Page 1
SIMPLE EQUATIONS 59
4.1 A MIND-READING GAME!
The teacher has said that she would be starting a new chapter in
mathematics and it is going to be simple equations. Appu, Sarita
and Ameena have revised what they learnt in algebra chapter in
Class VI. Have you? Appu, Sarita and Ameena are excited because
they have constructed a game which they call mind reader and they
want to present it to the whole class.
The teacher appreciates their enthusiasm and invites them to present their game. Ameena
begins; she asks Sara to think of a number , multiply it by 4 and add 5 to the product. Then,
she asks Sara to tell the result. She says it is 65. Ameena instantly declares that the number
Sara had thought of is 15. Sara nods. The whole class including Sara is surprised.
It is Appu’s turn now . He asks Balu to think of a number, multiply it by 10 and subtract
20 from the product. He then asks Balu what his result is? Balu says it is 50. Appu
immediately tells the number thought by Balu. It is 7, Balu confirms it.
Everybody wants to know how the ‘mind reader’ presented by Appu, Sarita and
Ameena works. Can you see how it works? After studying this chapter and chapter 12,
you will very well know how the game works.
4.2 SETTING UP OF AN EQUATION
Let us take Ameena’s example. Ameena asks Sara to think of a number . Ameena does not
know the number. For her, it could be anything 1, 2, 3, . . ., 11, . . . , 100, . . . . Let us
denote this unknown number by a letter, say x. Y ou may use y or t or some other letter in
place of x. It does not matter which letter we use to denote the unknown number Sara has
thought of. When Sara multiplies the number by 4, she gets 4x. She then adds 5 to the
product, which gives 4x + 5. The value of (4x + 5) depends on the value of x. Thus
if x = 1, 4x + 5 = 4 ×1 + 5 = 9. This means that if Sara had 1 in her mind, her result would
have been 9. Similarly, if she thought of 5, then for x = 5, 4x + 5 = 4 × 5 + 5 = 25; Thus
if Sara had chosen 5, the result would have been 25.
Chapter 4
Simple
Equations
Rationalised 2023-24
Page 2
SIMPLE EQUATIONS 59
4.1 A MIND-READING GAME!
The teacher has said that she would be starting a new chapter in
mathematics and it is going to be simple equations. Appu, Sarita
and Ameena have revised what they learnt in algebra chapter in
Class VI. Have you? Appu, Sarita and Ameena are excited because
they have constructed a game which they call mind reader and they
want to present it to the whole class.
The teacher appreciates their enthusiasm and invites them to present their game. Ameena
begins; she asks Sara to think of a number , multiply it by 4 and add 5 to the product. Then,
she asks Sara to tell the result. She says it is 65. Ameena instantly declares that the number
Sara had thought of is 15. Sara nods. The whole class including Sara is surprised.
It is Appu’s turn now . He asks Balu to think of a number, multiply it by 10 and subtract
20 from the product. He then asks Balu what his result is? Balu says it is 50. Appu
immediately tells the number thought by Balu. It is 7, Balu confirms it.
Everybody wants to know how the ‘mind reader’ presented by Appu, Sarita and
Ameena works. Can you see how it works? After studying this chapter and chapter 12,
you will very well know how the game works.
4.2 SETTING UP OF AN EQUATION
Let us take Ameena’s example. Ameena asks Sara to think of a number . Ameena does not
know the number. For her, it could be anything 1, 2, 3, . . ., 11, . . . , 100, . . . . Let us
denote this unknown number by a letter, say x. Y ou may use y or t or some other letter in
place of x. It does not matter which letter we use to denote the unknown number Sara has
thought of. When Sara multiplies the number by 4, she gets 4x. She then adds 5 to the
product, which gives 4x + 5. The value of (4x + 5) depends on the value of x. Thus
if x = 1, 4x + 5 = 4 ×1 + 5 = 9. This means that if Sara had 1 in her mind, her result would
have been 9. Similarly, if she thought of 5, then for x = 5, 4x + 5 = 4 × 5 + 5 = 25; Thus
if Sara had chosen 5, the result would have been 25.
