## Algebra Chapter 3 Laws Exercise 3 Solved Example Problems

**Algebra Introduction
**

In this chapter, we shall discuss some phenomena known to you in a different manner.

Actually, out of the four basic operations namely addition, subtraction, multiplication, and division only addition and multiplication (and in some special cases division also) abide by some laws.

**Math Solution Of Class 7 Wbbse**

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These laws are:

- Commutative law
- Associative law and
- Distributive law.

These laws play an important role in higher mathematics, especially in classical and modern algebra and also in the theory of sets. Let us have a look at these laws, starting with very simple examples.

**Wbbse Class 7 Maths Solutions**

**Commutative law **

You know that, 7 + 8 = 15

Again, 8 + 7 = 15

Therefore, 7 + 8 = 8 + 7.

So, it may be said that, if a and 6 be any two integers then, a+b = b+a.

In the above equation, we see that on the left-hand side a is in the first position and b is after a while on the right-hand side b is in the first position and a is after b.

This law, where a and b interchange their positions is known as commutative law.

Also, you know that, 7 x 4 = 28 and also, 4 x 7 = 28

Therefore, 7 x 4 = 4 x 7.

**Math Solution Of Class 7 Wbbse**

So, it may be said that, if a and b are two integers then, axb= bxa.

Thus, the commutative law is also applicable in the case of multiplication.

**Illustration: **1. Suppose, you have 6 mangoes and your brother has 4 mangoes. Thus the total number of mangoes can be obtained by adding your mangoes with your brother’s share i.e., 4 + 6 = 10.

Again we can find the total number of mangoes by adding your brother’s mangoes with your mangoes i.e., 6 + 4 = 10. Therefore, 4 + 6 = 6 + 4.

Hence, we can say that the addition of 6 with 4 and the addition of 4 with 6 give the same result. Thus if 4 and 6 change their places, their sum remains unaltered.

2. Suppose, in a garden, there are 4 rows of trees and in each row, there are 3 trees.

Then the total number of trees in the garden = 3 x 4 = 12.

Also in another garden, there are 3 rows of trees and in each row, there are 4 trees.

Then the total number of trees in the garden is 4 x 3 = 12. Thus we see that, 3×4 = 4×3.

**Associative law
**

The sum 4 + 5 + 7 may be written in different ways.

If, at first 5 is added with 4 and then 7 is added with this result of addition then it may be written as (4 + 5) + 7 = 9 + 7=16; again, if at first 7 is added with 5 and then the result of this addition is added with 4 then it may be written as 4 + (5 + 7) = 4 + 12 = 16.

So, it is clear that (4 + 5) + 7 = 4 + (5 + 7).

Therefore, in general, if a, 6 and c be three integers then, (a + 6) + c = a + (6 + c).

This is known as the associative law of addition.

The associative law is also applicable in the case of multiplication.

For example, we see that (2 x 3) x 4 = 6 x 4 = 24 and 2 x (3 x 4) = 2×12 = 24.

Therefore, in general, if a, b and c be three integers then it may be written in general that, (a x b) x c = a x (b x c).

**Math Solution Of Class 7 Wbbse**

**Illustration: **1. Suppose, you have 4 pens. Your brother has 3 pens and your sister has 2 pens.

At first, adding your and your brother’s pens the total number of pens becomes 4 + 3 = 7.

Now adding this number the number of your sister’s pens the total number of pens becomes (4 + 3) + 2 = 7 + 2 = 9.

Again, adding your brother’s and sister’s pens, the total number of pens = 3+2 = 5.

Now adding with your pens, the sum of your brother’s and sister’s pens the total number of pens becomes 4+(3+2) = 4+5 – 9.

Thus counting the total number of pens in both ways we get, (4 + 3) + 2 = 4 + (3 + 2).

2. Suppose, a boy can do 5 sums per day and we are to find the number of sums that can be done by the boy in 4 weeks.

Since in 1 day the boy can do 5 sums therefore in 1 week (i.e., 7 days) the boy can do 5 x 7 = 35 sums.

