Rational Numbers

(i) 5x = 30

X =

X = 6

x = 6.

(ii) = 5

5x – 1 = 10

5x = 11

X =

X = 2.2

x = 2.2

(iii) = 0

7x +20 = 0

7x = -20

X = -

x = -

(iv) + - = 0

X = 0

x = 0

This is an algebraic equation which is univariate as it contains only one variable, that is, x.

We have

5x = 30

We need to solve this equation.

That is, we need to find the value of variable x.

We will use any kind of mathematical operation (addition, subtraction, multiplication or division) in order to find x.

Note that, 5 is multiplied to x on L.H.S of the equation.

So, divide 5 on both sides of the equation so that we have x idle on L.H.S.

⇒ x = 6

Thus, we have solved the equation and x = 6.

This is an algebraic equation which is univariate as it contains only one variable, that is, x.

We have

We need to solve this equation.

That is, we need to find the value of variable x.

We will use any kind of mathematical operation (addition, subtraction, multiplication or division) in order to find x.

Here, we can simplify the equation by taking L.C.M of the denominators on L.H.S of the equation.

Denominator of 2x = 1

Denominator of = 2

L.C.M (1, 2) = 2

Therefore, multiplying 2 by 2x and 1 by (x – 1), we get

Now start simplifying it as much as possible. For instance, multiply 2 on both sides,

⇒ (2 × 2x) + (1 × (x – 1)) = 10

⇒ 4x + (x – 1) = 10

⇒ 4x + x – 1 = 10

Constants of the variable can be operated, so

⇒ (4x + x) – 1 = 10

⇒ 5x – 1 = 10

Take the constant 1 on R.H.S. The negative sign on L.H.S will become positive on R.H.S.

⇒ 5x = 10 + 1

⇒ 5x = 11

Divide 5 on both sides to leave x idle.

Thus, we have solved the equation and .

This is an algebraic equation which is univariate as it contains only one variable, that is, x.

We have

We need to solve this equation.

That is, we need to find the value of variable x.

We will use any kind of mathematical operation (addition, subtraction, multiplication or division) in order to find x.

Here, we can simplify the equation by taking L.C.M of the denominators on L.H.S of the equation.

Denominator of = 5

Denominator of = 7

L.C.M (5, 7) = 35

Therefore, multiplying 7 by x and 5 by 2, we get

Now, simplify it as much as possible. Cross multiplying, 35 by x and 10 by (7 × x) + (5 × 2), we get

⇒ 10 × ((7 × x) + (5 × 2)) = 35 × x

⇒ 10 × (7x + 10) = 35x

By distributive law, multiply 10 by 7x and by 10.

⇒ (10 × 7x) + (10 × 10) = 35x

⇒ 70x + 100 = 35x

Take x’s on one side and constants on the other. Here, shift 35x on L.H.S (positive sign of 35 will become negative) and 100 on R.H.S (positive sign of 100 will become negative).

⇒ 70x – 35x = -100

⇒ 35x = -100

Divide 35 on both sides to leave x idle.

Thus, we have solved the equation and .

This is an algebraic equation which is univariate as it contains only one variable, that is, x.

We have

We need to solve this equation.

That is, we need to find the value of variable x.

We will use any kind of mathematical operation (addition, subtraction, multiplication or division) in order to find x.

Here,

Shift the constant � from L.H.S to R.H.S, it’s positive sign will become negative.

Multiply by 4 on both sides so that x can be left idle.

⇒ x = 0

Thus, we have solved the equation and x = 0.

According to the question,

Let’s assume the number (to be calculated) to be ‘x’.

So,

X + = 0

X = -

x = -

Let x be the number to which when 2/9 is added, we shall get 0.

We need to find x.

So, formulate it into a mathematical equation using operators and numbers.

x is added to 2/9 can be written as,

Now, when x is added to 2/9, we get 0. That is,

To solve for x, shift 2/9 from L.H.S to R.H.S, positive sign of 2/9 will become negative.

Thus, we shall get 0 by adding with .

Let x be the number to which when -9/8 is added, we shall get 0.

We need to find x.

