# 3n 4 19 ## Unit 10 Section 2 : Finding the Formula for a Linear Sequence

### Example 1

Determine a formula for this sequence: First consider the differences between the terms,

As the difference is always 6, we can write,

u_n = 6n + c

As the first term is 7, we can write down the equation:

 7 = 6 × 1 + c = 6 + c c = 1

So the formula will be,

u_n = 6n + 1

We can check that this formula is correct by testing it on other terms, for example,

the 4th term = 6 × 4 + 1 = 25

which is correct.

### Example 2

Determine a formula for this sequence: First consider the differences between the terms,

The difference between each term is always 5, so the formula will be,

u_n = 5n + c

The first term can be used to form an equation to determine c:

 2 = 5 × 1 + c 2 = 5 + c c = –3

So the formula will be,

u_n = 5n – 3

Note that the constant term, c, is given by

c = first term – first difference

### Example 3

Determine a formula for the sequence:

28, 25, 22, 19, 16, 13, ... First consider the differences between the terms,

Here the difference is negative because the terms are becoming smaller.
Using the difference as –3 gives,

u_n = –3n + c

The first term is 28, so

 28 = –3 × 1 + c 28 = –3 + c c = 31

The general formula is then,

u_n = –3n + 31

or

u_n = 31 – 3n

### Exercises

Sours: https://www.cimt.org.uk/projects/mepres/book9/bk9i10/bk9_10i2.html

## Linear equations with one unknown

### Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

3*n-4-(19)=0

### Step  1  :

#### Solving a Single Variable Equation :

1.1      Solve  :    3n-23 = 0

Add  23  to both sides of the equation :
3n = 23
Divide both sides of the equation by 3:
n = 23/3 = 7.667

### One solution was found :

n = 23/3 = 7.667
Sours: https://www.tiger-algebra.com/drill/3n-4=19/

Exercise 2.1

Question 1:

Solve the following:      x – 2= 7

Given, x -2 = 7

=> x – 2 + 2= 7 + 2    [Adding 2 both sides]

=> x = 9

Question 2:

Solve of the following:    y + 3 = 10

Solution:

Given, y + 3 = 10

=> y + 3 – 3 = 10 – 3 [subtracting 3 on both side]

=> y = 7

Question 3:

Solve the following:     6 = z + 2

Given, 6 = 2 + z

=> 6 – 2 = z + 2 – 2     [Subtracting 2 both sides]

=> 4 = z

=> z = 4

Question 4:

Solve the following:     3/7 + x = 17/7

Given, 3/7 + x = 17/7

=> 3/7 + x – 3/7 = 17/7 – 3/7               [Subtracting 3/7 on both sides]

=> x = (17 - 3)/7

=> x = 14/7

=> x = 2

Question 5:

Solve the following:      6x = 12

Given, 6x = 12

=> 6x/6 = 12/6   [Dividing 6 on both side]

=> x = 2

Question 6:

Solve the following:      t/5 = 10

Given, t/5 = 10

=> t/5 * 5 = 10 * 5    [Multiply by 5 on both side]

=> t = 50

Question 7:

Solve the following:    2x/3 = 18

Given, 2x/3 = 18

=> 2x = 3 * 18

=> 2x = 54

=> x = 54/2

=> x = 27

Question 8:

Solve the following:    1.6 = y/1.5

Given, 1.6 = y/1.5

=> y = 1.6 * 1.5

=> y = 2.40

Question 9:

Solve the following:    7x – 9 = 16

Given, 7x – 9 = 16

=> 7x = 16 + 9

=> 7x = 25

=> x = 25/7

Question 10:

Solve the following:     14y – 8 = 13

Given, 14y – 8 = 13

=> 14y = 13 + 8

=> 14y = 21

=> y = 21/14

=> y = 3/2               [21 and 14 are divided by 7]

Question 11:

Solve the following:    17 + 6p = 9

Given, 17 + 6p = 9

=> 6p = 9 – 17

=> 6p = -8

=> p = -8/6

=> p = -4/3      [8 and 6 are divided 2]

Question 12:

Solve the following:     x/3 + 1 = 7/15

Given, x/3 + 1 = 7/15

=> x/3 = 7/15 – 1

=> x/3 = (7 - 15)/15            [LCM (15, 1) = 15]

=> x/3 = -8/15

=> x = 3 * (-8/15)

=> x = (-3 * 8)/15

=> x = -24/15

=> x = -8/5          [24 and 15 are divided by 3]

Exercise 2.2

Question 1:

If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?

