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# Class 9 SELINA Solutions Maths Chapter 6 - Simultaneous (Linear) Equations (Including Problems)

## Rational and Irrational Numbers Exercise Ex. 6(E)

### Solution 5

Let the two numbers be x and y.

Then,

x + y = 8 …..(i)

x2 - y2 = 32 …..(ii)

(x - y)(x + y) = 32

x - y = 4 ….. (iii)

Adding (i) and (iii), we get

2x = 12 x = 6

From (i), y = 8 - x = 8 - 6 = 2

Therefore, the numbers are 6 and 2.

### Solution 1

Let the two numbers be x and y

According to the question,

3x - 2y = 0 ...(1)

Also,

2x - 3y = -20 ...(2)

Multiplying equation no. (1) by 2 and (2) by 3and substracting

From (1), we get

3x - 2(12) = 0

x = 8

Thus, the numbers are 8 and 12.

### Solution 2

Let the smaller number be x

and the larger number be y.

According to the question,

7x -; 4y = 0...(1)

and,3y + 2x = 59...(2)

Multiplying equation no. (1) by 3 and (2) by 4.and adding them

x =

From (1)

Hence, the number are

### Solution 3

Let the two numbers be a and b respectively such that b > a.

According to given condition,

### Solution 4

Two numbers are x and y such that x > y.

Now,

x + y = 50 ….(i)

And,

y2 - x2 = 720

(y - x)(y + x) = 720

(y - x)(50) = 720

y - x = 14.4 ….(ii)

Adding (i) and (ii), we get

2y = 64.4

y = 32.2

Substituting the value of y in (i), we have

x + 32.2 = 50

x = 17.8

Thus, the two numbers are 17.8 and 32.2 respectively.

### Solution 6

Two numbers are x and y respectively such that x > y.

Then,

x - y = 4 ….(i)

x = 4 + y

And,

### Solution 8

Let the numerator and denominator a fraction be x and y respectively .

According to the question,

3x - 2y = -8...(1)

And,

Now subtracting,

From (1) ,

3x - 2 (7) = -8

3x = -8 + 14

x = 2

Required fraction =

### Solution 9

Let the numerator and denominator of a fraction be x and y respectively .Then the fraction will be

According to the question,

x + y = 7...(1)

5y - 4x = 8...(2)

Multiplying equation no. (1) by 4 and add with (2),

From (1)

x + 4 = 7

x = 3

Required fraction =

### Solution 12

Let the digit at unit’s place be x and the digit at ten’s place y.

Required no. = 10y + x

If the digit’s are reversed,

Reversed no. = 10y + x

According to the question,

x + y = 5...(1)

and,

(10y + x) - (10x + y) = 27

From (1)

x + 4 = 5

x = 1

Require no is

10 (4) + 1 = 41

### Solution 13

Let the digit at unit’s place be x and the digit at ten’s place be y.

Required no. = 10y + x

If the digit’s are reversed

Reversed no. = 10x + y

According to the question,

x + y = 7...(1)

and,

10x + y - 2 = 2(10y + x).

8x - 19y = 2...(2)

Multiplying equation no. (1) by 19.

x = 5

From (1)

5 +y = 7

y = 2

Required number is

10(2) + 5

= 25.

### Solution 14

Let the digit at unit’s place be x and the digit at ten’s place be y.

Required no. = 10y + x

According to the question

y = 3x 3x - y = 0...(1)

and,10y + x + x = 32

10y + 2x = 32...(2)

Multiplying equation no. (1) by 10

From (1),we get

y = 3(1) = 3

Required no is

10(3) + 1 = 31

### Solution 15

Let the digit a unit’s place be x and the digit at ten’s place be y.

Required no. = 10y + x.

According to the question,

y - 2x = 2

-2x + y = 2...(1)

and,

(10x + y) -3 (y + x) = 5

7x - 2y = 5...(2)

Multiplying equation no. (1) by 2.

From (1) ,we get

-2(3) + y = 2

y = 8

Required number is

10(8) + 3 = 83.