So, formulate it into a mathematical equation using operators and numbers.

x is added to -9/8 can be written as,

Or,

Now, when x is added to -9/8, we get 0. That is,

To solve for x, we need to shift 9/8 to R.H.S. Negative sign of 9/8 will become positive.

Thus, we shall get 0 by adding with .

According to the question,

Let’s assume the number (to be calculated) to be ‘x’.

So,

X +( - ) = 0

X =

x =

Let x be the number to which when is added, we shall get 0.

We need to find x.

So, formulate it into a mathematical equation using operators and numbers.

x is added to can be written as,

Or,

Or,

Now, when x is added to , we get 0. That is,

To solve for x, we need to shift 5/2 to R.H.S. Positive sign of 9/8 will become negative.

Thus, we shall get 0 by adding with .

According to the question,

Let’s assume the number (to be calculated) to be ‘x’.

So,

X +( - ) = 0

X =

x =

Let x be the number to which when is multiplied, we shall get 1.

We need to find x.

So, formulate it into a mathematical equation using operators and numbers.

x is multiplied with can be written as,

Or,

Now, when x is multiplied with , we get 1. That is,

To solve for x, multiply 8 on both sides.

⇒ 5x = 8

Dividing 5 on both sides to leave x idle, we get

Thus, we shall get 1 by multiplying with .

According to the question,

Let’s assume the number (to be calculated) to be ‘x’.

So,

x = 1

X =

x =

According to the question,

Let’s assume the number (to be calculated) to be ‘x’.

So,

x(-) = 1

X = -

x = - 3

Let x be the number to which when is multiplied, we shall get 1.

We need to find x.

So, formulate it into a mathematical equation using operators and numbers.

x is multiplied with can be written as,

Or,

Or,

Now, when x is multiplied with , we get 1. That is,

To solve for x, multiply -3 on both sides.

⇒ x = -3

Thus, we shall get 1 by multiplying -3 with .

As per the Commutative Law of multiplication,

a b = b a = ab = ba

The product is , which will result in calculating either of the following expressions :-

•

•

As per the Associative Law of multiplication,

a (b c d) = (ab c) d = abcd = abcd

The product is , which will result in calculating either of the following expressions :-

•

•[{

Hence proved, both the laws (Associative & Commutative) of multiplication.

We must understand that,

Commutative law, in mathematics, either of two laws relating to number operations of addition and multiplication, stated symbolically: a + b = b + a and ab = ba.

Associative law, in mathematics, either of two laws relating to number operations of addition and multiplication, stated symbolically: (a + b) + c = a + (b + c), and a(bc) = (ab)c.

We have,

We can write as,

According to the commutative law of multiplication,

So,

Also, by associative law of multiplication,

We can solve it by any way.

So,

Thus, the answer is .

A. 7x=14

⇒ x = 2

It can be written in rational form i.e. form as:

B. 4p = -32

It can be written in rational form i.e. form as:

C. 11x = 0

x = 0

It can be written in rational form i.e. form as:

D. 5m=3

It can be written in rational form i.e. form as:

E. 9y + 18 =0

9y = -18

Y = -2

It can also be written in a rational form as

F. t = 8 - 12

t=-4

It can be written in rational form i.e. form as:

G. 6y-y=5

⇒ 5y=5

It can be written in rational form i.e. form as:

H. 2x + = ______

Let ? = 1/2

So 2x + 1/2 = 1/2

2x = 1/2 - 1/2

2x = 0

x = 0

It can be written in rational form i.e. form as:

A. 7x=14

⇒ x = 2

It can be written in rational form i.e. form as:

B. 4p = -32

It can be written in rational form i.e. form as:

C. 11x = 0

x = 0

It can be written in rational form i.e. form as:

D. 5m=3

It can be written in rational form i.e. form as:

E. 9y + 18 =0

9y = -18

Y = -2

It can also be written in a rational form as

F. t = 8 - 12

t=-4

It can be written in rational form i.e. form as:

G. 6y-y=5

⇒ 5y=5

It can be written in rational form i.e. form as:

H. 2x + = ______

Let ? = 1/2

So 2x + 1/2 = 1/2

2x = 1/2 - 1/2

2x = 0

x = 0

It can be written in rational form i.e. form as:

Consider –(-y)

= y

= RHS

Hence –(-y)=y

Consider –(-y)

= y

= RHS

Hence –(-y)=y

A. Put in 2x + 5.

The L.C.M of 1 and 4 is 4.

