Algebraic Expressions - Mathematics Form 1 Notes

Introduction

• An algebraic expression is a mathematical expression that consists of variables, numbers and operations. The value of this expression can change Clarify the definitions and have students take notes on their graphic organizer.

Note:

• Algebraic Expression—contains at least one variable, one number and one operation. An example of an algebraic expression is n + 9.
• Variable—a letter that is used in place of a number. Sometimes, the variable will be given a value. This value will replace the variable in order to solve the equation. Other times, the variable is not assigned a value and the student is to solve the equation to determine the value of the variable.
• Constant—a number that stands by itself. The 9 in our previous vocabulary is an example of a constant.
• Coefficient—a number in front of and attached to a variable. For example, in the expression 5x + 3, the 5 is the coefficient.
• Term—each part of an expression that is separated by an operation. For instance, in our earlier example n + 9, the terms are n and 7.

Examples

Write each phrase as an algebraic expression.

Nine increased by a number r 9 + r
Fourteen decreased by a number x
1 4 − x
Six less than a number t
t− 6
The product of 5 and a number n
5 × n or 5n
Thirty-two divided by a number x 32 ÷ x or
32/x

Example

An electrician charges sh 450 per hour and spends sh 200 a day on gasoline. Write an algebraic expression to represent his earnings for one day.

Solution

Let x represent the number of hours the electrician works in one day. The electrician's earnings can be represented by the following algebraic expression:

450x − 200

Simplification of Algebraic Expressions

Note:

Basic steps to follow when simplify an algebraic expression:

• Remove parentheses by multiplying factors.
• Use exponent rules to remove parentheses in terms with exponents.
• Combine like terms by adding coefficients.
• Combine the constants.

Like and Unlike Terms

• Like terms have the same variable /letters raised to the same power i.e. 3 b + 2b = 5b or a + 5a = 6a and they can be simplified further into 5b and 6a respectively(a2 and 3a2 are also like terms).While unlike terms have different variables i.e. 3b + 2 c or 4b + 2x and they cannot be simplified further.

Example

3a + 12b + 4a − 2b = 7a + 10b (collect the like terms )
2x − 5y + 3x − 7y + 3w = 5x − 12y + 3w

Example

Simplify: 2x – 6y – 4x + 5z – y

Solution

2x – 6y – 4x + 5z – y = 2x – 4x – 6y – y + 5z
= (2x – 4x) – (6y + y) + 5z
= − 2x – 7y + 5z

Note:
–6y – y = −( 6y + y)

Example

Simplify: 1/2 a − 1/3b + 1/4a

Solution

The L.C.M of 2, 3 and 4 is 1 2.
Therefore
1a − 1b + 1a = 6a − 4b + 3a
2      3      4               12
6a − 4b + 3a
12
=9a−4 b
12

Example

Simplify:

a+b2a − b
2          3

Solution

a+b − 2a − b = 3(a + b) − 2(2a − b)
2          3                      6

3a + 3b − 4a + 2b
6
3a + 3b − 4a + 2b
6

= −a+5b
6

Example

1. 5x² − 2x² = 3x²
2. 4a²bc − 2a²bc = 2a²bc
3. a²b – 2b3c + 3a²b +b3c = 4a²b – b3c

Note:

• Capital letter and small letters are not like terms.

Brackets

• Brackets serve the same purpose as they do in arithmetic.

Example

Remove the brackets and simplify:

1. 3(a + b ) – 2(a – b)
2. 1/3a + 3 (5a + b – c )
3. 2b + 3{3 – 2 (a – 5)}

Solution

1. 3(a + b ) – 2(a – b) = 3a + 3b – 2a + 2b
= 3a - 2a + 3b + 2b
= a + 5b
2. 1/3a + 3 (5a + b – c ) = 1/3a + 15 a + 3b – 3c
= 151/3 a + 3b − 3c
3. 2b + a { 3 – 2(a – 5 )} = 2b + a {3 – 2a + 1 0}
=2b + 3a – 2a2 + 1 0a
= 2b + 3a + 10a − 2a2
= 2b + 13a − 2a2

The process of remaining the brackets is called expansion while the reverse process of inserting the brackets is called factorization.

Example

Factorize the following:

1. 3m + 3n
2. ar3 + ar4 + ar5
3. 4x2y + 20x4y2 − 36x3y

Solution

1. 3(m + n) (the common term is 3 so we put it outside the bracket)
2. ar3 + ar4 + ar(ar3 is common)
ar3(1 + r + r2)
3. 4x2y + 20x4y2 − 36x3(4x2y is common)
= 4x2y( 1 + 5x2y − 9x)

Factorization by Grouping

• When the terms of an expression which do not have a common factor are taken pairwise, a common factor can be found. This method is known as factorization by grouping.