## Simultaneous Linear Equations Exercise Ex. 6(A)

### Solution 1

8x + 5y = 9...(1)

3x + 2y = 4...(2)

(2)y =

Putting this value of y in (2)

3x + 2

15x + 18 - 16x = 20

x = -2

From (1) y = =

y = 5

### Solution 2

2x - 3y = 7...(1)

5x + y = 9...(2)

(2)y = 9 - 5x

Putting this value of y in (1)

2x - 3 (9 - 5x) = 7

2x - 27 + 15x = 7

17x = 34

x = 2

From (2)

y = 9 - 5(2)

y = -1

### Solution 3

2x + 3y = 8...(1)

2x = 2 + 3y...(2)

(2) 2x = 2 + 3y

Putting this value of 2x in (1)

2 + 3y + 3y = 8

6y = 6

y = 1

From (2) 2x = 2 + 3 (1)

x =

x = 2.5

### Solution 5

6x = 7y + 7...(1)

7y - x = 8...(2)

(2) x = 7y - 8

Putting this value of x in (1)

6(7y - 8) = 7y + 7

42y - 48 = 7y + 7

35y = 55

From (2) x =

x = 3

### Solution 6

y = 4x -7...(1)

16x- 5y = 25...(2)

(1) y = 4x - 7

Putting this value of y in (2)

16x - 5 (4x - 7) = 25

16x - 20x + 35 = 25

-4x = -10

From (1)

y = 10-7=3

Solution is .

### Solution 7

2x + 7y = 39...(1)

3x + 5y = 31...(2)

(1) x =

Putting this value of x in (2)

117 - 21y + 10y = 62

-11y = -55

y = 5

From (1) x =

## Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(B)

### Solution 1

13 + 2y = 9x...(1)

3y = 7x...(2)

Multiplying equation no. (1) by 3 and (2) by 2, we get,

From (2)

3y = 7(3)

y = 7

### Solution 2

3x - y = 23...(1)

4x + 3y = 48...(2)

Multiplying equation no. (1) by 3

From (1)

3(9) - y = 23

y = 27 - 23

y = 4

### Solution 4

Multiplying equation no. (1) by 3 and(2) by 5.

From (1)

y = 2x - 6

y = 0

x = 3 ; y = 0

### Solution 7

3 - (x - 5) = y + 2

2(x + y) = 4 - 3y

-x - y = -6

x + y = 6...(1)

2x + 5y = 4...(2)

Multiplying equation no. (1) by 2.

From (1)

### Solution 8

2x -3y - 3 = 0

2x - 3y = 3...(1)

4x + 24y = -3...(2)

Multiplying equation no. (1) by 8.

From (1)

### Solution 9

13x + 11y = 70...(1)

11x+ 13y = 74...(2)

24x + 24y = 144

x + y = 6...(3)

subtracting (2) from (1)

2x - 2y = -4

x - y = -2...(4)

x + y = 6...(3)

From (3)

2 + y = 6y = 4

### Solution 10

41x + 53y = 135...(1)

53x + 41y = 147...(2)

94x + 94y = 282

x + y = 3...(3)

Subtracting (2) from (1)

From (3)

x + 1 = 3x = 2

### Solution 11

2x + y = 23...(1)

4x - y = 19...(2)

Adding equation (1) and (2) we get,

2x + y = 23

4x - y = 19

From (1)

2(7) + y = 23

y = 23 - 14

y = 9

x - 3y = 7 - 3(9) = -20

And 5y - 2x = 5(9) - 2(7) = 45- 14 = 31

### Solution 12

10 y = 7x - 4

-7x + 10y = -4...(1)

12x + 18y = 1...(2)

Multiplying equation no. (1) by 12 and (2) by 7.

From (1)

-7x + 10

-7x = -4 +

4and 8y - x = 8

(i)

(ii)

## Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(D)

### Solution 1

Multiplying equation no. (1) by 7 and (2) by 4.

From (1)

### Solution 5

Multiplying equation no. (1) by 5 and (2) by 2.

From (1) 3

y = ax + 3

(ii)

### Solution 7

(i)

From (1)

(ii)

x + y = 7xy...(1)

2x - 3 = -xy...(2)

Multiplying equation no. (1) by 3.

From (1)

## Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(F)

### Solution 1

Let present age of A = x years

And present age of B = y years

According to the question,

Five years ago,

x - 5 = 4(y - 5)

x - 4y = -15...(1)

Five years later,

x + 5 = 2(y + 5)

Now subtracting (1)from(2)

From (1)

x - 4 (10) = -15

x = 25

Present ages of A and B are 25 years and 10 years respectively.

### Solution 2

Let A’s presentage be x years

and B’s present age be y years

According to the question

x = y + 20

x - y = 20...(1)

Five years ago,

x - 5 = 3(y - 5)

Subtracting (1)from(2),

y = 15

From (1)

x = 15 + 20

x = 35

Thus, present ages of A and B are 35 years and 15 years.