Thus in 4 weeks the boy can do (5 x 7) x 4 = 35 x 4 = 140 sums.

Also we may think the issue from another point of view. The total number of days in 4 weeks = 7 x 4 = 28.

Since in 1 day the boy can do 5 sums therefore in 7 x 4 days the boy can do 5 x (7 x 4) = 5 x 28 = 140 sums.

Thus the total number of sums done by the boy is same in both cases.

**Math Solution Of Class 7 Wbbse**

**Distributive law
**

You know that, 15 x 6 = 90

Again, since 6 = 4 + 2

∴ 15 x 6 = 15 x (4 + 2) = 15 x 4 + 15 x 2 = 60 + 30 = 90.

So, it may be said in general that, if a, b, and c are three integers then,

a x (b+c) = a x b + a x c and a x (b-c) = a x b -a x c.

So, in the case of multiplication, the distributive law is always applicable.

The distributive law of multiplication may also have the following forms:

(a+b) x c=a x c+b x c and (a-b) x c=a x c- b x c.

Let us now see that, how the distributive law may be applied in the case of division.

You know that, 35 ÷ 5 = 7 and 35 = 25 + 10

Now, (25+10) ÷ 5 = 25÷5 + 10 ÷ 5 = 5 + 2 = 7.

So, it may be said in general that, if a, b, and c are three integers then,

(a+b) ÷ c= a ÷ c + b ÷ c and (a-b) ÷ c= a ÷ c- b ÷c.

This distributive law of division is only applicable for the right-hand side.

If a, b, and c be three integers then it may be shown that,

c ÷ (a + b) ≠ c ÷ a + c ÷ b and c ÷ (a-b) ≠ c ÷a – c ÷ b.

Therefore, the distributive law of division is not applicable in this case.

**Illustration:** Suppose you can do 12 sums daily and your brother can do 8 sums daily.

In 10 days what will be the total number of sums done by you? We may solve this problem in two ways.

In one day the total sum done by you and your brother is (12 + 8) = 20.

Hence, the number of sums done in 10 days = {(12 + 8) x 10} = 20 x 10 = 200.

Again in 10 days, you can do 12 x 10 – 120 sums and in 10 days your brother can do 8 x 10 = 80 sums.

Hence, the total number of sums done by you and your brother in 10 days = 12 x 10 + 8 x 10 = 120 + 80 = 200.

Thus, we can say that, (12 + 8) x 10 = 12 x 10 + 8 x 10.

**Use of brackets: **In algebra brackets are used in the same manner as in arithmetic. In general, four types of brackets are used.

**They are:**

- Line bracket —
- First bracket ( )
- Second bracket { }
- Third bracket [ ]

If in a question all the brackets are present then first of all, the operation under the line bracket is performed. After that, the operation under first bracket, second bracket and third bracket are done in series after finding values inside the corresponding brackets.

## Algebra Chapter 3 Laws Exercise 3 Some Problems With Commutative, Associative, And Distributive Laws And The Use Of Brackets

**Example 1. Prove on the number axis that 4 + 3 = 3 + 4.**

Solution:

Moving 4 units to the right of 0

we get to point 4. After this, moving 3 more units in the same direction we get point 7.

Thus, 4 + 3 = 7.

Again, at first moving 3 units to the right of 0 we get point 3. After this, moving 4 more units in the same direction we get to point 7. Thus, 3 + 4 = 7.

Hence, 4 + 3 = 3 + 4 (proved).

**Example 2. Prove that : 4 x 5 = 5 x 4.
**

Solution:

4 x 5 = 5 + 5 + 5 + 5 = 20

5 x 4=4+4+4+4+4=20

Hence, 4 x 5 = 5 x 4 (proved).

**Wbbse Class 7 Maths Solutions**

**Example 3. Rewrite 2 + 3 + 5 in two different ways such that a single bracket is used in each case.**

Solution: 2 + 3 + 5 = (2 + 3) + 5 and 2 + 3 + 5 = 2 + (3 + 5).

**Example 4. Rewrite 5 x 6 x 7 in two different ways such that a single bracket is used in each case.
**

Solution: 5 x 6 x 7 = (5 x 6) x7 and 5 x 6 x 7 = 5 x (6 x 7).

**Example 5. In each of the following cases, determine which law of which operation has been used.**

- a + b = b + a.
- a x b – b x a,
- (a + y) + z – x + (y + z).
- m x (n x p) = (m x n) x p.
- a x (6 + c) = a x b + a x c.
- (a + b) ÷ c = a ÷ c + b ÷ c.

Solution:

- Commutative law of addition.
- Commutative law of multiplication.
- Associative law of addition.
- Associative law of multiplication.
- Distributive law of multiplication,
- Distributive law of division.

**Example 6. Add the sum of 3 and 7 with 5 and find the result.
**

Solution:

**Given:**

5 + (3 + 7) = 5 + 10 = 15

∴ 15.

The Result is 15.

**Example 7. From 50, subtract the sum of 25 and 15 and find the result.
**

Solution:

**Given:**

From 50, subtract the sum of 25 and 15

50 – (25 + 15) = 50 – 40 = 10

∴ 10.

The Result is 10.

**Wbbse Class 7 Maths Solutions**

**Example 8. Using the result (a + b). c = a.c + b.c prove that,**

- (a – b).c=a.c -b.c.
- a. (b-c) = a.b- a.c.

Solution:

**Given:**

Using the result (a + b). c = a.c + b.c

1. (a – b).c + b.c = x.c + b.c [Assuming x = a – b]= (x + b).c [Using the given relation]

= {(a – b) + b}.c [Putting the value of x] = [a + {(-b + b)}].c [By associative property]

= [a + o].c [According to the definition of zero]

= a.c + o.c [Using the given relation]

= a.c + o = a.c [According to the definition of zero]

Now, a.c – b.c = {(a – b).c + b.c} – b.c [Putting the value of a.c]

= (a – b).c + {b.c – b.c} [By associative property]

= (a – b).c + o = (a – b).c

Hence, it is proved that, (a – b).c = a.c – b.c.

2. a.(b – c)= (b – c).a [By commutative property]

= b.a – c.a [From the proof of 1st part]= a.b – a.c [By commutative property]

Hence, it is proved that a.(b – c) = a.b – a.c.

**Example 9.****Prove that,**

**(a + b + c).x = a.x+ x + c.x.****(a + b + c) = x.a+ x.b + x.c.**

**Solution:**

**Wbbse Class 7 Maths Solutions**

**Example **10. Prove that, (a + b) ÷ x = (a ÷ x) + (b ÷ x).

**Solution:** [(a ÷ x) + (b ÷ x)].x

Hence, (a + b) ÷ x = (a ÷ x) + (b ÷ x) [Proved].

**Example 11. Simplify xyz (x + y + z).**

Solution:

**The given expression = xyz (x + y + z)**

= xyz X x + xyz X y + xyz X z

∴ x^{2}yz + xy^{2}z + xyz^{2}

xyz (x + y + z) = x^{2}yz + xy^{2}z + xyz^{2}

**Example 12. Simplify : a(a ^{2} + 5a – 6).**

Solution:

**The given expression**

= a(a^{2} + 5a – 6)

= a x a^{2} + a x 5a – a x 6 = a^{3} + 5a^{2} – 6a

∴ a^{3} ** + **5a

^{2}

**6a.**

**–**a(a^{2} + 5a – 6) = a^{3} ** + **5a

^{2}

**6a.**

**–****Wbbse Class 7 Maths Solutions**

**Example 13. Simplify: \(\frac{x}{x-y}\)+\(\frac{y}{y-x}\)**

Solution:

**The given expression**

**Example 14. Simplify **

Solution:

**The given expression**

∴ 9

**Example 15. Simplify **

**Wbbse Class 7 Maths Solutions**

**Solution :**

**The given expression**

∴ – 3

**Class Vii Math Solution Wbbse**

**Example 16. Simplify: **

Solution:

**The given expression**

∴ y-x