Chapter 4
Simple
Equations
Rationalised 2023-24
MATHEMATICS 60
To find the number thought by Sara let us work backward from her answer 65. We
have to find x such that
4x + 5 = 65 (4.1)
Solution to the equation will give us the number which Sara held in her mind.
Let us similarly look at Appu’s example. Let us call the number Balu chose as y. Appu
asks Balu to multiply the number by 10 and subtract 20 from the product. That is, from y,
Balu first gets 10y and from there (10y – 20). The result is known to be 50.
Therefore, 10y – 20 = 50 (4.2)
The solution of this equation will give us the number Balu had thought of.
4.3 REVIEW OF WHAT WE KNOW
Note, (4.1) and (4.2) are equations. Let us recall what we learnt about equations in
Class VI. An equation is a condition on a variable. In equation (4.1), the variable is x;
in equation (4.2), the variable is y.
The word variable means something that can vary, i.e. change. A variable takes on
different numerical values; its value is not fixed. Variables are denoted usually by
letters of the alphabets, such as x, y, z, l, m, n, p, etc. From variables, we form
expressions. The expressions are formed by performing operations like addition, subtraction,
multiplication and division on the variables. From x, we formed the expression (4x + 5).
For this, first we multiplied x by 4 and then added 5 to the product. Similarly, from y, we
formed the expression (10y – 20). For this, we multiplied y by 10 and then subtracted 20
from the product. All these are examples of expressions.
The value of an expression thus formed depends upon the chosen value of the variable.
As we have already seen, when x = 1, 4x + 5 = 9; when x = 5, 4x + 5 = 25. Similarly,
when x = 15, 4 x + 5 = 4×15 + 5 = 65;
when x = 0, 4 x + 5 = 4 × 0 + 5 = 5; and so on.
Equation (4.1) is a condition on the variable x. It states that the value of the expression
(4x + 5) is 65. The condition is satisfied when x = 15. It is the solution to the equation
4x + 5 = 65. When x = 5, 4x + 5 = 25 and not 65. Thus x = 5 is not a solution to the
equation. Similarly, x = 0 is not a solution to the equation. No value of x other than 15
satisfies the condition 4x + 5 = 65.
The value of the expression (10y – 20) depends on the value of y. V erify this by
giving five different values to y and finding for each y the value of (10 y – 20). From
the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50?
If there is no solution, try giving more values to y and find whether the condition
10y – 20 = 50 is met.
TRY THESE
Rationalised 2023-24
Page 3
SIMPLE EQUATIONS 59
4.1 A MIND-READING GAME!
The teacher has said that she would be starting a new chapter in
mathematics and it is going to be simple equations. Appu, Sarita
and Ameena have revised what they learnt in algebra chapter in
Class VI. Have you? Appu, Sarita and Ameena are excited because
they have constructed a game which they call mind reader and they
want to present it to the whole class.
The teacher appreciates their enthusiasm and invites them to present their game. Ameena
begins; she asks Sara to think of a number , multiply it by 4 and add 5 to the product. Then,
she asks Sara to tell the result. She says it is 65. Ameena instantly declares that the number
Sara had thought of is 15. Sara nods. The whole class including Sara is surprised.
It is Appu’s turn now . He asks Balu to think of a number, multiply it by 10 and subtract
20 from the product. He then asks Balu what his result is? Balu says it is 50. Appu
immediately tells the number thought by Balu. It is 7, Balu confirms it.
Everybody wants to know how the ‘mind reader’ presented by Appu, Sarita and
Ameena works. Can you see how it works? After studying this chapter and chapter 12,
you will very well know how the game works.
