# RS Aggarwal Solutions Chapter 3 Linear Equations in Two Variables Exercise - 3F Class 10 Maths

 Chapter Name RS Aggarwal Chapter 3 Linear Equations in Two Variables Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 3AExercise 3BExercise 3CExercise 3DExercise 3EMCQ Related Study NCERT Solutions for Class 10 Maths

### Exercise 3F Solutions

1. Write number of solutions of the following pair of linear equations:

x + 2y – 8 = 0

2x + 4y = 16

Solution

The given equations are

x + 2y - 8 = 0 ...(i)

2x + 4y – 16 = 0 …(ii)

Which of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where,

a1 = 1, b1 = 2, c1 = -8, a2 = 2, b2 = 4 and c2 = - 18

Now,

a1/a2 = 1/2

b1/b2 = 2/4 = 1/2

c1/c2 = -8/-16 = 1/2

⇒ a1/a2 = b1/b2 = c1/c2 = 1/2

Thus, the pair of linear equations are coincident and therefore, has infinitely many solutions.

2. Find the value of k for which the system of linear equations has an infinite number of solutions.

2x + 3y – 7 = 0,

(k – 1)x + (k + 2)y = 3k

Solution

The given equations are

2x + 3y – 7 = 0 ...(i)

(k – 1)x + (k + 2)y – 3k = 0 ...(iii)

Which is of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where,

a1 = 2, b1 = 3, c1 = -7, a2 = k – 1, b2 = k + 2 and c2 = -3k

For the given pair of linear equations to have infinitely many solutions, we must have

a1/a2 = b1/b2 = c1/c2

⇒ 2/(k – 1) = 3/(k + 2), 3/(k + 2) = -7/-3k and 2/(k – 1) = -7/-3k

⇒ 2(k + 2) = 3(k – 1), 9k = 7k + 14 and 6k = 7k – 7

⇒ k = 7, k = 7 and k = 7

Hence, k = 7

3. Find the value of k for which the system of linear equations has an infinite number of solutions.

10x + 5y – (k – 5) = 0

20x + 10y – k = 0

Solution

The given pair of linear equations are

10x + 5y – (k – 5) = 0 ...(i)

20x + 10y – k = 0 ...(ii)

Which is of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where,

a1 = 10, b1 = 5, c1 = -(k – 5), a2 = 20, b2 = 10 and c2 = -k

For the given pair of linear equations to have infinitely many solutions, we must have

a1/a2 = b1/b2 = c1/c2

⇒ 10/20 = 5/10 = -(k – 5)/-k

⇒ 1/2 = (k – 5)/k

⇒ 2k – 10 = k

⇒ k = 10

Hence, k = 10.

4. Find the value of k for which the system of linear equations has an infinite number of solutions.

2x + 3y = 9,

6x + (k – 2)y = 3k – 2

Solution

The given pair of linear equations are

2x + 3y – 9 = 0 ...(i)

6x + (k – 2)y – (3k – 2) = 0 …(ii)

Which is of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where,

a1 = 2, b1 = 3, c1 = -9, a2 = 6, b2 = k – 2 and c2 = -(3k – 2)

For the given pair of linear equations to have infinitely many solutions, we must have

a1/a2 = b1/b2 ≠ c1/c2

⇒ 2/6 = 3/(k – 2) ≠ -9/(-3k – 2)

⇒ 2/6 = 3/(k – 2), 3/(k – 2) ≠ -9/-(3k – 2)

⇒ k = 11, 3/(k – 2) ≠ 9/(3k – 2)

⇒ k = 11, 3(3k – 2) ≠ 9(k – 2)

⇒ k = 11, 1 ≠ 3 (true)

Hence, k = 11.

5. Write the number of solutions of the following pair of linear equations:

x + 3y – 4 = 0, 2x + 6y – 7 = 0

Solution

The given pair of linear equations are

x + 3y – 4 = 0 ...(i)

2x + 6y – 7 = 0 ...(ii)

which is of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where,

a1 = 1, b1 = 3, c1 = -4, a2 = 2, b2 = 6 and c2 = - 7

Now

a1/a2 = 1/2

b1/b2 = 3/6 = 1/2

c1/c2 = -4/-7 = 4/7

⇒ a1/a2 = b1/b2 ≠ c1/c2

Thus, pair of the given linear equations has no solution.

6. Find the values of k for which the system of equations 3x + ky = 0

2x – y = 0 has a unique solution.