Let the number be x.

According to question,

(1/2) * (x – 1/2) = 1/8

=> (x – 1/2) = (1/8) * (2/1)

=> x – 1/2 = 2/8

=> x -1/2 = 1/4

=> x = 1/4 + 1/2

=> x = (1 + 2)/4

=> x = 3/4

Hence, the number is 3/4

Question 2:

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 breadth?

Let the breadth of the pool be x m.

Then, the length of the pool = 2x + 2 m

Perimeter = 2(l + b)

=> 154 = 2(2x + 2 + x)

=> 154 =2(3x + 2)

=> 154 = 6x + 4

=> 6x = 154 – 4

=> 6x = 150

=> x = 150/6

=> x = 25

Length of the pool = 2x + 2 = 2 * 25 + 2 = 50 + 2 = 52 m

Breadth of the pool = 25 m

Hence, the length of the pool is 52 m and breadth is 25 m

Question 3:

The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 cm. What is the length of either of the remaining equal sides?

Let each of equal sides of an isosceles triangle be x cm.

Perimeter of a triangle = Sum of all three sides

=> 4 = 4/3 + x + x

=> 2x + 4/3 = 62/15

=> 2x = 62/15 – 4/3

=> 2x = (62 * 1 – 4 * 5)/15

=> 2x = (62 - 20)/15

=> 2x = 42/15

=> x = 42/(15 * 2)   =>  x = 42/30   => x = 14/10

=> x = 1cm.

Hence, each equal side of an isosceles triangle is 1 cm.

Question 4:

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

Sum of two number = 95

Let the first number be x.

Then, another number be x +15.

According to the question,

x + x + 15 = 95

=>  2x + 15 = 95

=> 2x = 95 – 15

=> 2x = 80

=> x = 80/2

=> x = 40

So, the first number = 40

Another number = 40 + 15 = 55

Hence, the two numbers are 40 and 55

Question 5:

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

Let the two numbers be 5x and 3x

According to question,

5x – 3x = 18

=> 2x =18

=> x = 18/2

=> x = 9

Hence, first number = 5 * 9 = 45

and, second number = 3 * 9 = 27.

Question 6:

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

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

According to the question,

x + x + 1 + x + 2 = 51

=>  3x + 3 = 51

=> 3x = 51 – 3

=> 3x = 48

=> x = 48/3

=> x = 16

Hence, first integer = 16, second integer = 16 + 1 = 17 and third integer = 16 + 2 = 18

Question 7:

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

Let the three consecutive multiples of 8 be x, x + 8 and x +16.

According to question,

x + x + 8 + x + 16 = 888

=>  3x + 24 = 888

=> 3x = 888 – 24

=> 3x = 864

=> x = 864/3

=> x = 288

Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296

and third multiple of 8 = 288 + 16 = 304.

Question 8:

Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74.

Find these numbers.

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

According to the question,

2x + 3(x + 1) + 4(x + 2) = 74

=> 2x + 3x + 3 + 4x + 8 = 74

=> 9x + 11 = 74

=> 9x = 74 – 11

=> 9x = 63

=> x = 63/9

=> x = 7

Hence first integer = 7, second integer = 7 + 1 = 8 and third integer = 7 + 2 = 9.

Question 9:

The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Let the present ages of Rahul and Haroon be 5x years and 7x years respectively.

According to condition,

=> 5x + 4 + 7x + 4 = 56

=> 12x + 8 = 56

=> 12x = 56 – 8

=> 12x = 48

=> x = 48/12

=> x = 4

Hence, present age of Rahul = 5 * 4 = 20 years

and present age of Haroon = 7 * 4 = 28 years

Question 10:

The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls.

What is the total class strength?

Let the number of girls be x.

Then, the number of boys = x + 8.

According to the question,

(x + 8)/x = 7/5

=> 5(x + 8) = 7x

=> 5x + 40 = 7x

=> 7x – 5x = 40

=> 2x = 40

=> x = 40/2

=> x = 20

Hence, the number of girls = 20 and number of boys = 20 + 8 = 28.

Question 11:

Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years.

What is the age of each one of them?

Let Baichung’s age be x years, then Baichung’s father’s age = x + 29 years and

Baichung’s granddaughter’s age = x + 29 + 26 = x + 55 years

According to condition,

=> x + x + 29 + x + 55 = 135

=> 3x + 84 = 135

=> 3x = 135 – 84

=> 3x = 51

=> x = 51/3

=> x = 17

Hence, Baichung’s age = 17 years, Baichung’s father’s age = 17 + 29 = 46 years and

Baichung’s granddaughter’s age = 17 + 29 + 26 = 72 years.

Question 12:

Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Let Ravi’s present age be x years.

After fifteen years, Ravi’s age = 4x years.

Fifteen years from now, Ravi’s age = x + 15 years.

According to question,

4x – x = 15

=> 3x = 15

=> x = 15/3

=> x = 5

Hence, Ravi’s present age is 5 years.

Question 13:

A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

Let the rational number be x.

According to the question

5x/2 + 2/3 = -7/12

=> 5x/2 = -7/12 – 2/3

=> 5x/2 = -(7/12 + 2/3)

=> 5x/2 = -{(7 * 1 + 2 * 4)/12}

=> 5x/2 = -(7 + 8)/12

=> 5x/2 = -15/12

=> 5x = -(15 * 2)/12

=> 5x = -30/12

=> 5x = -5/2

=> x = -5/(2 *5)

=> x = -5/10

=> x = -1/2

Hence, the number is -1/2

Question 14:

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10 respectively. The ratio of the number of these notes is 2:3:5.

The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?

Let the number of notes be 2x, 3x and 5x.

According to question,

100 * 2x + 50 * 3x + 10 * 5x = 400000

=> 200x + 150x + 50x = 400000

=> 400x = 400000

=> x = 400000/400  => x = 1000

Hence, the number of denominations of Rs 100 notes = 2 * 1000 = 2000

Number of denominations of Rs 50 notes = 3 * 1000 = 3000

Number of denominations of Rs 10 notes = 5 * 1000 = 5000

Therefore, required denominations of notes of Rs 100, Rs 50 and Rs 10 are 2000, 3000 and

5000  respectively.

Question 15:

I have a total of Rs 300 in coins of denomination Rs 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins.

The total number of coins is 160. How many coins of each denomination are with me?

Total sum of money = Rs 300

Let the number of Rs 5 coins be x, number of Rs 2 coins be 3x and number of Rs 1 coins be 160

– (x + 3x) = 160 – 4x

According to question,

5 * x + 2 * 3x + 1 * (160 – 4x) = 300

=> 5x + 6x + 160 – 4x = 300

=> 11x + 160 – 4x = 300

=> 7x + 160 = 300

=> 7x = 300 – 160

=> 7x = 140

=> x = 140/7

=> x = 20

Hence, the number of coins of Rs 5 denomination = 20

Number of coins of Rs 2 denomination = 3 * 20 = 60

Number of coins of Rs 1 denomination = 160 – 4 * 20 = 160 – 80 = 80

Question 16:

The organizers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win, gets a prize of Rs 25.

The total prize money distributed is Rs 3,000. Find the number of participants is 63.

