# Equation with Binomial Expression

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This pdf includes the following topics-
Constant
Terms and expressions
Coefficient of a term
Writing patterns in geometry
Variable
Examples
1. CHRIST SENIOR SECONDARY SCHOOL
CLASS 7
Notes: Chapter 12 ALGEBRAIC EXPRESSIONS
Introduction to Algebraic Expressions
Constant is a quantity which has a fixed value.
Terms of Expression
Parts of an expression which are formed separately first and then added are known as terms. They are
Example: Terms 4x and 5 are added to form the expression (4x +5).
Coefficient of a term
The numerical factor of a term is called coefficient of the term.
Example: 10 is the coefficient of the term 10xy in the expression 10xy+4y
Writing Patterns in Geometry
 Algebraic expressions are used in writing patterns followed by geometrical figures.
Example: Number of diagonals we can draw from one vertex of a polygon of n sides is (n – 3).
Definition of Variables
 Any algebraic expression can have any number of variables and constants.
2.  Variable
 A variable is a quantity that is prone to change with the context of the situation.
 a,x,p,… are used to denote variables.
 Constant
 It is a quantity which has a fixed value.
 In the expression 5x+4, the variable here is x and the constant is 4.
 The value 5x and 4 are also called terms of expression.
 In the term 5x, 5 is called the coefficient of x. Coefficients are any numerical factor of a term.
Factors of a term
Factors of a term are quantities which can not be further factorised. A term is a product of its factors.
Example: The term –3xy is a product of the factors –3, x and y.
Like and Unlike Terms
Like terms
 Terms having same algebraic factors are like terms.
Example: 8xy and 3xy are like terms.
Unlike terms
 Terms having different algebraic factors are unlike terms.
Example: 7xy and −3x are unlike terms.
Monomial, Binomial, Trinomial and Polynomial Terms
Types of expressions based on the number of terms
Based on the number of terms present, algebraic expressions are classified as:
 Monomial: An expression with only one term.
Example: 7xy, −5m, etc.
 Binomial: An expression which contains two, unlike terms.
Example: 5mn+4, x+y, etc
 Trinomial: An expression which contains three terms.
Example: x+y+5, a+b+ab, etc.
 An expression with one or more terms.
Example: x+y, 3xy+6+y, etc.
3. Exercise 12.1 Page: 234
1. Get the algebraic expressions in the following cases using variables, constants and arithmetic
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
= ½ (x + y)
= (x + y)/2
(iii) The number z multiplied by itself.
= z2
(iv) One-fourth of the product of numbers p and q.
= ¼ (p × q)
= pq/4
(v) Numbers x and y both squared and added.
= x2 + y2
(vi) Number 5 added to three times the product of numbers m and n.
= 3mn + 5
(vii) Product of numbers y and z subtracted from 10.
= 10 – (y × z)
= 10 – yz
(viii) Sum of numbers a and b subtracted from their product.
= (a × b) – (a + b)
= ab – (a + b)
2. (i) Identify the terms and their factors in the following expressions
Show the terms and factors by tree diagrams.
(a) x – 3
4. Expression: x – 3
Terms: x, -3
Factors: x; -3
(b) 1 + x + x2
Expression: 1 + x + x2
Terms: 1, x, x2
Factors: 1; x; x,x
(e) – ab + 2b2 – 3a2
Expression: -ab + 2b2 – 3a2
Terms: -ab, 2b2, -3a2
Factors: -a, b; 2, b, b; -3, a, a
(ii) Identify terms and factors in the expressions given below:
(a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2
(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼
(h) 0.1 p2 + 0.2 q2
Expressions is defined as, numbers, symbols and operators (such as +. – , × and ÷) grouped together that
show the value of something.
5. In algebra a term is either a single number or variable, or numbers and variables multiplied together.
Terms are separated by + or – signs or sometimes by division.
Factors is defined as, numbers we can multiply together to get another number.
4. (a) Identify terms which contain x and give the coefficient of x.
(i) y2x + y (ii) 13y2 – 8yx (iii) x + y + 2
(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25
(vii) 7x + xy2
Sl.No. Expression Terms Coefficient of x
(i) y2x + y y2x y2
(ii) 13y2 – 8yx – 8yx -8y
(iii) x+y+2 x 1
(iv) 5 + z + zx x 1
zx z
(v) 1 + x + xy xy y
(vi) 12xy2 + 25 12xy2 12y2
(vii) 7x + xy2 7x 7
xy2 y2
(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2
Sl.No. Expression Terms Coefficient of y2
(i) 8 – xy2 – xy2 –x
(ii) 5y2 + 7x 5y2 5
(iii) 2x2y – 15xy2 + 7y2 – 15xy2 – 15x
7y2 7
5. Classify into monomials, binomials and trinomials.
(i) 4y – 7z
6. An expression which contains two unlike terms is called a binomial.
(ii) y2
An expression with only one term is called a monomial.
(iii) x + y – xy
An expression which contains three terms is called a trinomial.
(iv) 100
An expression with only one term is called a monomial.
(v) ab – a – b
An expression which contains three terms is called a trinomial.
