Linear Equations In One Variable

Note: - Our Main Motto is to bring variable to one side of the equation and find the value

(i) x + 3 = 11

Taking 3 to right side of equation,

⇒ x = 11 - 3 = 8

x = 8

(ii) y - 9 = 21

Taking 9 to the right side of equation,

⇒ y = 21 + 9 = 30

y = 30

(iii) 10 = z + 3

Bring 3 to left side of equation,

⇒ z = 10 - 3 = 7

z = 7

(iv)

⇒

(v) 10x = 30

Take 10 to right hand side of the equation,

⇒

x = 3

(vi)

Take 7 to right hand side of equation,

⇒ s = 7 × 4

s = 28

(vii)

⇒

Taking 2 to right side of equation,

⇒ x = 10 × 2

x = 20

(viii)

Take 1.5 to left side of equation,

⇒ x = 1.6 × 1.5

x = 2.4

(ix) 8x – 8 = 48

⇒ 8x = 48 + 8

⇒8x = 56

⇒ x = (56/8) = 7

x = 7

(x)

⇒

⇒

⇒ x = - 8/5

(xi)

⇒ x = 5 × 12 = 60

x = 60

(xii)

⇒ 3x = 15 × 5

⇒ x = (15 × 5)/3

⇒ x = 5 × 5 = 25

x = 25

(xiii) 3(x + 6) = 24

⇒

⇒ x + 6 = 8

⇒ x = 8 - 6 = 2

x = 2

(xiv)

⇒ x = 9 × 4 = 36

x = 36

(xv) 3(x + 2) – 2(x – 1) = 7

Expanding the expressing we get,

3x + 6 - 2x + 2 = 7

Place Variable in one side and numbers in the other side of equation,

⇒ 3x - 2x = 7 - 6 - 2

⇒ x = - 1

x = - 1

Note: In This part, our main objective would be to bring variables to one side of the equation and numbers to another side.

(i) 5x = 3x + 24

⇒ 5x - 3x = 24

⇒ 2x = 24

⇒ x = 12

(ii) 8t + 5 = 2t – 31

⇒ 8t - 2t = - 31 - 5

⇒ 6t = - 36

⇒ t = - 6

(iii) 7x – 10 = 4x + 11

⇒ 7x - 4x = 10 + 11

⇒ 3x = 21

⇒ x = 7

(iv) 4z + 3 = 6 + 2z

⇒ 4z - 2z = 6 - 3

⇒ 2z = 3

(v) 2x – 1 = 14 – x

⇒ 2x + x = 14 + 1

⇒ 3x = 15

⇒ x = 5

(vi) 6x + 1 = 3(x – 1) + 7

⇒ 6x + 1 = 3x - 3 + 7

⇒ 6x - 3x = - 1 - 3 + 7

⇒ 3x = 3

⇒ x = 1

(vii)

We can make it simple by eliminating fractions and making it as a equation of integers only

For this we should multiply all terms of the equation with an appropriate numbers

Our Purpose is to eliminate the denominators, so the number we choose to multiply has to be a multiple of the denominator, this has to satisfy all denominators. Hence, we choose that Number to be the L.C.M. of all numbers in denominator.

So. in this question, it has to be L.C.M. of 5 & 2 that is 10.

So, multiply all terms of equation with 10

⇒ 4x - 15 = 5x + 10

⇒ 4x - 5x = 15 + 10

⇒ - x = 25

⇒ x = - 25

(viii)

Multiply with L.C.M. of 5 & 1 that is 5

⇒ x - 3 - 10 = 2x

⇒ x - 2x = 13

⇒ - x = 13

⇒ x = - 13

(ix)3(x + 1) = 12 + 4(x – 1)

⇒ 3x + 3 = 12 + 4x - 4

⇒ 3x - 4x = 12 - 4 - 3

⇒ - x = 5

⇒ x = - 5

(x) 2x – 5 = 3(x – 5)

⇒ 2x - 5 = 3x - 15

⇒ 2x - 3x = 5 - 15

⇒ - x = - 10

⇒ x = 10

(xi) 6(1 – 4x) + 7(2 + 5x) = 53

⇒ 6 - 24x + 14 + 35x = 53

⇒ 35x - 24x = 53 - 6 - 14

⇒ 11x = 33

⇒ x = 3

(xii) 3(x + 6) + 2(x + 3) = 64

⇒ 3x + 18 + 2x + 6 = 64

⇒ 5x + 24 = 64

⇒ 5x = 64 - 24

⇒ 5x = 40

⇒ x = 8

(xiii)