4.2 SETTING UP OF AN EQUATION
Let us take Ameena’s example. Ameena asks Sara to think of a number . Ameena does not
know the number. For her, it could be anything 1, 2, 3, . . ., 11, . . . , 100, . . . . Let us
denote this unknown number by a letter, say x. Y ou may use y or t or some other letter in
place of x. It does not matter which letter we use to denote the unknown number Sara has
thought of. When Sara multiplies the number by 4, she gets 4x. She then adds 5 to the
product, which gives 4x + 5. The value of (4x + 5) depends on the value of x. Thus
if x = 1, 4x + 5 = 4 ×1 + 5 = 9. This means that if Sara had 1 in her mind, her result would
have been 9. Similarly, if she thought of 5, then for x = 5, 4x + 5 = 4 × 5 + 5 = 25; Thus
if Sara had chosen 5, the result would have been 25.
Chapter 4
Simple
Equations
Rationalised 2023-24
MATHEMATICS 60
To find the number thought by Sara let us work backward from her answer 65. We
have to find x such that
4x + 5 = 65 (4.1)
Solution to the equation will give us the number which Sara held in her mind.
Let us similarly look at Appu’s example. Let us call the number Balu chose as y. Appu
asks Balu to multiply the number by 10 and subtract 20 from the product. That is, from y,
Balu first gets 10y and from there (10y – 20). The result is known to be 50.
Therefore, 10y – 20 = 50 (4.2)
The solution of this equation will give us the number Balu had thought of.
4.3 REVIEW OF WHAT WE KNOW
Note, (4.1) and (4.2) are equations. Let us recall what we learnt about equations in
Class VI. An equation is a condition on a variable. In equation (4.1), the variable is x;
in equation (4.2), the variable is y.
The word variable means something that can vary, i.e. change. A variable takes on
different numerical values; its value is not fixed. Variables are denoted usually by
letters of the alphabets, such as x, y, z, l, m, n, p, etc. From variables, we form
expressions. The expressions are formed by performing operations like addition, subtraction,
multiplication and division on the variables. From x, we formed the expression (4x + 5).
For this, first we multiplied x by 4 and then added 5 to the product. Similarly, from y, we
formed the expression (10y – 20). For this, we multiplied y by 10 and then subtracted 20
from the product. All these are examples of expressions.
The value of an expression thus formed depends upon the chosen value of the variable.
As we have already seen, when x = 1, 4x + 5 = 9; when x = 5, 4x + 5 = 25. Similarly,
when x = 15, 4 x + 5 = 4×15 + 5 = 65;
when x = 0, 4 x + 5 = 4 × 0 + 5 = 5; and so on.
Equation (4.1) is a condition on the variable x. It states that the value of the expression
(4x + 5) is 65. The condition is satisfied when x = 15. It is the solution to the equation
4x + 5 = 65. When x = 5, 4x + 5 = 25 and not 65. Thus x = 5 is not a solution to the
equation. Similarly, x = 0 is not a solution to the equation. No value of x other than 15
satisfies the condition 4x + 5 = 65.
The value of the expression (10y – 20) depends on the value of y. V erify this by
giving five different values to y and finding for each y the value of (10 y – 20). From
the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50?
If there is no solution, try giving more values to y and find whether the condition
10y – 20 = 50 is met.
TRY THESE
Rationalised 2023-24
SIMPLE EQUATIONS 61
4.4 WHAT EQUATION IS?
In an equation there is always an equality sign. The equality sign shows that the value of
the expression to the left of the sign (the left hand side or LHS) is equal to
the value of the expression to the right of the sign (the right hand side or RHS). In
equation (4.1), the LHS is (4x + 5) and the RHS is 65. In equation (4.2), the LHS is
(10y – 20) and the RHS is 50.
If there is some sign other than the equality sign between the LHS and the RHS, it is
not an equation. Thus, 4x + 5 > 65 is not an equation.
It says that, the value of (4x + 5) is greater than 65.
Similarly, 4x + 5 < 65 is not an equation. It says that the value of (4x + 5) is smaller
than 65.
In equations, we often find that the RHS is just a number. In Equation (4.1), it is 65
and in equation (4.2), it is 50. But this need not be always so. The RHS of an equation may
be an expression containing the variable. For example, the equation
4x + 5 = 6x – 25
has the expression (4x + 5) on the left and (6x – 25) on the right of the equality sign.