Solution

The given pair of linear equations are

3x + ky = 0 ...(i)

2x – y = 0 ...(ii)

Which is of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where,

a1 = 3, b1 = k, c1 = 0, a2 = 2, b2 = -1 and c2 = 0

For the system to have a unique solution, we must have

a1/a2 = b1/b2

⇒ 3/2 ≠ k/-1

⇒ k ≠ -3/2

Hence, k ≠ -3/2.

7. The difference of two numbers is 5 and the difference between their squares is 65. Find the numbers.

Solution

Let the numbers be x and y, where x > y.

Then as per the question

x – y = 5 ...(i)

x2 – y2 = 65 ...(ii)

Dividing (ii) by (i), we get

(x2 – y2)/(x – y) = 65/5

⇒ (x – y)(x + y)/(x – y) = 13

⇒ x + y = 13 ...(iii)

Now, adding (i) and (ii), we have

2x = 18 ⇒ x = 9

Substitute x = 9 in (iii), we have

9 + y = 13

⇒ y = 4

Hence, the numbers are 9 and 4.

8. The cost of 5 pens and 8 pencils together cost Rs 120 while 8 pens and 5 pencils together cost Rs. 153. Find the cost of a 1 pen and that of a 1 pencil.

Solution

Let the cost of 1 pen and 1 pencil are ₹x and ₹y respectively.

then as per the question

5x + 8y = 120 ...(i)

8x + 5y = 153 ...(ii)

Adding (i) and (ii), we get

13x + 13y = 273

⇒ x + y = 21 ...(iii)

Subtracting (i) from (ii), we get

3x – 3y = 33

⇒ x – y = 11 ...(iv)

Now, adding (iii) and (iv), we get

2x = 32

⇒ x = 16

Substituting x = 16 in (iii), we have

16 + y = 21 ⇒ y = 5

Hence, the cost of 1 pen and 1 pencil are respectively ₹16 and ₹5.

9. The sum of two numbers is 80. the large number exceeds four times the smaller one by 5. Find the numbers.

Solution

Let the larger number be x and the smaller number be y.

then as per the question

x + y = 80 ...(i)

x = 4y + 5

x – 4y = 5 ...(ii)

Subtracting (ii) from (i), we get

5y = 75

⇒ y = 15

Now, putting y = 15 in (i), we have

x + 15 = 80

⇒ x = 65

Hence, the numbers are 65 and 15.

10. A number consists of two digits whose sum is 10. If 8 is subtracted form the number, its digits are reversed. Find the number.

Solution

Let the ones digit and tens digit be x and y respectively.

Then as per the question

x + y = 10 ...(i)

(10y + x) – 18 = 10x + y

x – y = -2 ...(ii)

Adding (i) and (ii), we get

2x = 8

⇒ x = 4

Now, putting x = 4 in (i), we have

4 + y = 10

⇒ y = 6

Hence, the number is 64.

11. A man purchased 47 stamps of 20p and 25p for ₹10. Find the number of each type of stamps.

Solution

Let the number of stamps of 20p and 25p be x and y respectively.

Then as per the question

x + y = 47 ...(i)

0.20x + 0.25y = 10

4x + 5y = 200 ...(ii)

From (i), we get

y = 47 – x

Now, substituting y = 47 – x in (ii), we have

4x + 5(47 – x) = 200

⇒ 4x – 5x + 253 = 200

⇒ x = 235 – 200 = 35

Putting x = 35 in (i), we get

35 + y = 47

⇒ y = 47 – 35 = 12

Hence, the number of 20p stamps and 25p stamps are 35 and 12 respectively.

12. A man has some hens and cows. If the number of heads be 48 and number of feet by 140. How many cows are there.

Solution

Let the number of hens and cow be x and y respectively.

As per the question

x + y = 48 ...(i)

2x + 4y = 140

x + 2y = 70 ...(ii)

Subtracting (i) from (ii), we have

y = 22

hence, the number of cows is 22.

13. If 2/x + 3/y = -9/xy and 4/x + 9/y = 21/xy, find the values of x and y.

Solution

The given pair of equation is

2/x + 3/y = 9/xy ...(i)

4/x + 9/y = 21/xy ...(ii)

Multiplying (i) and (ii) by xy, we have

3x + 2y = 9 ...(iii)

9x + 4y = 21 ...(iv)

Now, multiplying (iii) by 2 and subtracting from (iv), we get

9x – 6x = 21 – 8

⇒ x = 3/3 = 1

Putting x = 1 in (iii), we have

3 × 1 + 2y = 9

⇒ y = (9 – 3)/2 = 3

Hence, x = 1 and y = 3.