B. Put

C. Put

D. Let x = 5

Now –(-5) = 5

Hence x = -(-x)

A. Put in 2x + 5.

The L.C.M of 1 and 4 is 4.

B. Put

C. Put

D. Let x = 5

Now –(-5) = 5

Hence x = -(-x)

A. Let the number in the box be x.

B. Let the number be x.

C. Let the number be x.

D. Let the number be x.

E. Let the number be x.

F. Let the number be x.

A. Let the number in the box be x.

B. Let the number be x.

C. Let the number be x.

D. Let the number be x.

E. Let the number be x.

F. Let the number be x.

Reciprocal of

∴ the product will be-

Reciprocal of

∴ the product will be-

(i) (Using Commutative law)

(Using Associative law)

(ii) (Using Commutative law)

(i) (Using Commutative law)

(Using Associative law)

(ii) (Using Commutative law)

We know,

Now, multiplying and dividing both by 5, we get-

Therefore, the four in-between rational numbers can be-

We know,

Now, multiplying and dividing both by 5, we get-

Therefore, the four in-between rational numbers can be-

To find the in between rational numbers we’ll first make the denominator same.

So, here (multiplying and dividing by 2)

Similarly, (multiplying and dividing by 5)

Now, we can easily find the between rational numbers by taking the minimum and maximum values of numerator from given values. Here minimum value is -6 and maximum is 5.

∴ the rational numbers, which lies in between are-

To find the in between rational numbers we’ll first make the denominator same.

So, here (multiplying and dividing by 2)

Similarly, (multiplying and dividing by 5)

Now, we can easily find the between rational numbers by taking the minimum and maximum values of numerator from given values. Here minimum value is -6 and maximum is 5.

∴ the rational numbers, which lies in between are-

To find the in between numbers, we’ll first make the denominator same which will be the LCM of denominator of both the values.

Now, LCM(3,5)=15

∴ on multiplying and dividing by 5, we get-

And similarly,

On multiplying and dividing by 3, we get-

Now, the rational numbers in between will be-

But here we are getting 3 numbers only

To get more number we can now make the denominator different, likewise if we again multiply and divide the no. by 2, we get-

Therefore, the five numbers can be-

To find the in between numbers, we’ll first make the denominator same which will be the LCM of denominator of both the values.

Now, LCM(3,5)=15

∴ on multiplying and dividing by 5, we get-

And similarly,

On multiplying and dividing by 3, we get-

Now, the rational numbers in between will be-

But here we are getting 3 numbers only

To get more number we can now make the denominator different, likewise if we again multiply and divide the no. by 2, we get-

Therefore, the five numbers can be-

To find the in between numbers, we’ll first make the denominator same which will be the LCM of denominator of both the values.

Now, LCM(3,7)=21

∴ on multiplying and dividing by 7, we get-

And similarly,

On multiplying and dividing by 3, we get-

Now, the five rational numbers in between can be-

To find the in between numbers, we’ll first make the denominator same which will be the LCM of denominator of both the values.

Now, LCM(3,7)=21

∴ on multiplying and dividing by 7, we get-

And similarly,

On multiplying and dividing by 3, we get-

Now, the five rational numbers in between can be-

To find the in between numbers, we’ll first make the denominator same which will be the LCM of denominator of both the values.

Now, LCM(4,2)= 4

∴ on multiplying and dividing by 1, we get-

And similarly,

On multiplying and dividing by 2, we get-

To get more number we can now make the denominator different from LCM, likewise if we again multiply and divide the no. by 6, we get-

Therefore, the five numbers can be-

To find the in between numbers, we’ll first make the denominator same which will be the LCM of denominator of both the values.

Now, LCM(4,2)= 4

∴ on multiplying and dividing by 1, we get-

And similarly,

On multiplying and dividing by 2, we get-

To get more number we can now make the denominator different from LCM, likewise if we again multiply and divide the no. by 6, we get-

Therefore, the five numbers can be-