In short, an equation is a condition on a variable. The condition is that two
expressions should have equal value. Note that at least one of the two expressions
must contain the variable.
We also note a simple and useful property of equations. The equation 4x +5 = 65 is
the same as 65 = 4x + 5. Similarly, the equation 6x – 25 = 4x +5 is the same as
4x + 5 = 6x – 25. An equation remains the same, when the expressions on the left
and on the right are interchanged. This property is often useful in solving equations.
EXAMPLE 1 Write the following statements in the form of equations:
(i) The sum of three times x and 11 is 32.
(ii) If you subtract 5 from 6 times a number, you get 7.
(iii) One fourth of m is 3 more than 7.
(iv) One third of a number plus 5 is 8.
SOLUTION
(i) Three times x is 3x.
Sum of 3x and 11 is 3x + 11. The sum is 32.
The equation is 3x + 11 = 32.
(ii) Let us say the number is z; z multiplied by 6 is 6z.
Subtracting 5 from 6z, one gets 6z – 5. The result is 7.
The equation is 6z – 5 = 7
Rationalised 2023-24
Page 4
SIMPLE EQUATIONS 59
4.1 A MIND-READING GAME!
The teacher has said that she would be starting a new chapter in
mathematics and it is going to be simple equations. Appu, Sarita
and Ameena have revised what they learnt in algebra chapter in
Class VI. Have you? Appu, Sarita and Ameena are excited because
they have constructed a game which they call mind reader and they
want to present it to the whole class.
The teacher appreciates their enthusiasm and invites them to present their game. Ameena
begins; she asks Sara to think of a number , multiply it by 4 and add 5 to the product. Then,
she asks Sara to tell the result. She says it is 65. Ameena instantly declares that the number
Sara had thought of is 15. Sara nods. The whole class including Sara is surprised.
It is Appu’s turn now . He asks Balu to think of a number, multiply it by 10 and subtract
20 from the product. He then asks Balu what his result is? Balu says it is 50. Appu
immediately tells the number thought by Balu. It is 7, Balu confirms it.
Everybody wants to know how the ‘mind reader’ presented by Appu, Sarita and
Ameena works. Can you see how it works? After studying this chapter and chapter 12,
you will very well know how the game works.
4.2 SETTING UP OF AN EQUATION
Let us take Ameena’s example. Ameena asks Sara to think of a number . Ameena does not
know the number. For her, it could be anything 1, 2, 3, . . ., 11, . . . , 100, . . . . Let us
denote this unknown number by a letter, say x. Y ou may use y or t or some other letter in
place of x. It does not matter which letter we use to denote the unknown number Sara has
thought of. When Sara multiplies the number by 4, she gets 4x. She then adds 5 to the
product, which gives 4x + 5. The value of (4x + 5) depends on the value of x. Thus
if x = 1, 4x + 5 = 4 ×1 + 5 = 9. This means that if Sara had 1 in her mind, her result would
have been 9. Similarly, if she thought of 5, then for x = 5, 4x + 5 = 4 × 5 + 5 = 25; Thus
if Sara had chosen 5, the result would have been 25.
Chapter 4
Simple
Equations
Rationalised 2023-24
MATHEMATICS 60
To find the number thought by Sara let us work backward from her answer 65. We
have to find x such that
4x + 5 = 65 (4.1)
Solution to the equation will give us the number which Sara held in her mind.
Let us similarly look at Appu’s example. Let us call the number Balu chose as y. Appu
asks Balu to multiply the number by 10 and subtract 20 from the product. That is, from y,
Balu first gets 10y and from there (10y – 20). The result is known to be 50.
Therefore, 10y – 20 = 50 (4.2)
The solution of this equation will give us the number Balu had thought of.
4.3 REVIEW OF WHAT WE KNOW
Note, (4.1) and (4.2) are equations. Let us recall what we learnt about equations in
Class VI. An equation is a condition on a variable. In equation (4.1), the variable is x;
in equation (4.2), the variable is y.
