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 
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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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where,
a_{1} = 1, b_{1 }= 2, c_{1} = 8, a_{2} = 2, b_{2} = 4 and c_{2} =  18
Now,
a_{1}/a_{2} = 1/2
b_{1}/b_{2} = 2/4 = 1/2
c_{1}/c_{2} = 8/16 = 1/2
⇒ a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2} = 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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where,
a_{1} = 2, b_{1} = 3, c_{1} = 7, a_{2} = k – 1, b_{2} = k + 2 and c_{2} = 3k
For the given pair of linear equations to have infinitely many solutions, we must have
a_{1}/a_{2} = b_{1}/b_{2 }= c_{1}/c_{2}
⇒ 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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where,
a_{1} = 10, b_{1} = 5, c_{1} = (k – 5), a_{2 }= 20, b_{2} = 10 and c_{2} = k
For the given pair of linear equations to have infinitely many solutions, we must have
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
⇒ 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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where,
a_{1} = 2, b_{1} = 3, c_{1} = 9, a_{2} = 6, b_{2} = k – 2 and c_{2} = (3k – 2)
For the given pair of linear equations to have infinitely many solutions, we must have
a_{1}/a_{2 }= b_{1}/b_{2} ≠ c_{1}/c_{2}
⇒ 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 a_{1}x + b_{1}y + c_{1 }= 0 and a_{2}x + b_{2}y + c_{2} = 0
where,
a_{1} = 1, b_{1} = 3, c_{1} = 4, a_{2} = 2, b_{2} = 6 and c_{2 }=  7
Now
a_{1}/a_{2} = 1/2
b_{1}/b_{2 }= 3/6 = 1/2
c_{1}/c_{2} = 4/7 = 4/7
⇒ a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where,
a_{1} = 3, b_{1} = k, c_{1} = 0, a_{2} = 2, b_{2} = 1 and c_{2 }= 0
For the system to have a unique solution, we must have
a_{1}/a_{2} = b_{1}/b_{2}
⇒ 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)
x^{2} – y^{2} = 65 ...(ii)
Dividing (ii) by (i), we get
(x^{2} – y^{2})/(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 nonzero solution, it must have infinitely many solutions.
For this, we have
a_{1}/a_{2} = b_{1}/b_{2}
⇒ 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, a_{1} = k, b_{1} =  1, c_{1} = 2, a_{2} = 6, b_{2} =  2 and c_{2} =  3
For the system, to have a solution, we must have
a_{1}/a_{2} ≠ b_{1}/b_{2}
⇒ 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, a_{1} = 2, b_{1} = 3, c_{1} =  5, a_{2} = 4, b_{2} = k and c_{2} = 10
For the system, to have an infinite number of solutions, we must have
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
⇒ 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, a_{1} = 2, b_{1} = 3, c_{1} =  1, a_{2} = 4, b_{2} = 6 and c_{2} = 4
Now,
a_{1}/a_{2} = 2/4 = 1/2
b_{1}/b_{2} = 3/6 = 1/2
c_{1}/c_{2} = 1/4 = 1/4
Then, a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2} 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, a_{1} = 1, b_{1} = 2, c_{1} =  3, a_{2} = 5, b_{2} = k and c_{2} = 7
For the system, to be consistent, we must have
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
⇒ 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.