Total sum of money = Rs 3000

Let the number of winners of Rs 100 be x.

And those who are not winners = 63 - x

According to the question,

100 * x + 25 * (63 - x) = 3000

=> 100x + 1575 – 25x = 3000

=> 75x + 1575 = 3000

=> 75x = 3000 – 1575

=> 75x = 1425

=> x = 1425/75

=> x = 19

Hence the number of winner is 19.

Exercise 2.3

Question 1:

Solve the following equations and check your results:   3x = 2x + 18

Given, 3x = 2x + 18

=> 3x – 2x = 18

=> x = 18

To check:

3x = 2x + 18

=> 3 * 18 = 2 *1 8 + 18

=> 54 = 36 + 18

=> 54 = 54

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 2:

Solve the following equations and check your results:   5t – 3 = 3t – 5

Given, 5t – 3 = 3t – 5

=> 5t = 3t – 5 + 3

=> 5t = 3t – 2

=> 5t – 3t = -2

=> 2t = -2

=> t = -2/2

=> t = -1

To check:

5t – 3 = 3t – 5

=> 5 * (-1) - 3 = 3 * (-1) – 5

=> -5 – 3 = -3 – 5

=> -8 = -8

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 3:

Solve the following equations and check your results: 5x + 9 = 5 + 3x

Given, 5x + 9 = 5 + 3x

=> 5x + 9 – 3x = 5

=> 2x + 9 = 5

=> 2x = 5 – 9

=> 2x = -4

=> x = -4/2

=> x = -2

To check:      5x + 9 = 5 + 3x

=> 5 * (-2) + 9 = 5 + 3 * (-2)

=> -10 + 9 = 5 – 6

=> -1 = -1

=> L.H.S. = R.H.S. Hence, it is correct.

Question 4:

Solve the following equations and check your results: 4z + 3 = 6 + 2z

Given, 4z + 3 = 6 + 2z

=> 4z + 3 – 2z = 6

=> 2z + 3 = 6

=> 2z = 6 – 3

=> 2z = 3

=> z = 3/2

To check:

4z + 3 = 6 + 2z

=> 4 * 3/2 + 3 = 6 + 2 * 3/2

=> 2 * 3 + 3 = 6 + 3

=> 6 + 3 = 9

=> 9 = 9

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 5:

Solve the following equations and check your results:  2x – 1 = 14 – x

Given, 2x – 1 = 14 – x

=> 2x – 1 + x = 14

=> 3x – 1 = 14

=> 3x = 14 + 1

=> 3x = 15

=> x = 15/3

=> x = 5

To check:

2x – 1 = 14 – x

=> 2 * 5 – 1 = 14 – 5

=> 10 – 1 = 9

=> 9 = 9

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 6:

Solve the following equations and check your results: 8x + 4 = 3(x - 1) + 7

Given, 8x + 4 = 3(x - 1) + 7

=> 8x + 4 = 3x – 3 + 7

=> 8x + 4 = 3x + 4

=> 8x + 4 – 4 = 3x

=> 8x = 3x

=> 8x – 3x = 0

=> 5x = 0

=> x = 0/5

=> x = 0

To check:

8x + 4 = 3(x – 1) + 7

=> 8 * 0 + 4 = 3(0 - 1) + 7

=> 0 + 4 = -3 + 7

=> 4 = 4

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 7:

Solve the following equations and check your results: x = 4(x  +10)/5

Given, x = 4(x  +10)/5

=> 5 * x = 4(x  +10)

=> 5x = 4x + 40

=> 5x – 4x = 40

=> x = 40

To check:

x = 4(x  +10)/5

=> 40 = 4(40 + 10)/5

=> 40 = (4 * 50)/5

=> 40 = 200/5

=> 400 = 40

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 8:

Solve the following equations and check your results:   2x/3 + 1 = 7x/15 + 3

Given, 2x/3 + 1 = 7x/15 + 3

=> 2x/3 = 7x/15 + 3 – 1

=> 2x/3 = 7x/15 + 2

=> 2x/3 - 7x/15 = 2

=> (2x * 5 - 7x * 1)/15 = 2

=> (10x – 7x)/15 = 2

=> 3x/15 = 2

=> x/5 = 2

=> x = 5 * 2

=> x = 10

To check:

2x/3 + 1 = 7x/15 + 3

=> (2 * 10)/3 + 1 = (7 * 10)/15 + 3

=> 20/3 + 1 = 70/15 + 3

=> 20/3 + 1 = 14/3 + 3

=> (20 + 3)/3 = (14 + 3 * 3)/3

=> 23/3 = (14 + 9)/3

=> 23/3 = 23/3

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 9:

Solve the following equations and check your results:  2y + 5/3 = 26/3 – y

Given, 2y + 5/3 = 26/3 – y

=> 2y + 5/3 + y = 26/3

=> 3y + 5/3 = 26/3

=> 3y = 26/3 – 5/3

=> 3y = (26 - 5)/3

=> 3y = 21/3

=> 3y = 7

=> y = 7/3

To check:

2y + 5/3 = 26/3 – y

=> 2 * 7/3 + 5/3 = 26/3 – 7/3

=> 14/3 + 5/3 = 26/3 – 7/3

=> (14 + 5)/3 = (26 - 7)/3

=> 19/3 = 19/3

=> L.H.S. = R.H.S.

Hence, it is correct.

Question 10:

Solve the following equations and check your results:  3m = 5m – 8/5

Given, 3m = 5m – 8/5

=> 3m - 5m = -8/5

=> -2m = -8/5

=> 2m = 8/5

=> m = 8/(5 * 2)

=> m = 8/10

=> m = 4/5                             [8 and 10 are divided by 2]

To check:

3m = 5m – 8/5

=> 3 * 4/5 = 5 * 4/5 – 8/5

=> 12/5 = 20/5 – 8/5

=> 12/5 = (20 - 8)/5

=> 12/5 = 12/5

=> L.H.S. = R.H.S.

Hence, it is correct.

Exercise 2.4

Question 1:

Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of.

What is the number?

Let the number be x.

According to question,

8(x – 5/2) = 3x

=> 8x – 8 * 5/2 = 3x

=> 8x – 4 * 5 = 3x

=> 8x – 20 = 3x

=> 8x = 3x + 20

=> 8x – 3x = 20

=> 5x = 30

=> x = 20/5

=> x = 4

Hence, the number is 4

Question 2:

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number.

What are the numbers?

Let another number be x.

Then positive number = 5x

According to the question,

5x + 21 = 2(x + 21)

=> 5x + 21 = 2x + 42

=> 5x – 2x + 21 = 42

=> 3x + 21 = 42

=> 3x = 42 – 21

=> 3x = 21

=> x = 21/3

=> x = 7

Hence another number = 7 and positive number = 5 * 7 =35

Question 3:

Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27.

What is the two digit number?

Let the unit place of a two digit number be x.

Therefore, the tens place = 9 – x

Since 2-digit number = 10 * tens place digit + unit place digit

So, original number = 10(9 - x) + x

According to question,

New number = original number + 27

=> 10x + (9 - x) = 10(9 - x) + x + 27

=> 10x + 9 – x = 90 – 10x + x + 27

=> 9x + 9 = 117 – 9x

=> 9x + 9 + 9x = 117

=> 18x + 9 = 117

=> 18x = 117 – 9

=> 18x = 108

=> x = 108/18

=> x = 6

Hence, the two digit number = 10(9 - x) + x

= 10(9 - 6) + 6

= 10 * 3 + 6

= 30 + 6

= 36

Question 4:

One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this two-digit number and add

the resulting number to the original number, you get 88. What is the original number?

Let the unit place digit of a two-digit number be x.