14. If x/4 + y /3 = -15/12 and x/2 + y = 1, then find the value of (x + y).

Solution

The given pair of equations is

x/4 + y/3 = 5/12 ...(i)

x/2 + y = 1 ...(ii)

Multiplying (i) by 12 and (ii) by 4, we have

3x + 4y = 5 ...(iii)

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

Now, subtracting (iv) from (iii), we get

x = 1

Putting x = 1 in (iv), we have

2 + 4y = 4

⇒ 4y = 2

⇒ y = 1/2

∴ x + y = 1 + 1/2 = 3/2

Hence, the value of x + y is 3/2.

15. If 12x + 17y = 53 and 17x + 12y = 63 then find the value of (x + y)

Solution

The given pair of equations is

12x + 17y = 53 ...(i)

17x + 12y = 63 ...(ii)

Adding (i) and (ii), we get

29x + 29y = 116

⇒ x + y = 4 (Dividing by 4)

Hence, the value of x + y is 4.

16. Find the value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has infinite nonzero solutions.

Solution

The given system is

3x + 5y = 0 …(i)

kx + 10y = 0 …(ii)

This is a homogeneous system of linear differential equation, so it always has a zero solution i.e., x = y = 0

But to have a non-zero solution, it must have infinitely many solutions.

For this, we have

a1/a2 = b1/b2

⇒ 3/k = 5/10 = 1/2

⇒ k = 6

Hence, k = 6

17. Find the value of k for which the system of equations kx – y = 2 and 6x – 2y = 3 has a unique solution.

Solution

The given system is

kx – y – 2 = 0 ...(i)

6x – 2y – 3 = 0 ...(ii)

Here, a1 = k, b1 = - 1, c1 = -2, a2 = 6, b2 = - 2 and c2 = - 3

For the system, to have a solution, we must have

a1/a2 ≠ b1/b2

⇒ k/6 ≠ -1/-2 = 1/2

⇒ k ≠ 3

Hence, k ≠ 3.

18. Find the value of k for which the system of equations 2x + 3y – 5 = 0 and 4x + ky – 10 = 0 has infinite number of solutions.

Solution

The given system is

2x + 3y – 5 = 0 ...(i)

4x + ky – 10 = 0 ...(ii)

Here, a1 = 2, b1 = 3, c1 = - 5, a2 = 4, b2 = k and c2 = -10

For the system, to have an infinite number of solutions, we must have

a1/a2 = b1/b2 = c1/c2

⇒ 2/4 = 3/k = -5/-10

⇒ 1/2 = 3/k = 1/2

⇒ k = 6

Hence, k = 6.

19. Show that the system 2x + 3y – 1 = 0 and 4x + 6y – 4 = 0 has no solution.

Solution

The given system is

2x + 3y – 1 = 0 ...(i)

4x + 6y – 4 = 0 ...(ii)

Here, a1 = 2, b1 = 3, c1 = - 1, a2 = 4, b2 = 6 and c2 = 4

Now,

a1/a2 = 2/4 = 1/2

b1/b2 = 3/6 = 1/2

c1/c2 = -1/-4 = 1/4

Then, a1/a2 = b1/b2 ≠ c1/c2 and therefore the given system has no solution.

20. Find the value of k for which the system of equations x + 2y – 3 = 0 and 5x + ky + 7 = 0 is inconsistent.

Solution

The given system is

x + 2y – 3 = 0 ...(i)

5x + ky + 7 = 0 ...(ii)

Here, a1 = 1, b1 = 2, c1 = - 3, a2 = 5, b2 = k and c2 = 7

For the system, to be consistent, we must have

a1/a2 = b1/b2 ≠ c1/c2

⇒ 1/5 = 2/k ≠ -3/7

⇒ 1/5 = 2/k

⇒ k = 10

Hence, k = 10.

21. Solve for x and y: 3/(x + y) + 2/(x – y) = 2, 9/(x + y) – 4/(x – y) = 1

Solution

The given system is equations is

3/(x + y) – 2/(x – y) = 2 ...(i)

9/(x + y) – 4/(x – y) = 1 ...(ii)

Substituting 1/(x + y) = u and 1/(x – y)= v in (i) and (ii), the give equations are changed to

3u + 2v = 2 ...(iii)

9u – 4v = 1 ...(iv)

Multiplying (i) by 3 and subtracting (ii) from it, we get

6u + 4v = 6 – 1

⇒ u = 5/10 = 1/2

Therefore

x + y = 3 ...(v)

x – y = 2 ...(vi)

Now, adding (v) and (vi) we have

2x = 5

⇒ x = 5/2

Substituting x = 5/2 in (v), we have

5/2 + y = 3

⇒ y = 3 - 5/2 = 1/2

Hence, x = 5/2 and y = 1/2.