Linear Equations In One Variable

Taking 2 to right hand side of the equation we write,

x - 2 = 7
x = 7 + 2

x = 9

Taking 3 to right hand side of the equation we write,

y = 10-3

y = 7

6 = z + 2

Sending Variables on the LHS and Constants on the RHS.

z = 6 - 2

z = 4

Taking to right hand side of the equation we write,

6 x = 12

Taking 6 from L.H.S to R.H.S, multiplication changes into division

x = .

Taking 5 from L.H.S to R.H.S, division changes into multiplication.

t = 510 = 50

Taking 1.5 from R.H.S to L.H.S, division changes into multiplication,

1.5 x 1.6 = y

y = 2.4

On taking 9 to the RHS,
7x = 16 + 9
7x=25

x =

14 y - 8 = 13

Transposing 8 to RHS.

⇒ 14 y = 13 + 8

⇒ 14 y = 21

Taking 1 from left hand side to the right side of the equation we write,

Let the number is x.

Then, on subtracting ½ from x, we get, x - ½.

On multiplying, (x - ½) by ½, we get 1/8.

Thus, solving we get,
(x - ½) × ½ = 1/8
Now, we will take L.C.M and cross-multiplying, we get,



Multiplying both side by 4, we have

⇒ 2(2x-1) = 1
⇒ 4x - 2 = 1
⇒ 4x = 3

Let the length be “l” and breadth be “b”

Given:
Perimeter of rectangle = 154 m
Length is two more than twice of breadth,
twice of breadth means 2b, and 2 more than this means : 2b + 2
Therefore,
l = (2b + 2) m
Then, perimeter = 2(l + b) = 154 m
Putting the value of l in above equation,

2{(2b + 2) + b} = 154

2{2b + b + 2} = 154
2{3b + 2} = 154
6b + 4 = 154
6b = 154 - 4
6b = 150

Hence, Breadth of the rectangle = 25 m
So, length of the rectangle = 2b + 2 = 2(25) + 2 = 52 m

The isosceles triangle is the one which has two side equal, say side is x. And it is given that the base side = 4/3 cm

Now, the perimeter = sum of three sides = 2 x + base

Therefore, each side is of length cm.

Let the number be x.

The other number is 15 more than the number x.

Therefore, other number is x + 15. [ given second number exceeds first by 15]

Now, x + (x + 15) = 95 [ given sum of numbers to be 95]

2 x = 95-15

2 x = 80

x =

Therefore, one number is 40 and other number is 40+15 = 55

Let the common ratio be x.

Then, the numbers are 5 x and 3 x.

Given: Difference of numbers is 18

Now,

Difference of 5x and 3x = 18

5 x – 3 x = 18

2 x = 18

x =

Therefore, one number is 5 × 9 = 45 and other number is 3 × 9 = 27.

Let the three consecutive numbers be x+1, x, x-1

Then,
x+1 + x + x - 1 =51

3x = 51

x = 17
Also,

x - 1 = 17 - 1 = 16
x + 1 = 17 + 1 = 18

Three consecutive multiples of 8 means, three consecutive numbers that are divisible by 8.
Let the three consecutive multiples of 8 are
= 8(x - 1), 8x, 8(x + 1)
= 8 x - 8, 8 x, 8 x + 8

Then,
Sum of three consecutive multiples of 8 = 888
8 x - 8 + 8 x + 8 x + 8 = 888

24 x = 888

Therefore, the numbers are, 288,296 and 304.

As the numbers are consecutive they will be one after another
Let the numbers are x, x + 1, x + 2

According to the question,
2x + 3(x + 1) + 4(x + 2) = 74
Opening brackets we get,

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

9x = 74 - 8 - 3
9x = 63

Also,
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9
Thus, the numbers are, 7, 8 and 9.

Let the common ratio be x.
So their present ages be 5x and 7x.

4 years later their individual ages will be (5x+4) and (7x+4)

According to question,

5x+4+7x+4 = 56

12x = 56-8=48

Given ratio 7:5
Let the number of boy and girls be 7x and 5x respectively.

Then, 7x = 5x + 8

7x - 5x = 8

2x = 8

x = 4

Therefore, the number of girls is 20 and the number of boys is 28. The total class strength is 48.