Multiplying with L.C.M. of 3 & 2, that is 6

⇒ 4m + 48 = 3m - 6

⇒ 4m - 3m = - 48 - 6

⇒ m = - 54

(xiv) (3/4) ( x - 1) = (x - 3)

⇒3 (x - 1) = 4 (x - 3)

⇒ 3x - 3 = 4x - 12

⇒ x = 9

Let the Number be x,

4 added to number, ⇒ x + 4

Summultiplied by 3, result is 30

⇒ 3(x + 4) = 30

⇒ x + 4 = 10

⇒ x = 6.

Sum of three consecutive odd numbers is 219

Odd number is of the form 2k - 1

Let the numbers be 2k - 3,2k - 1,2k + 1,

⇒ 6k - 3 = 219

⇒ 6k = 222

⇒ k = 37

⇒ numbers are 71,73,75.

Let the number be x,

Subtracted by 30, ⇒ x - 30

⇒ x - 30 = 14 - 3x

⇒ 4x = 44

⇒ x = 11

Let number be x

3x - 5 = 16

⇒ 3x = 21

⇒ x = 7

One of them exceeds the other by 9, so let the numbers be x & x + 9

Sum of both the numbers is 81

⇒ 2x + 9 = 8 1 ⇒ 2x = 72

⇒ x = 36

They are 36 and 45

Let Age of Sahil be x. ⇒ Age of Prakruthi is 6x

After 15 years, x + 15 & 6x + 15 are their ages

⇒ 6x + 15 = 3 ( x + 15)

⇒ 6x + 15 = 3x + 45

⇒ 3x = 30

⇒ x = 10

The Present ages of Sahil and Prakruthi are 10 and 60 respectively.

Let Present age of Ahmed be x. ⇒ Age of his father is 3x

After 12 years, their ages will be x + 12 & 3x + 12

3x + 12 = 2(x + 12)

⇒ x = 12

The Present Ages of Ahmed and his father are 12 and 36 respectively.

Let Age of Nishu be x. ⇒ Age of Sanju = x + 6

Given, Sum of their ages = 28

⇒ 2x + 6 = 28

⇒ 2x = 22

⇒ x = 11

The Present Ages of Nishu and Sanju are 11 and 17 respectively.

Let age of Deepu be x. ⇒ Age of Viji is 2x

Difference of their ages is 11

⇒ 2x - x = 11

⇒ x = 11

The Present ages of Viji and Deepu are 22 and 11 respectively.

Let the Present age of Bindu be x. ⇒ Present age of Mrs. Joseph is x + 27

After 8 years, their ages are x + 8 & x + 35

x + 35 = 2 (x + 8)

⇒ x = 35 - 16 = 19

The Present ages of Bindu and Mrs. Joseph are 19 and 46 respectively.

Let Present age of Leena be x,

Given, x + 16 = 3x

⇒ 2x = 16

⇒ x = 8

Let b be the breadth of the rectangle, ⇒ length = 2b - 5

Given, 2(2b - 5 - 5 + b + 2) = 74

⇒ 3b - 8 = 74/2 = 37

⇒ b = 15

The Length and Breadth of the rectangle are 15 cm and 25 cm respectively.

Let b be the breadth of the rectangle, ⇒ length = 2b

Given, Perimeter = 288

⇒ 2(Length + breadth) = 288

⇒ 2b + b = 288/2

⇒ 3b = 144

⇒ b = 48

The Length and breadth of the rectangle are 48 and 96 respectively.

Let Salary of Azar be x, ⇒ Salary of Sristi = 4x

Given, x + 4x = 3750

⇒ 5x = 3750

⇒ x = 750

The Salaries of Azar and Sristi are 750 and 3000 respectively.