(vi) 5 – 3t
An expression which contains two unlike terms is called a binomial.
(vii) 4p2q – 4pq2
An expression which contains two unlike terms is called a binomial.
(viii) 7mn
An expression with only one term is called a monomial.
(ix) z2 – 3z + 8
An expression which contains three terms is called a trinomial.
(x) a2 + b2
7. An expression which contains two unlike terms is called a binomial.
(xi) z2 + z
An expression which contains two unlike terms is called a binomial.
(xii) 1 + x + x2
An expression which contains three terms is called a trinomial.
6. State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
Like term.
When term have the same algebraic factors, they are like terms.
(ii) –7x, (5/2)x
Like term.
When term have the same algebraic factors, they are like terms.
(iii) – 29x, – 29y
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(iv) 14xy, 42yx
Like term.
When term have the same algebraic factors, they are like terms.
Exercise 12.2 Page: 239
1. Simplify combining like terms:
(i) 21b – 32 + 7b – 20b
8. When term have the same algebraic factors, they are like terms.
= (21b + 7b – 20b) – 32
= b (21 + 7 – 20) – 32
= b (28 – 20) – 32
= b (8) – 32
= 8b – 32
(ii) – z2 + 13z2 – 5z + 7z3 – 15z
When term have the same algebraic factors, they are like terms.
= 7z3 + (-z2 + 13z2) + (-5z – 15z)
= 7z3 + z2 (-1 + 13) + z (-5 – 15)
= 7z3 + z2 (12) + z (-20)
= 7z3 + 12z2 – 20z
(iii) p – (p – q) – q – (q – p)
When term have the same algebraic factors, they are like terms.
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
When term have the same algebraic factors, they are like terms.
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)
= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)
= a (1) + b (0) + ab (1)
= a + ab
(i) 3mn, – 5mn, 8mn, – 4mn
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
9. = 3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= mn (3 – 5 + 8 – 4)
= mn (11 – 9)
= mn (2)
= 2mn
(ii) t – 8tz, 3tz – z, z – t
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= t – 8tz + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= t – t – 8tz + 3tz – z + z
= t (1 – 1) + tz (- 8 + 3) + z (-1 + 1)
= t (0) + tz (- 5) + z (0)
= – 5tz
(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3)
= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3
= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)
= mn (- 9 + 21) + (7 – 11)
= mn (12) – 4
= 12mn – 4
3. Subtract:
(i) –5y2 from y2
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= y2 – (-5y2)
= y2 + 5y2
= 6y2
(ii) 6xy from –12xy
10. When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= -12xy – 6xy
= – 18xy
(iii) (a – b) from (a + b)
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= (a + b) – (a – b)
= a (1 – 1) + b (1 + 1)
= a (0) + b (2)
= 2b
(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2
= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2
= 10ab – 7a2 – 7b2
(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq
= 8p2 + 8q2 – 5pq
4. (a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?
Let us assume p be the required term
p + (x2 + xy + y2) = 2x2 + 3xy
11. p = (2x2 + 3xy) – (x2 + xy + y2)
p = 2x2 + 3xy – x2 – xy – y2
p = 2x2 – x2 + 3xy – xy – y2
p = x2 + 2xy – y2
5. What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20?
Let us assume a be the required term
3x2 – 4y2 + 5xy + 20 – a = -x2 – y2 + 6xy + 20
a = 3x2 – 4y2 + 5xy + 20 – (-x2 – y2 + 6xy + 20)
a = 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20
a = 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20
a = 4x2 – 3y2 – xy
6. (a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
First we have to find out the sum of 3x – y + 11 and – y – 11
= 3x – y + 11 + (-y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
Now, subtract 3x – y – 11 from 3x – 2y
= 3x – 2y – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11
(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and
–x2 + 2x + 5.