Example

Factorize:

1. 3ab + 2b + 3ca + 2c
2. ab + bx – a - x

Solution

1. 3ab + 2b + 3ca + 2c = b(3a + 2) + c(3a +2 )
= (3a + 2) (b + c)
2. ab + bx – a – x = b (a + x) – 1 (a + x)
= (a + x) (b – 1)

Algebraic Fractions

• In algebra, fractions can be added and subtracted by finding the L.C.M of the denominators.

Examples

Express each of the following as a single fraction:

1. x − 1 + x + 2 + x
2          4        5
2. a + b − b − a
b           a
3.     1    +     3      +
3(a+b)    8(a+b)   12a

Solution

1. x − 1 + x + 2 + x10(x − 1+ 5 (x+2 + 4x
2          4        5                     20
(10x − 10 + 5x + 10 + 4x)
20
19 x
20

2. a + b b − a = b(a+b) − a(b−a)
b          a                   ab
= a² + b²
ab
3.   1    +     3      +    (find the L.C.M. of 3 , 8 and 12 which is 24)
3(a+b)    8(a+b12a
(find the L.C.M. of a and a+bis a( a+b) )
(
The L.C.M. of 3(a+b) , 8(a+b )and 12 is 24a(a+b ) )
=    1        3      +  5
3(a+b)    8(a +b)   12a
= 8a + 9a + 10a + 10b
24a (a+b)
=27a+10b
24a( a+b)

Simplification by Factorization

• Factorization is used to simplify expressions

Examples

Simplify
– 2pq + q²
2p²-3pq + q²

Solution

Numerator is solved first.

Pq − pq + q²
p(p − q) − q(p + q)
(p−q)²

Then solve the denominator

2p² − 2pq − pq − q²
(2p − q) (p − q)
=       (p−q)²
(2p − q)(p − q)
(p − q)²= (p − q)(p − q) hence it cancels with the denominator

Example

Simplify

16m² − 9n²
4m² − mn − 3n²

Solution

Num. (4m – 3n) (4m + 3n)
Den. 4m²
– 4mn + 3mn – 3n²
= (4m + 3n) (m – n)
(4m – 3n)(4m + 3n)
(4m + 3n)(m – n)

4m – 3n
m – n

Example

Simplify the expression.

18xy – 18xr
9xr – 9xy

Solution

Numerator

18x(y – r)

Denominator

9x (r – y)

Therefore
18x(y − r)
9x(r − y)
(y − r)
(r − y)

Example

Simplify

12x² + ax − 6a²
9x² 4a²

Solution

(3x − 2a) (4x + 3a)
(3x+2a)(3x − 2a

= 4x + 3a
3x + 2a

Example

Simplify the expression completely.

ay − ax
bx − by

Solution

a(y − x)a(y − x) = a = −a
b(x − y)    −b(x − y)   −b    b

Note:

x − y = −(y − x)

Substitution

• This is the process of giving variables specific values in an expression

Example

Evaluate the expression
x² + y²   if x =2 and y = 1
y + 2

Solution

x² + y²  = 2² + 1² = 4 + 1
y + 2         1 + 2           3
= 5/3 = 12/3

Past KCSE Questions on the Topic

1. Given that y = 2x – z  express x in terms of y and z
x + 3z
2. Simplify the expression
x − 1 − 2x + 1
x           3x
Hence solve the equation
x − 1 − 2x + 1 = 2
x           3x       3
3. Factorize a2 – b2
Hence find the exact value of 25572 - 25472
4. Simplify
p²
– 2pq + q²
P3 - pq²+ p²q – q3
5. Given that y = 2x – z, express x in terms of y and z.
Four farmers took their goats to a market. Mohammed had two more goats as Koech had 3 times as many goats as Mohammed, whereas Odupoy had 1 0 goats less than both Mohammed and Koech.
1. Write a simplified algebraic expression with one variable, representing the total number of goats.
2. Three butchers bought all the goats and shared them equally. If each butcher got 1 7 goats, how many did odupoy sell to the butchers?
6. Solve the equation
1  = 5  − 7
4x    6x
7. Simplify
a      +    b
2(a+b)     2(a−b
8. Three years ago, Juma was three times as old as Ali. In two years time, the sum of their ages will be 62. Determine their ages

• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students

Related items

.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp