**Expressions and equations : **

This is one of the important basic stuff of algebra in math.

Students who would like to learn solving word problems in math must study this basic stuff.

Let us see, how to write verbal phrase into expressions.

If there is any unknown value, in the given verbal phrase, replace it by some English alphabet.

**Question 1 :**

Write the given verbal phrase into mathematical expression

"The sum of 5 times a number and 8"

**Answer :**

5x + 8

**Question 2 :**

Write the given verbal phrase into mathematical expression

"2 times the sum of 7 times a number and 4"

**Answer :**

2(7x + 4)

**Question 3 :**

Write the given verbal phrase into mathematical expression

"7 less than 4 times a number"

**Answer :**

4x - 7

**Question 4 :**

Write the given verbal phrase into mathematical expression

"One fifth of sum of 3 times a number and 9"

**Answer :**

( 3x + 9 ) / 5

**Question 5 :**

Write the given verbal phrase into mathematical expression

"7 less than 3 times the sum of a number and 6"

**Answer :**

3(x + 6) - 7

**Question 1 :**

Evaluate the given expression for x = 3 and y = 5.

3x + 2y

**Answer :**

Plug x = 3 and y = 5 in the given expression

3(3) + 2(5) = 9 + 10 = 19

Hence, the answer is 19

**Question 2 :**

Evaluate the given expression for m = 5 and n = 2.

m² + 2mn²

**Answer :**

Plug m = 5 and n = 2 in the given expression

5² + 2(5)(2)² = 25 + 40 = 65

Hence, the answer is 65

**Question 3 :**

Evaluate the given expression for s = 5.

s² + 7s - 2

**Answer :**

Plug s = 5 in the given expression

5² + 7(5) - 2 = 25 + 35 - 2 = 58

Hence, the answer is 58

**Question 4 :**

Evaluate the given expression for m = 1/3.

18m² + 3m + 7

**Answer :**

Plug m = 1/3 in the given expression

18(1/3)² + 3(1/3) + 7 = 2 + 1 + 7 = 10

Hence, the answer is 10.

**Question 5 :**

Evaluate the given expression for m = 13.

m² + m - 54

**Answer :**

Plug m = 13 in the given expression

13² + 13 -54 = 169 + 13 - 54 = 128

Hence, the answer is 128

**Example 1 :**

The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool ?

**Solution :**

Let the breadth be x m.

The length will be (2x + 2) m.

Perimeter of swimming pool = 2(l + b) = 154 m

2(2x + 2 + x) = 154

2(3x + 2) = 154

Dividing both sides by 2,

3x + 2 = 77

Subtracting 2 on both sides, we get

3x + 2 - 2 = 77 - 2

3x = 75

On dividing 3 on both sides

x = 25

2x + 2 = 2 × 25 + 2 = 52

Hence, the breadth and length of the pool are 25 m and 52 m respectively.

Let us look at the next word problem on"Expressions and equations".

**Example 2 :**

Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

**Solution :**

Let one number be x.

Therefore, the other number will be x+ 15.

According to the question,

x + x + 15 = 95

2x + 15 = 95

Subtract 15 on both sides

2x = 95 - 15

2x = 80

Divide by 2 on both sides

x = 40

x + 15 = 40 + 15 = 55

Hence, the numbers are 40 and 55.

Let us look at the next word problem on"Expressions and equations".

**Example 3 :**

Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

**Solution :**

Let the common ratio between these numbers be x. Therefore, the numbers will be 5x and 3x respectively.

Difference between these numbers = 18

5x - 3x= 18

2x= 18

Divide by 2 on both sides

x = 9

First number = 5x= 5 × 9 = 45

Second number = 3x= 3 × 9 = 27

Hence the required numbers are 27, 45.

Let us look at the next word problem on"Expressions and equations".

**Example 4 :**

Three consecutive integers add up to 51. What are these integers?

**Solution :**

Let three consecutive integers be x, x + 1, and x+ 2.

Sum of these numbers = x + x + 1 + x + 2 = 51

3x+ 3 = 51

Subtract by 3 on both sides

3x = 51 - 3

3x = 48

Divide by 3 on both sides

x = 16

x + 1 = 17

x + 2 = 18

Hence, the consecutive integers are 16, 17, and 18.

Let us look at the next word problem on"Expressions and equations".

**Example 5 :**

The sum of three consecutive multiples of 8 is 888. Find the multiples.

**Solution :**

Let the three consecutive multiples of 8 be 8x, 8(x + 1), 8(x + 2).

Sum of these numbers = 8x + 8(x+ 1) + 8(x+ 2) = 888

8(x + x + 1 +x + 2) = 888

8 (3x + 3) = 888

Dividing by 8 both sides

3x+ 3 = 111

Subtract 3 on both sides

3x + 3 - 3 = 111 - 3

3x= 108

Divide by 3 on both sides

x = 36

First multiple = 8x = 8 × 36 = 288

Second multiple = 8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296

Third multiple = 8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304

Hence, the required numbers are 288, 296, and 304.

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