### Solution 5

Let A’s annual in come = Rs.x

and B’s annual income = Rs. y

According to the question,

4x - 3y = 0...(1)

and,

7x - 5y = 10000...(2)

Multiplying equation no. (1) by 7 and (2) by 4.and subtracting (4) from (3)

From (1)

4x - 3 (40000) = 0

x = 30000

Thus, A’s income in Rs. 30,000 and B’s income is Rs. 40,000.

### Solution 6

Let the no. of pass candidates be x

and the no. of fail candidates be y.

According to the question,

x - 4y = 0...(1)

and

x - 5y = -30...(2)

From (1)

- 4(30) = 0

x = 120

Total students appeared = x + y

= 120 + 30

= 150

### Solution 7

Let the numberof pencils with A = x

and the number of pencils with B = y.

If A gives 10 pencils to B,

y + 10 = 2(x - 10)

2x - y = 30...(1)

If B gives to pencils to A

y - 10 = x + 10

From (1)

2(50) - y = 30

y = 70

Thus, A has 50 pencils and B has 70 pencils.

### Solution 8

Let the number of adults = x

and the number of children = y

According to the question,

x + y = 1250...(1)

and 75x + 25y = 61250

x = 6000

From (1)

600 + y = 1250

y = 650

Thus, number of adults = 600

and the number of children = 650.

### Solution 9

Let the cost price of article A = Rs. x

and the cost price of articles B = Rs. y

According to the question,

(x + 5% of x) +(y + 7% of y) = 1167

105x + 107y = 1167...(1)

and

107x + 105y = 116500...(2)

212x + 212y = 233200

x + y = 1100...(3)

subtracting (2)from (1)

-2x + 2y = 200

y = 600

from (3)

x +600 = 1100

x = 500

Thus, cost price of article A is Rs. 500.

and that of article B is Rs. 600.

### Solution 10

Let Pooja’s 1 day work =

and Ritu’s 1 day work =

According the question,

and,

Using the value of y from (2) in (1)

From (2)

y = 30

Pooja will complete the work in 40 days and Ritu will complete the work in 30 days.

## Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(G)

### Solution 1

Let Rohit has Rs. x

and Ajay has Rs. y

When Ajay gives Rs. 100 to Rohit

x + 100 = 2(y - 100)

x - 2y = -300...(1)

When Rohit gives Rs. 10 to Ajay

6(x-10) = y + 10

6x - y = 70...(2)

Multiplying equation no. (2) By 2.

x = 40

From (1)

40 - 2y = -300

-2y = -340

y = 170

Thus, Rohit has Rs. 40

and Ajay has Rs. 170

### Solution 3

Let the digit at ten’s place be x

And the digit at unit’s place be y

Required number = 10x + y

When the digits are interchanged,

Reversed number = 10y + x

According to the question,

7(10x + y) = 4(10y + x)

66x = 33y

2x - y = 0...(1)

Also,

From (1) 2(3) - y = 0

y = 6

Thus, Required number = 10(3) + 6 = 36

### Solution 4

Let, the fare of ticket for station A be Rs. x

and the fare of ticket for station B be Rs. y

According, to the question

2x + 3y = 77....(1)

and3x+5y = 124...(2)

Multiplying equation no. (1) by 3 and (2) by 2.

y = 17

From (1) 2x + 3 (17) = 77

2x = 77 - 51

2x = 26

x = 13

Thus, fare for station A = Rs. 13

and, fare for station B = Rs. 17.

### Solution 6

Let the quantity of 90% acid solution be x litres and

The quantity of 97% acid solution be y litres

According to the question,

x + y = 21...(1)

and 90% of x + 97% of y = 95% of 21

90x + 97y = 1995...(2)

Multiplying equation no. (1) by90, we get,

From (1)x + 15 = 21

x = 6

Hence, 90% acid solution is 6 litres and 97% acid solution is 15 litres.

### Solution 8

Weight of Mr. Ahuja = x kg

and weight of Mrs. Ahuja = y kg.

After the dieting,

x - 5 = y

x - y = 5...(1)

and,

7x - 8y = -32...(2)

Multiplying equation no. (1) by 7, we get

Now subtracting (2) from (3)

From (1)

x - 67 = 5x = 72

Thus, weight of Mr. Ahuja = 72 kg.

and that of Mr. Anuja = 67 kg.