⇒ 5x – 35 = 0

⇒ 5x = 35

⇒ x =35/5

⇒ x = 7

∴Value of x is 7

Let the number is X

One fifth of the the number is X/5

According to the problem,

⇒ X/5 – 14 = 20

∴The equation expressing this statement is

(X/5) – 14 =20

Let the number is X

According to problem,

⇒ 5x + 8 = 53

⇒ 5x = 53 - 8

⇒ 5x = 45

⇒ x =45 / 5

⇒ x = 9

5(x - 2) = 3(x - 3)

⇒ 5x – 10 = 3x – 9

⇒ 5x – 3x = 10 – 9

⇒ 2x = 1

⇒ x = 1/2

Let the numbers are X and Y

According to the problem,

X + Y = 84 …. (1)

X – Y = 30 …..(2)

Now minus 2^nd equation from 1^st,

⇒ 2X =84 - 30

⇒ 2X = 54

⇒ X = 54/2

⇒ X = 27

Now put the value of X in the 1st equation,

⇒ X + Y = 84

⇒ 27 + Y = 84

⇒ Y = 84 – 27

⇒ Y = 57

∴ The value of X and Y are 27 and 57

Let, breadth = x m. and breadth = 2x

According to problem,

⇒ x × 2x = 800

⇒ 2x² = 800

⇒ x² = 400

⇒ x = 20

∴ Breadth = x = 20 m and Length = 2x = 2 × 20 = 40 m

Let, the three odd numbers = x, x + 2, x + 4

According to problem,

⇒ x + x + 2 + x + 4 = 249

⇒ 3x + 6 = 249

⇒ 3x = 243

⇒ x = 81

∴ The numbers are 81, (81 + 2) = 83, (81 + 4) = 85

According to the problem,

⇒ =0.85

⇒ (x + 0.7x) = 0.85 × 2

⇒ 1.7x = 1.7

⇒ x = 1.7/1.7

⇒ x = 1

⇒ 2x – (3x - 4) = 3x – 5

⇒ 4 – x = 3x – 5

⇒ 4 + 5 = 3x + x

⇒ 9 = 4x

⇒ x = 9/4

⇒ (3x + 24) ÷ (2x + 7) = 2:

⇒ (3x + 24) = 2 ×(2x + 7)

⇒ (3x + 24) = 4x + 14

⇒ 24 – 14 = 4x – 3x

⇒ 10 = x

∴The value of x is 10

⇒ (1 – 9y) ÷ (11 – 3y) = .

⇒ 8×(1 - 9y) = 5×(11 – 3y)

⇒ 8 – 72y = 55 - 15y

⇒ 8 – 55 = 72y -15y

⇒ 47 = 63y

⇒ y =47/63

∴ The value of y is 47/63

According to the problem the two numbers ratio is 7:8 and they are 7x and 8x

According to the problem,

7x + 8x = 45

15x = 45

X = 3

∴ The 1st number is 7x

⇒ 7x = 7 × 3 = 21

∴The 2nd number is 8x

⇒ 8 × 3 = 24

∴The two numbers are 21 and 24

Let, Shona’s age = x

Shona’s mother’s age = 4x

According to problem,

⇒ 4x + 5 = 3(x + 5)

⇒ 4x + 5 = 3x + 15

⇒ x = 10

∴ Shona’s age = x = 10 years

Shona’s mother’s age = 4x = 40 years

Let the consecutive numbers are x, x + 1,x + 2, x + 4

According to the questions,

Sum of the Consecutive numbers is 336,

⇒ (x + x + 2 + x + 4 ) = 336

⇒ 3x + 6 = 336

⇒ 3x =330

⇒ x = 110

∴ The Consecutive numbers are

X =110,

2^nd number =x + 2 =110 + 2 = 112

3^rd number = x + 4 =110 + 4 = 114

Let,B share X and A share 2X.

According to the problem,

⇒ X + 2X =60,000

⇒ 3X = 60,000

⇒ X = 20000

∴ B share X = 20000

∴ A share 2X = 2 × 20000 =40000

Let the number is X

According to the problem,

⇒ X/3 - 40 = X

⇒ X/3 – X = 40

⇒ = 40

⇒ -2X = 120

⇒ X = 60

∴ The Original number is 60.

Let, the number is = x

According to problem,

⇒ x/24 = 8

⇒ x = 24 × 8

⇒ x = 192

Let price of the house is X and price of the garden is

According to the problem,

⇒ x + = 8,40,000

⇒ =8,40,000

⇒ = 8,40,000

⇒ X =

⇒ X =70000 × 7

⇒ X = 49000

∴ Price of the house is X = 490000

∴ Price of the garden is = = 350000

Let, value of each bag = Rs. x

According to problem,

⇒ 92x – 8000 = 72x + 8000

⇒ 20x = 16000

⇒ x = 800

∴ Value of each bag = Rs. 800

Let, son’s age = x

Father’s age = 4x

According to problem,

⇒ 4x + 5 = 3(x + 5)

⇒ 4x + 5 = 3x + 15

⇒ x = 10

∴ son’s age = x = 10 years

∴ Father’s age = 4x = 40 years

Let, it will take y more years to be the father’s age twice than the son.