The word variable means something that can vary, i.e. change. A variable takes on
different numerical values; its value is not fixed. Variables are denoted usually by
letters of the alphabets, such as x, y, z, l, m, n, p, etc. From variables, we form
expressions. The expressions are formed by performing operations like addition, subtraction,
multiplication and division on the variables. From x, we formed the expression (4x + 5).
For this, first we multiplied x by 4 and then added 5 to the product. Similarly, from y, we
formed the expression (10y – 20). For this, we multiplied y by 10 and then subtracted 20
from the product. All these are examples of expressions.
The value of an expression thus formed depends upon the chosen value of the variable.
As we have already seen, when x = 1, 4x + 5 = 9; when x = 5, 4x + 5 = 25. Similarly,
when x = 15, 4 x + 5 = 4×15 + 5 = 65;
when x = 0, 4 x + 5 = 4 × 0 + 5 = 5; and so on.
Equation (4.1) is a condition on the variable x. It states that the value of the expression
(4x + 5) is 65. The condition is satisfied when x = 15. It is the solution to the equation
4x + 5 = 65. When x = 5, 4x + 5 = 25 and not 65. Thus x = 5 is not a solution to the
equation. Similarly, x = 0 is not a solution to the equation. No value of x other than 15
satisfies the condition 4x + 5 = 65.
The value of the expression (10y – 20) depends on the value of y. V erify this by
giving five different values to y and finding for each y the value of (10 y – 20). From
the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50?
If there is no solution, try giving more values to y and find whether the condition
10y – 20 = 50 is met.
TRY THESE
Rationalised 2023-24
SIMPLE EQUATIONS 61
4.4 WHAT EQUATION IS?
In an equation there is always an equality sign. The equality sign shows that the value of
the expression to the left of the sign (the left hand side or LHS) is equal to
the value of the expression to the right of the sign (the right hand side or RHS). In
equation (4.1), the LHS is (4x + 5) and the RHS is 65. In equation (4.2), the LHS is
(10y – 20) and the RHS is 50.
If there is some sign other than the equality sign between the LHS and the RHS, it is
not an equation. Thus, 4x + 5 > 65 is not an equation.
It says that, the value of (4x + 5) is greater than 65.
Similarly, 4x + 5 < 65 is not an equation. It says that the value of (4x + 5) is smaller
than 65.
In equations, we often find that the RHS is just a number. In Equation (4.1), it is 65
and in equation (4.2), it is 50. But this need not be always so. The RHS of an equation may
be an expression containing the variable. For example, the equation
4x + 5 = 6x – 25
has the expression (4x + 5) on the left and (6x – 25) on the right of the equality sign.
In short, an equation is a condition on a variable. The condition is that two
expressions should have equal value. Note that at least one of the two expressions
must contain the variable.
We also note a simple and useful property of equations. The equation 4x +5 = 65 is
the same as 65 = 4x + 5. Similarly, the equation 6x – 25 = 4x +5 is the same as
4x + 5 = 6x – 25. An equation remains the same, when the expressions on the left
and on the right are interchanged. This property is often useful in solving equations.
EXAMPLE 1 Write the following statements in the form of equations:
(i) The sum of three times x and 11 is 32.
(ii) If you subtract 5 from 6 times a number, you get 7.
(iii) One fourth of m is 3 more than 7.
(iv) One third of a number plus 5 is 8.
SOLUTION
(i) Three times x is 3x.
Sum of 3x and 11 is 3x + 11. The sum is 32.
The equation is 3x + 11 = 32.
(ii) Let us say the number is z; z multiplied by 6 is 6z.
Subtracting 5 from 6z, one gets 6z – 5. The result is 7.
The equation is 6z – 5 = 7
Rationalised 2023-24
MATHEMATICS 62
(iii) One fourth of m is
m
4
.
It is greater than 7 by 3. This means the difference (
m
4
– 7) is 3.
The equation is
m
4
– 7 = 3.
(iv) Take the number to be n. One third of n is
n
3
.
This one-third plus 5 is
n
3
+ 5. It is 8.
