Pairs Of Equations

Let the price of two hundred pages notebook be ‘x’ and price of hundred pages book be ‘y’.

According to the question,

Raju bought 7 notebooks of two hundred pages and five of hundred pages for 107 rupees

⇒ 7x + 5y = 107 ... (1)

Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees

⇒ 5x + 7y = 97… (2)

Equating (1) and (2)

7x + 5y = 107

5x + 7y = 97

Multiply Equation (1) by 7 and equation (2) by 5

Then, Subtract equation (2) from (1)

x = 11

Put x = 11 in Equation (3)

49 × 11 + 35 y = 749

⇒ 539 + 35y = 749

⇒ 35y = 749 – 539

⇒ 35y = 210

⇒

⇒ y = 6

Hence, price of two hundred pages notebook is 11rupees and price of hundred pages notebook is 6 rupees.

Let the first number be ‘x’ and second number be ‘y’.

According to the question,

Four times a number and three times a number added together make 43

⇒ 4x + 3y = 43 … (1)

Two times the second number, subtracted from three times the first give 11

⇒ 3x – 2y = 11 … (2)

Equating equation (1) and (2)

4x + 3y = 43 … (1)

3x – 2y = 11 … (2)

Multiply equation (1) by 2 and (2) by 3

x = 7

Put x = 7 in equation (4)

9 × 7 – 6y = 33

⇒ 63 – 6y = 33

⇒ 63 = 33 + 6y

⇒ 6y = 63 – 33

⇒ 6y = 30

⇒

⇒ y = 5

Hence the two numbers are 7 and 5.

Let the digit in the unit’s place be ‘x’ and the digit in the tens place be ‘y’.

According to the question,

x + y = 11 … (1)

Then, the number = 10y + x = 11

The number got by interchanging the digits is 27 more than the original number

⇒ 10x + y = 10y + x + 27

⇒ 10x + y – 10y – x = 27

⇒ 9x – 9y = 27

Divide by 9 both sides

⇒ x – y = 3 … (2)

Equating equation (1) and (2)

x = 7

Put x = 7 in Equation (1)

7 + y = 11

y = 11 – 7

y = 4

∴ the number = 10y + x

= 10 × 4 + 7

= 40 + 7

= 47

Let present age of Rahim be ‘x’ and present age of Ramu be ‘y’

According to the question,

Four years ago, Rahim’s age was three times Ramu’s age

⇒ (x – 4) = 3(y – 3)

⇒ x – 4 = 3y – 9

⇒ x – 3y = –9 + 4

⇒ x – 3y = –5 … (1)

After two years, it would just be double

⇒ (x + 2) = 2(y + 2)

⇒ x + 2 = 2y + 4

⇒ x – 2y = 4 – 2

⇒ x – 2y = 2 … (2)

Subtract equation (2) from (1)

y = 3

Put y = 3 in Equation (1)

x – 3 × 3 = –5

⇒ x – 9 = –5

⇒ x = –5 + 9

⇒ x = 4

Hence, Rahim is of 4years and Ramu is 3 years old.

Let the length be ‘x’ and breadth be ‘y’.

Area of rectangle = length × breadth

= x × y

= xy

According to the question,

The length of a rectangle is increased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres

⇒ (x + 5) × (y – 3) = xy – 5

⇒ x (y – 3) + 5(y – 3) = xy – 5

⇒ xy – 3x + 5y – 15 = xy – 5

⇒ xy – 3x + 5y – xy = –5 + 15

⇒ –3x + 5y = 10 … (1)

the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres

⇒ (x + 3) × (y + 2) = xy + 50

⇒ x (y + 2) + 3(y + 2) = xy + 50

⇒ xy + 2x + 3y + 6 = xy + 50

⇒ xy + 2x + 3y – xy = 50 – 6

⇒ 2x + 3y = 44 … (2)

Equating equation (1) and (2)

–3x + 5y = 10 … (1)

2x + 3y = 44 … (2)

Multiplying equation (1) by 2 and equation (2) by 3 and equate

y = 8

Put y = 8 in Equation (3)

–6x + 10 × 8 = 20

⇒ –6x + 80 = 20

⇒ –6x = 20 – 80

⇒ –6x = –60

⇒ 6x = 60

⇒

⇒ x = 10

∴ The length of rectangle is 10 metres and breadth are 8 metres.

Let the larger part of rope be ‘x’ and smaller part be ‘y’.

According to question,

x + y = 10 … (1)

Area of larger square – area of smaller square

⇒

⇒

⇒

⇒

⇒

⇒ 8(x – y) = 1

⇒ 8x – 8y = 1 … (2)

Equating equation (1) and (2)

x + y = 10

8x – 8y = 1

Multiply equation (1) by 8

8x + 8y = 80 … (3)

8x – 8y = 1 … (2)

16x = 81

Put in equation (2)

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ The length of the rope to after cutting should be respectively.

Let the length be ‘x’ and breadth be ‘y’.

Length = Breadth + 1

⇒ x = y + 1 … (1)

Area of rectangle = Length × breadth

⇒

⇒

⇒

⇒ 15 = 4(y² + y)

⇒ 15 = 4y² + 4y

⇒ 4y² + 4y – 15 = 0

⇒ 4y² – 6y + 10y – 15 = 0

⇒ 2y (2y – 3) + 5(2y – 3) = 0

⇒ (2y – 3) (2y + 5) = 0

2y – 3 = 0 or 2y + 5 = 0

Hence, side of a rectangle cannot be negative. Therefore,

Put in equation (1)

⇒

⇒

⇒

∴ The length is metre and breadth are metre.

Let the base be ‘x’ and perpendicular be ‘y’.

Area of triangle

⇒

⇒

⇒

⇒ 15 = xy … (1)

By Pythagoras Theorem,

(Base)² + (Height)² = (hypotenuse)²

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ … (2)

From equation (1),

⇒

⇒

⇒ 2(x² + 15) = 17 × x

⇒ 2x² + 30 = 17x

⇒ 2x² – 17x + 30 = 0

⇒ 2x² – 12x - 5x + 30 = 0

⇒ 2x (x – 6) – 5(x – 6) = 0

⇒ (2x – 5) (x – 6) = 0

2x – 5 = 0 or x – 6 = 0

2x = 5 or x = 6

or x = 6

Put in equation (2)

⇒

⇒

⇒

Put x = 6 in equation 2

⇒

⇒

⇒

∴ Base of triangle can be or 6

Perpendicular of triangle can be or 6.