According to problem,

⇒ 4x + 5 + y = 2(x + 5 + y)

⇒ 45 + y = 30 + 2y

⇒ y = 15

∴ 15 more years will take if father’s age is to be twice that of his son.

Let, the number is = x

According to problem,

⇒ 7x = x + 132

⇒ 6x = 132

⇒ x = 22

∴ The number is = 22

He will sell = 25 – 5 = 20

According to problem,

Selling price of 20 pens = 250

∴ Selling price of each pen = 250/20 = 12.5

Let, the unit’s digit = x

The ten’s digit = 12 – x

∴ The number = 10(12 – x) + x = 120 – 9x

After reversing,

Unit’s digit = 12 – x

Ten’s digit = x

∴The number after reversing = 10x + 12 – x = 9x + 12

According to problem,

⇒ 9x + 12 = 120 – 9x + 18

⇒ 18x = 126

⇒ x = 7

∴ The original number,

⇒ 120 – 9 × 7

⇒ 57

Let, the speed of the slower train = x km/h

Speed of the faster train = (x + 5) km/h

∴ In 2 hours slower train moved = 2x km

∴ In 2 hours faster train moved = 2(x + 5) = (2x + 10) km

According to problem,

⇒ 2x + 2x + 10 = 340 + 30

⇒ 4x = 370 – 10

⇒ x = 360/9

⇒ x = 90

∴ Speed of the slower train = x = 90 km/h

∴ Speed of the faster train = x + 5 = 90 + 5 = 95 km/h

Let, the speed of the steamer in still water = x km/hr

Speed of the stream = y km/hr

Speed of the steamer in upstream = 2 km/hr

Time taken by the steamer to cover the distance in upstream = 5 hrs.

∴ Distance = 2 × 5 = 10 km

Time taken by the steamer to cover the distance in downstream = 4 hrs.

∴ Speed of the steamer in downstream = 10/4 = 2.5 km/hr

∴ x – y = 2 …… (1)

∴ x + y = 2.5 …… (2)

From (1) + (2) we get,

⇒ 2x = 2 + 2.5

⇒ x = 2.25

∴ Speed of the steamer in still water = 2.25 km/hr

Let, the denominator = x

The numerator = x – 3

According to problem,

⇒ 24x – 72 = x + 20

⇒ 23x = 92

⇒ x = 4

∴ the original number,

Let, the digit at unit’s place = x

∴ digit at ten’s place = 3x

∴ The two digit number = 10 × 3x + x = 31x

After reversing the digits,

Digit at unit’s place = 3x

Digit at ten’s place = x

∴ Number formed after reversing the digits = 10 × x + 3x = 13x

According to problem,

⇒ 31x + 13x = 88

⇒ 44x = 88

⇒ x = 2

∴ The two digit number = 31 × 2 = 62

Let, base of the triangle = x cm

∴ altitude of the triangle = 5x/3 cm

∴ area of the triangle = 1/2 × x × 5x/3 = 5x²/6

In 2^nd case,

Base of the triangle = x – 2 cm

Altitude of the triangle = (5x/3) + 4

∴ area of the triangle = 1/2 × (x – 2) × (5x/3 + 4)

= 5x²/6 + 2x – 5x/3 – 4

According to problem,

⇒ 5x²/6 = 5x²/6 + 2x – 5x/3 – 4

⇒ x/3 = 4

⇒ x = 12

∴ Base = 12 cm

∴ Altitude = 5 × 12/3 = 20 cm

Let, the other two angles = 4x and 5x

∴ The biggest angle = 4x + 5x = 9x

According to problem,

⇒ 4x + 5x + 9x = 180

⇒ 18x = 180

⇒ x = 10

∴The angles of the triangle,

4x = 4 × 10° = 40°

5x = 5 × 10° = 50°

9x = 9 × 10° = 90°

AB is a straight line.

∴ x + 20 + x + 40 + x = 180

⇒ 3x + 60 = 180

⇒ 3x = 120

⇒ x = 40

∴ x = 40°