# Some Applications Solving Equations Which Have Linear Expressions on One Side and Numbers on the Other Side

#### Example

Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the numbers?

The sum of these two numbers x and x + 10 is 74.

This means that x + (x + 10) = 74.

or    2x + 10 = 74

Transposing 10 to RHS,

2x = 74 - 10

or 2x = 64

Dividing both sides by 2,

x = 32

This is one number (x) = 32

The other number is x + 10 = 32 + 10 = 42.

The desired numbers are 32 and 42.

#### Example

What should be added to twice the rational number (-7)/3 "to get" 3/7?

Twice the rational number ((-7)/3)  "is"  2 xx ((-7)/3) = (-14)/3.

Suppose x added to this number gives 3/7; i.e.,

x + ((-14)/3) = 3/7

or x - 14/3 = 3/7

or x = 3/7 + 14/3         .......(transposing 14/3 to RHS)

or x= ((3 xx 3) + (14 xx 7))/21

or x = (9 + 98)/21

or x = 107/21.

Thus, 107/21 "should be added to" 2 xx ((-7)/3) "to give" 3/7.

#### Example

The perimeter of a rectangle is 13 cm and its width is 2 3/4 cm. Find its length.

Assume the length of the rectangle to be x cm.
The perimeter of the rectangle = 2 × (length + width)

= 2 xx (x + 2 3/4)

= 2(x + 11/4)

The perimeter is given to be 13 cm. Therefore,

2(x + 11/4) = 13

or x + 11/4 = 13/2      ......(Divinding both sides by 2)

or x = 13/2 - 11/4

or x = 26/4 - 11/4

or x = 15/4

or x = 3 3/4

The length of the rectangle is 3 3/4 cm.

#### Example

The present age of Sahil's mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.

Let Sahil’s present age be x years.

 Sahil Mother Sum Present age x 3x Age 5 years later x + 5 3x + 5 4x + 10

It is given that this sum is 66 years.

Therefore,    4x + 10 = 66.

This equation determines Sahil's present age which is x years.

We transpose 10 to RHS,

4x = 66 - 10

or 4x = 56

or x = 56/4 = 14.

Thus, Sahil's present age is 14 years, and his mother's age is 42 years.

#### Example

Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of Rs. 77, how many coins of each denomination does he have?

Let the number of five-rupee coins that Bansi has be x. Then the number of two-rupee coins he has is 3 times x or 3x.

The amount Bansi has:
(i) from 5 rupee coins, Rs. 5 × x = Rs. 5x.
(ii) from 2 rupee coins, Rs. 2 × 3x = Rs. 6x

Hence the total money he has = Rs. 11x

But this is given to be Rs. 77; therefore,

11x = 77

x = 77/11 = 7

Thus, number of five-rupee coins = x = 7. and

number of two-rupee coins = 3x = 21.

#### Example

The sum of three consecutive multiples of 11 is 363. Find these multiples.
If x is a multiple of 11, the next multiple is x + 11. The next to this is x + 11 + 11 or x + 22.
So we can take three consecutive multiples of 11 as x, x+ 11, and x+ 22. It is given that the sum of these consecutive multiples of 11 is 363.

x + (x + 11) + (x + 22) = 363

or x + x + 11 + x + 22 = 363.

or 3x + 33 = 363.

or 3x = 363 - 33

or 3x = 330

or x = 330/3

or x = 110

hence, the three consecutive multiples are 110, 121, 132.

#### Example

The difference between two whole numbers is 66. The ratio of the two numbers is 2: 5. What are the two numbers?

Since the ratio of the two numbers is 2: 5, we may take one number to be 2x and the other to be 5x.
The difference between the two numbers is (5x – 2x). It is given that the difference
is 66.
Therefore,
5x - 2x = 66
or 3x = 66
or x = 22
Since the numbers are 2x and 5x, they are 2 × 22 or 44 and 5 × 22 or 110, respectively.
The difference between the two numbers is 110 – 44 = 66 as desired.

#### Example

Deveshi has a total of ₹ 590 as currency notes in the denominations of ₹ 50, ₹ 20, and ₹ 10. The ratio of the number of ₹ 50 notes and ₹ 20 notes is 3: 5. If she has a total of 25 notes, how many notes of each denomination she has?

Let the number of ₹ 50 notes and ₹ 20 notes be 3x and 5x, respectively. But she has 25 notes in total.
Therefore, the number of ₹ 10 notes = 25 - (3x + 5x) = 25 - 8x
The amount she has
from ₹ 50 notes: 3x × 50 = ₹ 150x
from ₹ 20 notes: 5x × 20 = ₹ 100x
from ₹ 10 notes: (25 - 8x) × 10 = ₹ (250 - 80x)

Hence the total money she has = 150x + 100x + (250 - 80x) = ₹ (170x + 250)
But she has ₹ 590. Therefore,

170x + 250 = 590
or 170x = 590 - 250 = 340
or x = 340/170 = 2

The number of ₹ 50 notes she has = 3x = 3 × 2 = 6.

The number of ₹ 20 notes she has = 5x = 5 × 2 = 10.

The number of ₹ 10 notes she has = 25 - 8x = 25 - (8 × 2) = 25 - 16 = 9.

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