First we have to find out the sum of 4 + 3x and 5 – 4x + 2x2
= 4 + 3x + (5 – 4x + 2x2)
= 4 + 3x + 5 – 4x + 2x2
= 4 + 5 + 3x – 4x + 2x2
= 9 – x + 2x2
= 2x2 – x + 9 … [equation 1]
Then, we have to find out the sum of 3x2 – 5x and – x2 + 2x + 5
= 3x2 – 5x + (-x2 + 2x + 5)
12. = 3x2 – 5x – x2 + 2x + 5
= 3x2 – x2 – 5x + 2x + 5
= 2x2 – 3x + 5 … [equation 2]
Now, we have to subtract equation (2) from equation (1)
= 2x2 – x + 9 – (2x2 – 3x + 5)
= 2x2 – x + 9 – 2x2 + 3x – 5
= 2x2 – 2x2 – x + 3x + 9 – 5
= 2x + 4
Exercise 12.3 Page: 242
1. If m = 2, find the value of:
(i) m – 2
From the question it is given that m = 2
Then, substitute the value of m in the question
= 2 -2
(ii) 3m – 5
From the question it is given that m = 2
Then, substitute the value of m in the question
= (3 × 2) – 5
(iii) 9 – 5m
From the question it is given that m = 2
Then, substitute the value of m in the question
= 9 – (5 × 2)
= 9 – 10
(iv) 3m2 – 2m – 7
From the question it is given that m = 2
Then, substitute the value of m in the question
13. = (3 × 22) – (2 × 2) – 7
= (3 × 4) – (4) – 7
= 12 – 4 -7
= 12 – 11
(v) (5m/2) – 4
From the question it is given that m = 2
Then, substitute the value of m in the question
= ((5 × 2)/2) – 4
= (10/2) – 4
2. If p = – 2, find the value of:
(i) 4p + 7
From the question it is given that p = -2
Then, substitute the value of p in the question
= (4 × (-2)) + 7
= -8 + 7
= -1
(ii) – 3p2 + 4p + 7
From the question it is given that p = -2
Then, substitute the value of p in the question
= (-3 × (-2)2) + (4 × (-2)) + 7
= (-3 × 4) + (-8) + 7
= -12 – 8 + 7
= -20 + 7
= -13
(iii) – 2p3 – 3p2 + 4p + 7
From the question it is given that p = -2
Then, substitute the value of p in the question
= (-2 × (-2)3) – (3 × (-2)2) + (4 × (-2)) + 7
14. = (-2 × -8) – (3 × 4) + (-8) + 7
= 16 – 12 – 8 + 7
= 23 – 20
3. Find the value of the following expressions, when x = –1:
(i) 2x – 7
From the question it is given that x = -1
Then, substitute the value of x in the question
= (2 × -1) – 7
(ii) – x + 2
From the question it is given that x = -1
Then, substitute the value of x in the question
= – (-1) + 2
(iii) x2 + 2x + 1
From the question it is given that x = -1
Then, substitute the value of x in the question
= (-1)2 + (2 × -1) + 1
(iv) 2x2 – x – 2
From the question it is given that x = -1
Then, substitute the value of x in the question
= (2 × (-1)2) – (-1) – 2
= (2 × 1) + 1 – 2
15. 4. If a = 2, b = – 2, find the value of:
(i) a2 + b2
From the question it is given that a = 2, b = -2
Then, substitute the value of a and b in the question
= (2)2 + (-2)2
5. When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 × 0) + (2 × -1)
= -2
(ii) 2a2 + b2 + 1
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 × 02) + (-1)2 + 1
(iii) 2a2b + 2ab2 + ab
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 × 02 × -1) + (2 × 0 × (-1)2) + (0 × -1)
= 0 + 0 +0
(iv) a2 + ab + 2
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (02) + (0 × (-1)) + 2
16. 6. Simplify the expressions and find the value if x is equal to 2
(i) x + 7 + 4 (x – 5)
From the question it is given that x = 2
We have,
= x + 7 + 4x – 20
= 5x + 7 – 20
Then, substitute the value of x in the equation
= (5 × 2) + 7 – 20
= 10 + 7 – 20
= 17 – 20
7. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.
(i) 3x – 5 – x + 9
From the question it is given that x = 3
We have,
= 3x – x – 5 + 9
= 2x + 4
Then, substitute the value of x in the equation
= (2 × 3) + 4
= 10
(ii) 2 – 8x + 4x + 4
From the question it is given that x = 3
We have,
= 2 + 4 – 8x + 4x
= 6 – 4x
Then, substitute the value of x in the equation
= 6 – (4 × 3)
= 6 – 12
17. (iii) 3a + 5 – 8a + 1
From the question it is given that a = -1
We have,
= 3a – 8a + 5 + 1
= – 5a + 6
Then, substitute the value of a in the equation
= – (5 × (-1)) + 6
= – (-5) + 6
= 11
(iv) 10 – 3b – 4 – 5b
From the question it is given that b = -2
We have,
= 10 – 4 – 3b – 5b
= 6 – 8b
Then, substitute the value of b in the equation
= 6 – (8 × (-2))
= 6 – (-16)
= 6 + 16
= 22
8. (i) If z = 10, find the value of z3 – 3(z – 10).
From the question it is given that z = 10
We have,
= z3 – 3z + 30
Then, substitute the value of z in the equation
= (10)3 – (3 × 10) + 30
= 1000 – 30 + 30
= 1000
(ii) If p = – 10, find the value of p2 – 2p – 100
18. From the question it is given that p = -10
We have,
= p2 – 2p – 100
Then, substitute the value of p in the equation
= (-10)2 – (2 × (-10)) – 100
= 100 + 20 – 100
= 20
9. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?
From the question it is given that x = 0
We have,
2x2 + x – a = 5
a = 2x2 + x – 5
Then, substitute the value of x in the equation
a = (2 × 02) + 0 – 5
a = -5
10. Simplify the expression and find its value when a = 5 and b = – 3.
2(a2 + ab) + 3 – ab
From the question it is given that a = 5 and b = -3
We have,
= 2a2 + 2ab + 3 – ab
= 2a2 + ab + 3
Then, substitute the value of a and b in the equation
= (2 × 52) + (5 × (-3)) + 3
= (2 × 25) + (-15) + 3
= 50 – 15 + 3
= 53 – 15
= 38
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