Let Baichung’s father age is x.
Given,
Baichung's father is 26 years younger than Baichung's grandfather
⇒ Age of Baichung's father = x + 26
and
Baichung's father is 29 years older than Baichung
⇒ Age of Baichiung = x - 29

Now,
Sum of ages = 135
⇒ x + (x+ 26) + (x- 29) = 135
⇒ 3x -3 = 135
⇒ 3x = 138

⇒ x = = 46

So, Baichung's Father Age = 46 years

Baichung's Age = x - 29 = 46 - 29 = 17 years

Baichung's grandfather Age = x + 26 = 46 + 26 = 72 years

Let present age is x years.

Then, Fifteen years later , his age will be 4x years.

Thus, x + 15 = 4x

3x = 15

x = 5

Thus, his present age is 5 years.

Let the rational number be x.
Now according to the question,

Let the number of notes of denominations Rs 100, Rs 50 and Rs 10, be 2x, 3x and 5x respectively as they are in the ratio 2:3:5.

Hence, Total Amount of 100 Rupees Notes are = 100 × 2x = 200x

Total Amount of 50 Rupees Notes are = 50 × 3x = 150x

Total Amount of 10 Rupees Notes are = 10 × 5x = 50x
Given: Total Amount = 400000
Therefore,

Thus, 200x + 150x + 50x = 400000

⇒ 400x = 400000

⇒ x = 1000

So, Number of 100 Rs Notes = 2 × 1000 = 2000

Number of 50 Rs Notes = 3 × 1000 = 3000

Number of 10 Rs Notes = 5 × 1000 = 5000

Let the number of Rs 5 coin be x

Then, No of Rs 2 coins = 3 x

And no. of 1 Re coin = 160 –(3 x + x) = 160 - 4 x
The, the total amount of rupee coin is: Rs(1 × (160 - 4x)) = Rs. 160 - 4x
Total amount of 2 rupee coins is: Rs. 2 × 3x = Rs. 6x
The, total amount of 5 rupee coins is: Rs. 5 × x = Rs. 5x
Thus, 160 - 4x + 6x + 5x = 300
160 + 7x = 300
7x = 300 - 160
x = 140/7 = 20
Therefore,
Number of 1 Rs coins = 160 - (4 × 20) = 160 - 80 = 80
Number of 2 Rs coins = 3x = 3 × 20 = 60
Number of 5 Rs coins = x = 20

Let the number of winners be x. therefore, the number of participants who did not win = 63 - x.

Each winner gets Rs. 100 as prize money, therefore
Total amount given to winners be Rs 100x

Then, Amount given to the participants who did not win = Rs 25(63 - x) = Rs 1575 – 25x
Total Amount distributed = Rs. 3000

100 x + 1575 – 25 x = 3000

75 x = 3000 – 1575

75 x = 1425

Thus, the number of winners are 19.

3 x = 2 x + 18
3 x - 2 x = 18
x = 18

Checking the result:
L.H.S :
= 3 x
= 3 (18)
= 54

R.H.S:
= 2 x + 18
= 2(18) + 18
= 36 +18
= 54

Since L.H.S. = R.H.S

Hence, the solution is correct.

5t – 3 = 3t - 5

Rearranging the terms, we get,

5t - 3t = -5 + 3

2t = -2

t = -1

L.H.S: 5(-1) -3 = -8

R.H.S: = 3(-1) -5 = - 8

Since L.H.S. = R.H.S

Hence, the solution is correct.

5x + 9 = 5 + 3x

5x – 3x = 5 - 9

2x = -4

x = -2

L.H.S: 5x + 9
= 5(-2) + 9
= -1

R.H.S: 5 + 3x
= 5 + 3(-2)
= - 1

Since L.H.S. = R.H.S

Hence, the solution is correct.

4x – 2x = 6 - 3

2x = 3

x = .

L.H.S: 4() + 3 = 9

R.H.S: = 6 + 2() = 9

Since L.H.S. = R.H.S

Hence, the solution is correct.

2x + x = 14 + 1

3x = 15

x = = 5.

L.H.S: 2(5) – 1 = 9

R.H.S: = 14 - 5= 9

Since L.H.S. = R.H.S

Hence, the solution is correct.

8 x + 4 = 3 x – 3 + 7

8 x + 4 = 3 x + 4

8x – 3x = 4 - 4

5 x = 0

x = 0

Check for the solution:


L.H.S: 8 (0) + 4 = 4

R.H.S: = 3 (0 -1) + 7 = 4

Since L.H.S. = R.H.S

Hence, the solution is correct.

5 x = 4 x + 40
5 x - 4 x = 40
x = 40

L.H.S: x = 40

R.H.S=

Since L.H.S. = R.H.S

Hence, the solution is correct.

L.H.S:

R.H.S: =

Since L.H.S. = R.H.S

Hence, the solution is correct.

2 y + y =

3 y = 21/3

3 y = 7

L.H.S:

R.H.S:

Since L.H.S. = R.H.S

Hence, the solution is correct.

3m – 5m =

L.H.S: 3() =

R.H.S:

Since L.H.S. = R.H.S

Hence, the solution is correct.

Let the number Amina thinks of be x.

According to question.
she subtracts from it, then she multiplies the result by 8 and the result equals 3 times the original number

Multiplying inside the bracket, we get,

8x - 5/2 × 8 = 3x

8x - 5 × 4 = 3x

⇒ 8x - 20 = 3x

⇒ 8x - 3x = 20

⇒ 5x = 20

⇒ x = 4

Thus, the number is 4.

Let the numbers be x and 5x (as one number is 5 times the other number)
Given,
If 21 is added to both the numbers, numbers become (x + 21) and (5 x + 21), then one of the new numbers becomes twice the other new number

Then,

21 + 5x = 2(x + 21)
21 + 5x = 2x + 42
5x - 2x = 42 - 21
3x = 21
x = 7

Thus, one number is 7 and the other number is 35.

Let the two digit number be = 10 x + y
Given: Sum of digits of two digit number = 9, x + y = 9
So, if one of the digit is x, second digit will be 9 - x
So we have our 2 digit number = 10 x + (9 - x)
Reversing the digits of this number = 10 (9 - x) + x
Now it is given that
resulting new number is greater than the original number by 27.
Therefore, 10(9 - x) + x - (10 x + 9 - x) = 27
90 - 10 x + x - 10 x - 9 + x = 27
81 - 18 x = 27
18 x = 81 - 27
18 x = 54
x = 54/18
x = 3
So the number is 10 x 3 + (9 - 3) = 36

Let the digit at ten’s place and one’s place be x and 3x.

Then original number =10(x) + 1(3x) = 13x

On interchanging the digits, the digits at ones place and tens place will be x and 3x

The number formed after interchanging = 10(3x) + x = 31x

Now, New Number + Original Number = 88

13x + 31x = 88

44x = 88

We get, x = 2

Thus, the two digit number is 13x = 13(2) = 26 or 62.

Let Shobo’s present age is x years.

Thus, his mother’s present age will be 6x years.

After 5 Years:
Shobo's Age = x + 5 years

Given,
Shobo’s age five years from now will be one third of his mother’s present age

⇒ 3(x + 5) = 6x

⇒ 3x + 15 = 6x

⇒ 15 = 6x – 3x

⇒ 3x = 15

⇒ x = 5

Hence, Age of Shobo = x= 5 years

Age of his mother = 6x = 6(5) = 30 years

Since, The ratio given is 11 : 4
Let the length and breadth of the plot be l = 11x and b = 4x respectively.

Since the plot is of rectangular shape and we know the perimeter of a rectangle is 2(l + b)

Also, rate of fencing the plot is Rs 100 for 1m

Now,

Perimeter of Rectangular plot = 2(l + b) = 2(11x + 4x)

Perimeter of rectangular plot = 2(15 x) = 30x

Cost of fencing one meter = Rs. 100

Therefore cost of fencing 30x m = 30 x × 100 = 75000

3000x = 75000

x = 25

Thus, the length = 11 x (25) = 275 m

And the breadth = 4 x (25) = 100 m

Cost of 1 m shirt material = Rs. 50

Cost of 1 m trouser material = Rs. 90

Now after 12% Profit, Selling Price of 1 m shirt material = 50 + 12% of 50





= 50 + 6

= Rs. 56

And after 10% Profit, Selling Price of 1 m shirt material = 90 + 10% of 90


= 90 + 9

= Rs. 99

It is given that if he buys 2 m of shirt material then he buys 3 m of shirt material.