Therefore, the tens place digit = 3x

Since 2-digit number = 10 * tens place digit + unit place digit

So, original number = 10 * 3x + x = 30x + x = 31x

According to question,

New number + original number = 88

=> 10x + 3x + 31x = 88  => 44x = 88   => x = 88/44

=> x = 2

Hence, the two digit number = 31x = 31 * 2 = 62

Question 5:

Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age.

What are their present age?

Let Shobo’s present age be x years.

And Shobo’s mother’s present age = 6x years

According to the question,

x + 5 = 6x * 1/3

=> x + 5 = 2x

=> 2x – x = 5

=> x = 5

Hence, Shobo’s present age = 5 years and Shobo’s mother’s present age = 6 * 5 = 30 years

Question 6:

There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4.

At the rate 100 per meter it will cost the village panchayat 75,000 to fence the plot. What are the dimensions of the plot?

Let the length and breadth of the rectangular plot be 11x and 4x respectively.

So, perimeter of the plot = Total cost/cost of 1 meter = 75000/100 = 750 m

We know that the perimeter of rectangle = 2(length + breadth)

Now, according to question,

750 = 2 * 15x

=> 750 = 30x

=> x = 750/30

=> x = 25

Hence, length of rectangular plot = 11 * 25 = 275 m

and breadth of rectangular plot = 4 * 25 = 100 m

Question 7:

Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per meter and trouser material that costs him Rs 90 per meter.

For every 2 meters of the trouser material he buys 3 meters of the shirt material. He sells the materials at 12% and 10% respectively.

His total sale is Rs 36,000. How much trouser material did he buy?

Let ratio between shirt material and trouser material be 3x : 2x

The cost of shirt material = 50 * 3x = 150x

The selling price at 12% gain = {(100 + P%)/100} * CP

= {(100 + 12)/100} * 150x

= (112/100) * 150 x

= 168x

The cost of trouser material = 90 * 2x = 180x

The selling price at 12% gain = {(100 + P%)/100} * CP

= {(100 + 12)/100} * 180x

= (112/100) * 180 x

= 198x

According to question,

168x + 198x = 36600

=> 366x = 36600

=> x = 36600/366

=> x = 100

Now, trouser material = 2x = 2 * 100 = 200 meters

Hence, Hasan bought 200 meters of the trouser material.

Question 8:

Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond.

Find the number of deer in the herd.

Let the total number of deer in the herd be x.

According to question,

x = x/2 + 3(x – x/2)/4 + 9

=> x = x/2 + 3{(2x – x)/2}/4 + 9

=> x = x/2 + 3{x/2}/4 + 9

=> x = x/2 + 3x/8 + 9

=> x – x/2 – 3x/8 = 9

=> (8x – 4x – 3x)/8 = 9

=> x/8 = 9

=> x = 9 * 8

=> x = 72

Hence, the total number of deer in the herd is 72.

Question 9:

A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Let present age of granddaughter be x years.

Therefore, Grandfather’s age = 10x years

According to question,

10x = x + 54

=> 10x – x = 54

=> 9x = 54

=> x = 54/9

=> x = 6

Hence, granddaughter’s age = 6 years and grandfather’s age = 10 * 6 = 60 years.

Question 10:

Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Let the present age of Amon’s son be x years.

Therefore, Aman’s age = 3x years

According to question,

3x – 10 = 5(x - 10)

=> 3x – 10 = 5x – 50

=> 3x – 10 – 5x = -50

=> -2x – 10 = -50

=> -2x = -50 + 10

=> -2x = -40

=> 2x = 40

=> x = 40/2

=> x = 20

Hence, Aman’s son’s age = 20 years and Aman’s age = 3 x 20 = 60 years

Exercise 2.5

Question 1:

Solve the following linear equation: x/2 – 1/5 = x/3 + 1/4

Given, x/2 – 1/5 = x/3 + 1/4

=> x/2 – x/3 = 1/5 + 1/4

=> (3x - 2x)/6 = (4 + 5)/20

=> x/6 = 9/20

=> x = (9 * 6)/20

=> x = 54/20

=> x = 27/10

Question 2:

