Equations In One Variable

⇒ x = 6 – 1

∴ x = 5

⇒ x = 3 – 4

∴ x = -1

⇒ x = 5 + 3

∴ x = 8

∴ x = 4

⇒ a = 7 × 3

∴ a = 21

∴ a = -4

⇒ 2x = 1 + 5

⇒ 2x = 6

∴ x = 3

⇒ 3x = -7 + 2

⇒ 3x = -5

⇒ 4x = 15 – 7

⇒ 4x = 8

∴ x = 2

⇒ 3x – x = 5

⇒ 2x = 5

⇒ 7x + x =20

⇒ 8x = 20

⇒ 6x – 6 = 5x + 5

Bringing all the variable terms to one side and constants to other,

⇒ 6x – 5x = 5 + 6

∴ x = 11

⇒ 2x + 3 = 3x – 3

⇒ 3x - 2x = 3 + 3

∴ x = 6

On cross multiplying the given equation, we get,

⇒ x + 5 = 2(1 – x)

⇒ x + 5 = 2 – 2x

⇒ x + 2x = 2 – 5

⇒ 3x = -3

∴ x = -1

On cross multiplying the given equation, we get,

⇒ 5x – 4 = 2 × 2x

⇒ 5x – 4 = 4x

⇒ 5x – 4x = 4

∴ x = 4

On cross multiplying the given equation, we get,

⇒ 2x = -3(3x + 1)

⇒ 2x = -9x – 3

⇒ 2x + 9x = -3

⇒ 11x = -3

On cross multiplying the given equation, we get,

⇒ 5a – 7 = 2 × 3a

⇒ 5a – 7 = 6a

⇒ 6a – 5a = -7

∴ a = -7

On cross multiplying, we get,

⇒ 9(5m – 3) = 8 × 2m

⇒ 45m – 27 = 16m

⇒ 29m = 27

On cross multiplying the given equation, we get,

⇒ 4x + 2 = 2(x – 4)

⇒ 4x + 2 = 2x – 8

⇒ 4x – 2x = -8 – 2

⇒ 2x = -10

∴ x = -5

On cross multiplying the given equation, we get,

⇒ 2x + 3 = 3(x – 1)

⇒ 2x + 3 = 3x – 3

⇒ 3x – 2x = 3 + 3

∴ x = 6

On cross multiplying, we get,

5(x + 1) = 6(x – 1)

⇒ 5x + 5 = 6x – 6

⇒ 6x – 5x = 5 + 6

∴ x = 11

On cross multiplying the given equation, we get,

⇒ 3x + 5 = 4(2x + 7)

⇒ 3x + 5 = 8x + 28

⇒ 8x – 3x = 5 – 28

⇒ 5x = - 23

On cross multiplying the given equation, we get,

⇒ 5x + 4 = 2(2x + 1)

⇒ 5x + 4 = 4x + 2

⇒ 5x – 4x = 2 – 4

∴ x = -2

On cross multiplying the given equation, we get,

⇒ 11m – 1 = 2(2m + 3)

⇒ 11m – 1 = 4m + 6

⇒ 11m – 4m = 6 + 1

⇒ 7m = 7

∴ m = 1

On cross multiplying the given equation, we get,

⇒ 2(x + 2) = 7(x – 3)

⇒ 2x + 4 = 7x – 21

⇒ 7x – 2x = 4 + 21

⇒ 5x = 25

∴ x = 5

On cross multiplying the given equation, we get,

⇒ 5(2 – m) = -3(m + 7)

⇒ 10 – 5m = -3m – 21

⇒ 3m – 5m = -21 – 10

⇒ -2m = -31

On cross multiplication, we get,

⇒ 4(2x – 1) = 9(x – 1)

⇒ 8x – 4 = 9x – 9

⇒ 9x – 8x = 9 – 4

∴ x = 5

On cross multiplying the given equation, we get,

⇒ 3(3y + 5) = 5(3 – 7y)

⇒ 9y + 15 = 15 – 35y

⇒ 9y + 35y = 15 – 15

⇒ 44y = 0

∴ y = 0

On cross multiplying the given equation, we get,

⇒ 9(2x + 1) = 5(3x – 2)

⇒ 18x + 9 = 15x – 10

⇒ 18x – 15x = -9 – 10

⇒ 3x = -19

On cross multiplying the given equation, we get,

⇒ 17(8 – 3y) = 5y + 2

⇒ 136 – 51y = 5y + 2

⇒ 5y + 51y = 136 – 2

⇒ 56y = 134

On cross multiplying the given equation, we get,

⇒ 3(18 – 2y) = 4(y – 4)

⇒ 54 – 6y = 4y – 16

⇒ 6y + 4y = 54 + 16

⇒ 10y = 70

∴ y = 7

Let the age of Poorva be x years.

Since Poorva is older than Durga by 6 years

Age of Durva = x-6

Now, age of Poorva after three years = x + 3

Also, age of Durga = (x - 6) + 3

It is given that, 3 years hence (later), their age will be in the

ratio 4:3

On cross multiplying, we get

⇒ 3(x + 3) = 4(x-3)

⇒ 3x + 9 = 4x – 12

⇒ x = 21

∴ Present age of Poorva is 21 years.

Let the smaller number be x

Since the difference of two numbers is 72,

⇒ The larger number is (x + 72)

According to the question, it is given that,

On cross multiplying, we get,

⇒ x + 72 = 4x

⇒ 3x = 72

⇒ x = 24

The other number is = x + 72 = 96

∴The two numbers are 24, 96

Let the number be x

On Subtracting 2 from thrice a number = 3x - 2

Sum of twice the number and 5 = 2x + 5

It is given that, their ratio is 13:15

On cross multiplying, we get,

⇒ 45x – 30 = 26x + 65

⇒ 19x = 95

⇒ x = 5

∴The number is 5

Let the age of younger brother be x years

Therefore, the age of elder brother is (x + 4) years

Now, Five years before, age of younger brother = x-5

Also, age of elder brother = (x + 4) -5

It is given that, 5 years ago(before), their age was in the

ratio 5:7

On cross multiplying, we get,

⇒ 7x – 35 = 5x – 5

⇒ 2x = 30

⇒ x = 15

Therefore, the age of younger brother is 15 years.

Let the value denominator be x

Since the numerator is 3 less than its denominator,

⇒ numerator = x-3

When numerator is tripled, new numerator = 3(x-3) = 3x-9

And, when denominator is increased by 2, new denominator = x + 2

Now it is given that,

On cross multiplying, we get,

6x – 18 = x + 2

⇒ 5x = 20

⇒ x = 4

∴Old Denominator = 4

⇒ Old Numerator = x – 3 = 1

Therefore, the original fraction was

Let the numerator be x.

Since, the denominator is bigger than numerator by 3

⇒ Denominator = x + 3

If 3 is subtracted from the numerator, it becomes = x-3

And, 2 added to the denominator, it becomes = (x + 3) + 2

It is given that :-

On cross multiplying, we get,

⇒ 5x – 15 = x + 5

⇒ 4x = 20

⇒ x = 5

Let the money that Gayatri initially has = Rs. x

Therefore, money with Sandeep = Rs. (x + 50)

When each of them is given Rs 15, the ratio of the money they have becomes 3 : 1

Now, Money with Sandeep = x + 50 + 15 = x + 65

Also, money with Gayatri = x + 15

_{On cross multiplying, we get,}

_{x + 65 = 3x + 45}

_{⇒2x = 20}

_{⇒x = 10}

_{Therefore, Gayatri initially had Rs. 10 with her}