Let that he buys 2x m of shirt material then he will buy 3x m of trouser material.

Total Selling Price = 56 x 3x + 99 x 2x = 168x + 198x = Rs. 366x

It is given that the total selling price = 36600

Therefore,

366 x = 36600

x = 100

So the trouser material he buys = 2x

= 2 x 100

= 200 m
Therefore, the trouser material is 200 m.

Let the number of deer be x

The no. of deer grazing in the field= half of x = x/2

No. of deer remaining = x - x/2

The no. of Deer playing in the field = three-fourth of remaining deer =

The no. of Deer drinking water = 9


∵ Total Deer= (Grazing Deer) + (Playing Deer) + (Deer drinking water)
∴

∴ the number of deer is 72.

Let granddaughter’s age is x years.
∴ grandfather’s age will be 10x years

Now, 10x = x + 54

9x =54

x = 6

Granddaughter’s age is 6 years and the grandfather’s age is 60

Let the son’s present age is x years.
Then, Aman’s present age is 3 x years.
10 years before,
Aman's age = 3 x - 10
Son's age = x - 10

3 x – 10 = 5 (x - 10) [ Age 10 years before]

3 x – 5 x = - 50 + 10

- 2 x = - 40

x = 20

thus, son’s age is 20 years and Aman’s age is 60.

L.C.M of 2 and 5 = 10 and L.C.M of 3 and 4 = 12
Therefore,

Cross Multiplying we get,
12(5 x - 2) = 10( 4 x + 3)
60 x - 24 = 40 x + 30
60 x - 40 x = 30 + 24
20 x = 54

L.C.M of 2,6, and 4 is 12

Thus,

As the equation contains variable as well as constants. First step should be taking variables at one side and constants at other side.
Therefore, the equation becomes

Method 1:


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

5 x - 25 = 3 x - 9

5 x - 3 x = 25 - 9

2 x = 16

x = 16/2

x = 8

L.C.M of 4 and 3 is 12

Thus,

Opening the brackets, we write,

3t – 9 =10t + 5

-9 – 5 = 10t-3t

-14 = 7t

t = -2

15 (y - 4) - 2 ( y - 9) + 5 ( y + 6) = 0
Opening brackets
15y - 15 x 4 - 2y + 2 x 9 + 5y + 5 x 6 = 0
15y - 60 - 2y + 18 + 5y + 30 = 0
(15y - 2y + 5y) + 30 + 18 - 60 = 0
18 y - 12 = 0
18 y = 12

Opening the brackets, we write,

15z – 21 – 18z + 22 = 32z -52 -17

-3z + 1 = 32z – 69
-3z - 32z = -69 - 1
- 35z = -70
35z = 70
z = 2

The given Equation can be written as:

On Solving this By Cross Multiplication,

The given equation can be written as:



Now, On doing Cross Multiplication we have,

9 x = 15 (7 - 6 x)

9 x = 105 – 90 x

9 x + 90 x = 105

99 x = 105

By Cross Multiplication,

9z = 4 (z + 15)

9z = 4z + 60

5z = 60

z = 60/5

z = 12

5(3y + 4) = -2(2-6y)

⇒ 15 y + 20 = - 4 + 12y

⇒ 15 y - 12 y = -4 - 20

⇒ 3 y = -24


⇒y = -8

Multiplying by 3(y+2) on both the sides we get,

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

21y + 12 = - 4y - 8

25y = -20

y = -

Let the present ages of Hari and Harry are 5 x and 7 x

Then, four years later, Hari age will be 5 x + 4 years and Harry age will be 7 x + 4 years

Now,

Thus, presently, Hari age is 20 and Harry age is 28

Let the numerator of rational number is x and the denominator of rational number is y.

∵ Denominator y = x+ 8

∴ The rational number is

Now according to question,

[Since, denominator, y = x + 8, denominator = 13 + 8 = 21]