The equation is
n
3
+ 5 = 8.
EXAMPLE 2 Convert the following equations in statement form:
(i) x – 5 = 9 (ii) 5p = 20 (iii) 3n + 7 = 1 (iv)
m
5
– 2 = 6
SOLUTION (i) Taking away 5 from x gives 9.
(ii) Five times a number p is 20.
(iii) Add 7 to three times n to get 1.
(iv) Y ou get 6, when you subtract 2 from one-fifth of a number m.
What is important to note is that for a given equation, not just one, but many statement
forms can be given. For example, for Equation (i) above, you can say:
Subtract 5 from x, you get 9.
or The number x is 5 more than 9.
or The number x is greater by 5 than 9.
or The difference between x and 5 is 9, and so on.
EXAMPLE 3Consider the following situation:
Raju’s father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years
old. Set up an equation to find Raju’s age.
SOLUTION We do not know Raju’s age. Let us take it to be y years. Three times
Raju’s age is 3y years. Raju’s father’s age is 5 years more than 3y; that
is, Raju’s father is (3y + 5) years old. It is also given that Raju’s father
is 44 years old.
Therefore, 3y + 5 = 44 (4.3)
This is an equation in y. It will give Raju’s age when solved.
EXAMPLE 4A shopkeeper sells mangoes in two types of boxes, one small and one
large. A large box contains as many as 8 small boxes plus 4 loose mangoes.
Set up an equation which gives the number of mangoes in each small box.
The number of mangoes in a large box is given to be 100.
SOLUTION Let a small box contain m mangoes. A large box contains 4 more than 8
times m, that is, 8m + 4 mangoes. But this is given to be 100. Thus
8m + 4 = 100 (4.4)
Y ou can get the number of mangoes in a small box by solving this equation.
Write atleast one other form for
each equation (ii), (iii) and (iv).
TRY THESE
Rationalised 2023-24
Page 5
SIMPLE EQUATIONS 59
4.1 A MIND-READING GAME!
The teacher has said that she would be starting a new chapter in
mathematics and it is going to be simple equations. Appu, Sarita
and Ameena have revised what they learnt in algebra chapter in
Class VI. Have you? Appu, Sarita and Ameena are excited because
they have constructed a game which they call mind reader and they
want to present it to the whole class.
The teacher appreciates their enthusiasm and invites them to present their game. Ameena
begins; she asks Sara to think of a number , multiply it by 4 and add 5 to the product. Then,
she asks Sara to tell the result. She says it is 65. Ameena instantly declares that the number
Sara had thought of is 15. Sara nods. The whole class including Sara is surprised.
It is Appu’s turn now . He asks Balu to think of a number, multiply it by 10 and subtract
20 from the product. He then asks Balu what his result is? Balu says it is 50. Appu
immediately tells the number thought by Balu. It is 7, Balu confirms it.
Everybody wants to know how the ‘mind reader’ presented by Appu, Sarita and
Ameena works. Can you see how it works? After studying this chapter and chapter 12,
you will very well know how the game works.
4.2 SETTING UP OF AN EQUATION
Let us take Ameena’s example. Ameena asks Sara to think of a number . Ameena does not
know the number. For her, it could be anything 1, 2, 3, . . ., 11, . . . , 100, . . . . Let us
denote this unknown number by a letter, say x. Y ou may use y or t or some other letter in
place of x. It does not matter which letter we use to denote the unknown number Sara has
thought of. When Sara multiplies the number by 4, she gets 4x. She then adds 5 to the
product, which gives 4x + 5. The value of (4x + 5) depends on the value of x. Thus
if x = 1, 4x + 5 = 4 ×1 + 5 = 9. This means that if Sara had 1 in her mind, her result would
have been 9. Similarly, if she thought of 5, then for x = 5, 4x + 5 = 4 × 5 + 5 = 25; Thus
if Sara had chosen 5, the result would have been 25.
