and the variable. The variable is the algebra i.e. unknown. The root word of the variable is “vary” i.e. it can change but the coefficient is fixed. 

Putting the coefficient and the variable together gives us an algebraic term. They do not have any operations next to them i.e. to the left or to the right. There is no plus, minus, multiply or divide. They are just on their own.

 

 

Algebraic Expressions and Linear Expressions

Algebraic Expressions

An algebraic expression consists of algebraic terms, operation symbols and, sometimes, brackets.

An algebraic expression has no equal sign. 

\(3x - 7 \) \(y^2 - 2x +1 \) \(\begin{align} -\frac{1}{a} + b^3 \end{align}\)
\(-2(a+b)\) \(\begin{align} 1 - \frac{1}{2}z \end{align}\) \(a - bc +d\)

 

Linear Expressions

In a linear expression, the power (or index) of each variable must be \(1\). Look at the variables, if the power is \(1\) and also all other variables have a power of \(1\), then it is considered linear. 

Let’s discuss the answers to the above-mentioned expressions one by one:

  1. \(\begin{align} 3x - 7 \end{align}\)

The first term \(3x\) has a coefficient of \(3\) and a variable of \(x\). The power of \(x\), in this case, would be \(1\). Similar to the coefficient, if we do not write the coefficient it means it is \(1\), like ways if no power is written that means the power is \(1\).

So, in this case, the only variable is \(x\) having power \(1\), so it would be considered a Linear expression.

To solve these kinds of expressions, we need to look at the variables and then we check the power, if the power is \(1\) and all the other variables also have a power of \(1\), then it is considered linear.

 

  1. \(\begin{align} y^2 - 2x +1 \end{align}\)

\(x\) has a power of \(1\) but \(y\) has a power of \(2\). The power of each variable must be \(1\). As this expression does not meet the condition, hence it is not a linear expression.

 

  1. \(\begin{align} -\frac{1}{a} + b^3 \end{align}\)

The power of \(b\) is not \(1\), hence it is not linear.

 

  1. \(\begin{align} -2(a+b) \end{align} \)

Both the variables “\(a\)” and “\(b\)” have no power which means they have a power of 1.

Hence, the above expression is linear.

 

  1. \(\begin{align} 1-\frac{1}{2}z \end{align}\)

As we know 1 is a constant and as there is no power of “\(z\)” which means it has a power of \(1\), hence the above-mentioned expression is linear.

 

  1. \(\begin{align} a-bc+d \end{align} \)

In this case, \(a, b, c \text{ and } d\) all have a power of \(1\). But if we have two variables that are multiplied together, it makes the expression non-linear.

 

Note: One variable to the power of \(1\) (\(b\) in this case) that is multiplied to another variable (\(c\) in this case) that is to the power of \(1\), the combined power is actually not \(1\).Hence, \(a-bc+d\) is not linear.

 

Let’s understand this with the help of some examples:

 

Question 1: 

Write down an algebraic expression for each of the following statements.

  1. Subtract \(6\) from the product of \(p\) and \(q\).
  2. Multiply \(3\) by the square of \(a\).

 

Solution: 

  1. \(\begin{align*} ​\text{Product of } p  \text{ and } q ​&= p \times q\\ &= pq \end{align*}\)

The answer would be \(=pq-6\)

 

  1. Square of \(a=a^2\)

The answer would be \(=3\times a^2\)

                                   \(=3a^2\)

 

 

Question 2: 

Divide the product of \(a\) and \(b\) by the cube of \(c\).

 

Solution: 

Product of \(a\) and \(b\) \(=a\times b\)

                              \(= ab\)

Cube of \(\begin{align} c= c^3 \end{align} \)

As per the above expression, it is written to divide the  product of \(a\) and \(b\) by the cube of \(c\)., so the words that are written after “by” become the denominator.

 i.e. \(\begin{align} \text{Product of a and b} \over \text{Cube of C} \end{align}\)

The correct answer would be \(\begin{align} ab \over c^3 \end{align}\)

 

 

Question 3: 

Divide \(7\) by the sum of \(m\) and twice the variable \(n\)

 

Solution:

The correct answer would be \(\begin{align} \frac{7}{m+2n} \end{align}\) .

 

 

Question 4: 

Given that \(a=2\) and \(b=-7\), evaluate each of the following expressions. 

\(4a-3b\)

 

Solution:

Substituting values of \(a=2\) and \(b=-7\) in the equation:

\(\begin{align*} &=4(2)-3(-7) \\ &=8+21\\ &=29 \end{align*}\)

 

 

Question 5:

Given that \(a=2\) and \(b=-7\), evaluate each of the following expressions:

  1. \(\begin{align} 5ab+a \end{align}\)
  2. \(\begin{align} \frac{-2a^2}b \end{align}\)

 

Solution: 

  1. Substituting values of \(a=2\) and \(b=-7\) in the equation:

\(\begin{align*} &=5ab+a \\ \\&=5\;(2)\;(-7)+2\\ \\&=-70+2\\ \\&=-68 \end{align*}\)

  1. Substituting values of \(a=2\) and \(b=-7\) in the equation:

\(\begin{align*} &=\frac{-2(2)^2}{-7}\\ \\ &=\frac87\\ \\&=1\frac17\\ \end{align*}\)

 

 

Question 6:

Given that \(x = -3\) and \(y = 9\), evaluate the expression \(\begin{align} \frac{4x + y}{\sqrt{y}} \end{align}\).

 

Solution:

Substituting values of \(x = -3\) and \(y = 9\) in the equation:

\(\begin{align*} \frac{4x \;+ \;y}{\sqrt{y}} &= \frac{4 (-3) \;+\; 9}{\sqrt{9}}\\ \\                       &= \frac{-12 + 9}3\\ \\                       &= \frac{- 3} 3 \\ \\                  &= -1 \end{align*}\)

 

 

Test Your Concepts

Answer the following questions based on the concepts we’ve covered in this article. If you get stuck, revisit the relevant section to revise the concepts.

 

Question 1: 

Given that \(a = 2\) and \(b = -7\), evaluate each of the following expressions. 

\(\begin{align} \frac{5a}{2b} \end{align}\)

 

Question 2: 

Given that \(p = 2\), \(q = -6\) and \(r = 3.5\), evaluate the expression \(\begin{align} \frac{p^3 - q^2}{3qr} \end{align}\).

Continue Learning
Basic Geometry Linear Equations
Number Patterns Percentage
Prime Numbers Ratio, Rate And Speed
Functions & Linear Graphs 1 Integers, Rational Numbers And Real Numbers
Basic Algebra And Algebraic Manipulation 1 Approximation And Estimation
Resources - Academic Topics
Primary
Primary 1
Primary 2
Primary 3
Primary 4
Primary 5
Primary 6
Secondary
Secondary 1
English
+ More
Maths
Basic Geometry
Linear Equations
Number Patterns
Percentage
Prime Numbers
Ratio, Rate And Speed
Functions & Linear Graphs 1
Integers, Rational Numbers And Real Numbers
Basic Algebra And Algebraic Manipulation 1
Approximation And Estimation
+ More
Science
+ More
Secondary 2
Secondary 3
Secondary 4
+ More
Sign up for a free demo
(P1 to S4 levels)
Our Education Consultants will get in touch to offer a complimentary product demo and Strength Analysis to your child.
Ready to power up your
child's academic success?
Let our Education Consultants show you how.
*By submitting your phone number, we have your permission to contact
you regarding Geniebook. See our Privacy Policy.