Solve the following linear equation: n/2 – 3n/4 + 5n/6 = 21

Given, n/2 – 3n/4 + 5n/6 = 21

=> (6n – 3n * 3 + 5n * 2)/12 = 21

=> (6n – 9n + 10n)/12 = 21

=> (16n – 9n)/12 = 21

=> 7n/12 = 21

=> 7n = 12 * 21

=> n = (12 * 21)/7

=> n = 12 * 3

=> n = 36

Question 3:

Solve the following linear equation: x + 7 – 8x/3 = 17/6 – 5x/2

Given, x + 7 – 8x/3 = 17/6 – 5x/2

=> x – 8x/3 + 5x/2 = 17/6 – 7

=> (6x – 8x * 2 + 5x * 3)/6 = (17 – 7 * 6)/6

=> (6x – 16x + 15x)/6 = (17 – 42)/6

=> (21x – 16x)/6 = -25/6

=> 5x/6 = -25/6

=> 5x = -25

=> x = -25/5

=> x = -5

Question 4:

Solve the following linear equation: (x - 5)/3 = (x - 3)/5

Given, (x - 5)/3 = (x - 3)/5

=> 5(x - 5) = 3(x - 3)

=> 5x – 25 = 3x – 9

=> 5x – 3x = -9 + 25

=> 2x = 16

=> x = 16/2

=> x = 8

Question 5:

Solve the following linear equation:  (3t - 2)/4 – (2t + 3)/3 = 2/3 – t

Given, (3t - 2)/4 – (2t + 3)/3 = 2/3 – t

=> (3t - 2)/4 – (2t + 3)/3 + t = 2/3

=> {3(3t - 2) – 4(2t + 3) + 12t}/12 = 2/3

=> (9t – 6 – 8t – 12 + 12t)/12 = 2/3

=> (13t - 18)/12 = 2/3

=> 3(13t - 18) = 2 * 12

=> 39t – 54 = 24

=> 39t = 24 + 54

=> 39t = 78

=> t = 78/39

=> t = 2

Question 6:

Solve the following linear equation: m – (m - 1)/2 = 1 – (m - 2)/3

Given, m – (m - 1)/2 = 1 – (m - 2)/3

=> m – (m - 1)/2 + (m - 2)/3 = 1

=> {6m – 3(m - 1) + 2(m - 2)}/6 = 1

=> {6m – 3m + 3 + 2m - 4}/6 = 1

=> (5m - 1)/6 = 1

=> 5m – 1 = 1 * 6

=> 5m – 1 = 6

=> 5m = 6 + 1

=> 5m = 7

=> m = 7/5

Question 7:

Simplify and solve the following linear equation: 3(t - 3) = 5(2t + 1)

Given, 3(t - 3) = 5(2t + 1)

=> 3t – 9 = 10t + 5

=> 3t – 10t = 5 + 9

=> -7t = 14

=> 7t = -14

=> t = -14/7

=> t = -2

Question 8:

Simplify and solve the following linear equation: 15(y - 4) – 2(y - 9) + 5(y + 6) = 0

Given, 15(y - 4) – 2(y - 9) + 5(y + 6) = 0

=> 15y – 60 – 2y + 18 + 5y + 30 = 0

=> 18y – 12 = 0

=> 18y = 12

=> y = 12/18

=> y = 2/3   [12 and 18 are divided by 6]

Question 9:

Simplify and solve the following linear equation: 3(5z - 7) – 2(9z - 11) = 4(8z - 13) – 17

Given, 3(5z - 7) – 2(9z - 11) = 4(8z - 13) – 17

=> 15z – 21 – 18z + 22 = 32z – 52 – 17

=> -3z + 1 = 32z – 69

=> -3z – 32z = -69 – 1

=> -35z = -70

=> 35z = 70

=> z = 70/35

=> z = 2

Question 10:

Simplify and solve the following linear equation: 0.25(4f - 3) = 0.05(10f - 9)

Given, 0.25(4f - 3) = 0.05(10f - 9)

=> 1.00f – 0.75 = 0.50f – 0.45

=> 1.00f – 0.50f = -.045 + 0.75

=> 0.50f = 0.30

=> f = 0.30/0.50

=> f = 3/5

=> f = 0.6

Exercise 2.6

Question 1:

Solve the following equation: (8x - 3)/3x = 2

Given, (8x - 3)/3x = 2

=> 8x – 3 = 2 * 3x

=> 8x – 3 = 6x

=> 8x – 6x = 3

=> 2x = 3

=> x = 3/2

Question 2:

Solve the following equation: 9x/(7 – 6x) = 15

Given, 9x/(7 – 6x) = 15

=> 9x = 15(7 – 6x)

=> 9x = 105 – 90x

=> 9x + 90x = 105

=> 99x = 105

=> x = 105/99

=> x = 35/33        [105 and 99 are divided by 3]

Question 3:

Solve the following equation: z/(z + 15) = 4/9

Given, z/(z + 15) = 4/9

=> z * 9 = 4(z + 15)

=> 9z = 4z + 60

=> 9z – 4z = 60

=> 5z = 60

=> z = 60/5

=> z = 12

Question 4:

Solve the following equation:  (3y + 4)/(2 – 6y) = -2/5

Given, (3y + 4)/(2 – 6y) = -2/5

=> 5(3y + 4) = -2(2 – 6y)

=> 15y + 20 = -4 + 12y

=> 15y – 12y = -4 – 20

=> 3y = -24

=> y = -24/3

=> y = -8

Question 5:

Solve the following equation: (7y + 4)/(y + 2) = -4/3

Given, (7y + 4)/(y + 2) = -4/3

=> 3(7y + 4) = -4(y + 2)

=> 21y + 12 = -4y – 8

=> 21y + 4y = -8 – 12

=> 25y = -20

=> y = -20/25

=> y = -4/5                      [20 and 25 are divided by 4]

Question 6:

The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Let the Ages of Hari and Harry be 5x years and 7x years.

According to question,

(5x + 4)/(7x + 4) = 3/4

=> 4(5x + 4) = 3(7x + 4)

=> 20x + 16 = 21x + 12

=> 20x – 21x = 12 – 16

=> -x = -4

=> x = 4

Hence, the age of Hari = 5x = 5 * 4 = 20 years

and the age of Harry = 7x = 7 * 4 = 28 years.

Question 7:

The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1,

the number obtained is 3/2. Find the rational number.

Let the numerator of a rational number be x,

then the denominator is x + 8.

Therefore, Rational number = x/(x + 8)

According to the question,

(x + 17)/(x + 8 - 1) = 3/2

=> (x + 17)/(x + 7) = 3/2

=> 2(x + 17) = 3(x + 7)

=> 2x + 34 = 3x + 21

=> 2x – 3x = 21 – 34

=> -x = -13

=> x = 13

Hence, the required rational number = x/(x + 8) = 13/(13 + 8) = 13/21

Sours: https://www.examfear.com/cbse-ncert-solution/Class-8/Maths/Linear-Equations-in-One-Variable/solutions.htm
3n+1 by Tom Spallone

## Linear equations with one unknown

### Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

3*n+4-(19)=0

### Step  1  :

#### Pulling out like terms :

1.1     Pull out like factors :

3n - 15  =   3 • (n - 5)

### Step  2  :

#### Equations which are never true :

2.1      Solve :    3   =  0

This equation has no solution.
A a non-zero constant never equals zero.

#### Solving a Single Variable Equation :

2.2      Solve  :    n-5 = 0

Add  5  to both sides of the equation :
n = 5

### One solution was found :

n = 5
Sours: https://www.tiger-algebra.com/drill/3n_4=19/

## 4 19 3n

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Mitosis vs. Meiosis: Side by Side Comparison

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