Chapter 4
Simple
Equations
Rationalised 2023-24
MATHEMATICS 60
To find the number thought by Sara let us work backward from her answer 65. We
have to find x such that
4x + 5 = 65 (4.1)
Solution to the equation will give us the number which Sara held in her mind.
Let us similarly look at Appu’s example. Let us call the number Balu chose as y. Appu
asks Balu to multiply the number by 10 and subtract 20 from the product. That is, from y,
Balu first gets 10y and from there (10y – 20). The result is known to be 50.
Therefore, 10y – 20 = 50 (4.2)
The solution of this equation will give us the number Balu had thought of.
4.3 REVIEW OF WHAT WE KNOW
Note, (4.1) and (4.2) are equations. Let us recall what we learnt about equations in
Class VI. An equation is a condition on a variable. In equation (4.1), the variable is x;
in equation (4.2), the variable is y.
The word variable means something that can vary, i.e. change. A variable takes on
different numerical values; its value is not fixed. Variables are denoted usually by
letters of the alphabets, such as x, y, z, l, m, n, p, etc. From variables, we form
expressions. The expressions are formed by performing operations like addition, subtraction,
multiplication and division on the variables. From x, we formed the expression (4x + 5).
For this, first we multiplied x by 4 and then added 5 to the product. Similarly, from y, we
formed the expression (10y – 20). For this, we multiplied y by 10 and then subtracted 20
from the product. All these are examples of expressions.
The value of an expression thus formed depends upon the chosen value of the variable.
As we have already seen, when x = 1, 4x + 5 = 9; when x = 5, 4x + 5 = 25. Similarly,
when x = 15, 4 x + 5 = 4×15 + 5 = 65;
when x = 0, 4 x + 5 = 4 × 0 + 5 = 5; and so on.
Equation (4.1) is a condition on the variable x. It states that the value of the expression
(4x + 5) is 65. The condition is satisfied when x = 15. It is the solution to the equation
4x + 5 = 65. When x = 5, 4x + 5 = 25 and not 65. Thus x = 5 is not a solution to the
equation. Similarly, x = 0 is not a solution to the equation. No value of x other than 15
satisfies the condition 4x + 5 = 65.
The value of the expression (10y – 20) depends on the value of y. V erify this by
giving five different values to y and finding for each y the value of (10 y – 20). From
the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50?
If there is no solution, try giving more values to y and find whether the condition
10y – 20 = 50 is met.
TRY THESE
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SIMPLE EQUATIONS 61
4.4 WHAT EQUATION IS?
In an equation there is always an equality sign. The equality sign shows that the value of
the expression to the left of the sign (the left hand side or LHS) is equal to
the value of the expression to the right of the sign (the right hand side or RHS). In
equation (4.1), the LHS is (4x + 5) and the RHS is 65. In equation (4.2), the LHS is
(10y – 20) and the RHS is 50.
If there is some sign other than the equality sign between the LHS and the RHS, it is
not an equation. Thus, 4x + 5 > 65 is not an equation.
It says that, the value of (4x + 5) is greater than 65.
Similarly, 4x + 5 < 65 is not an equation. It says that the value of (4x + 5) is smaller
than 65.
In equations, we often find that the RHS is just a number. In Equation (4.1), it is 65
and in equation (4.2), it is 50. But this need not be always so. The RHS of an equation may
be an expression containing the variable. For example, the equation
4x + 5 = 6x – 25
has the expression (4x + 5) on the left and (6x – 25) on the right of the equality sign.
In short, an equation is a condition on a variable. The condition is that two
expressions should have equal value. Note that at least one of the two expressions
must contain the variable.
We also note a simple and useful property of equations. The equation 4x +5 = 65 is
the same as 65 = 4x + 5. Similarly, the equation 6x – 25 = 4x +5 is the same as
4x + 5 = 6x – 25. An equation remains the same, when the expressions on the left
and on the right are interchanged. This property is often useful in solving equations.
EXAMPLE 1 Write the following statements in the form of equations:
(i) The sum of three times x and 11 is 32.
(ii) If you subtract 5 from 6 times a number, you get 7.
(iii) One fourth of m is 3 more than 7.
(iv) One third of a number plus 5 is 8.
SOLUTION
(i) Three times x is 3x.
Sum of 3x and 11 is 3x + 11. The sum is 32.
The equation is 3x + 11 = 32.
(ii) Let us say the number is z; z multiplied by 6 is 6z.
Subtracting 5 from 6z, one gets 6z – 5. The result is 7.
The equation is 6z – 5 = 7
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MATHEMATICS 62
(iii) One fourth of m is
m
4
.
It is greater than 7 by 3. This means the difference (
m
4
– 7) is 3.
The equation is
m
4
– 7 = 3.
(iv) Take the number to be n. One third of n is
n
3
.
This one-third plus 5 is
n
3
+ 5. It is 8.
The equation is
n
3
+ 5 = 8.
EXAMPLE 2 Convert the following equations in statement form:
(i) x – 5 = 9 (ii) 5p = 20 (iii) 3n + 7 = 1 (iv)
m
5
– 2 = 6
SOLUTION (i) Taking away 5 from x gives 9.
(ii) Five times a number p is 20.
(iii) Add 7 to three times n to get 1.
(iv) Y ou get 6, when you subtract 2 from one-fifth of a number m.
What is important to note is that for a given equation, not just one, but many statement
forms can be given. For example, for Equation (i) above, you can say:
Subtract 5 from x, you get 9.
or The number x is 5 more than 9.
or The number x is greater by 5 than 9.
or The difference between x and 5 is 9, and so on.
EXAMPLE 3Consider the following situation:
Raju’s father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years
old. Set up an equation to find Raju’s age.
SOLUTION We do not know Raju’s age. Let us take it to be y years. Three times
Raju’s age is 3y years. Raju’s father’s age is 5 years more than 3y; that
is, Raju’s father is (3y + 5) years old. It is also given that Raju’s father
is 44 years old.
Therefore, 3y + 5 = 44 (4.3)
This is an equation in y. It will give Raju’s age when solved.
EXAMPLE 4A shopkeeper sells mangoes in two types of boxes, one small and one
large. A large box contains as many as 8 small boxes plus 4 loose mangoes.
Set up an equation which gives the number of mangoes in each small box.
The number of mangoes in a large box is given to be 100.
SOLUTION Let a small box contain m mangoes. A large box contains 4 more than 8
times m, that is, 8m + 4 mangoes. But this is given to be 100. Thus
8m + 4 = 100 (4.4)
Y ou can get the number of mangoes in a small box by solving this equation.
Write atleast one other form for
each equation (ii), (iii) and (iv).
TRY THESE
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SIMPLE EQUATIONS 63
EXERCISE 4.1
1. Complete the last column of the table.
S. Equation V alue Say, whether the Equation
No. is Satisfied. (Y es/ No)
(i) x + 3 = 0 x = 3
(ii) x + 3 = 0 x = 0
(iii) x + 3 = 0 x = – 3
(iv) x – 7 = 1 x = 7
(v) x – 7 = 1 x = 8
(vi) 5x = 25 x = 0
(vii) 5x = 25 x = 5
(viii) 5x = 25 x = – 5
(ix)
m
3
= 2 m = – 6
(x)
m
3
= 2 m = 0
(xi)
m
3
= 2 m = 6
2. Check whether the value given in the brackets is a solution to the given equation
or not:
(a) n + 5 = 19 (n = 1) (b) 7n + 5 = 19 (n = – 2) (c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1) (e) 4p – 3 = 13 (p = – 4) (f) 4p – 3 = 13 (p = 0)
3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4
4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9. (ii) 2 subtracted from y is 8.
(iii) Ten times a is 70. (iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15. (vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
5. Write the following equations in statement forms:
(i) p + 4 = 15 (ii) m – 7 = 3 (iii) 2m = 7 (iv)
m
5
= 3
(v)
3
5
m
= 6 (vi) 3p + 4 = 25 (vii) 4p – 2 = 18 (viii)
p
2
+ 2 = 8
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