RS Aggarwal Solutions Chapter 3 Linear Equations in Two Variables MCQ Class 10 Maths
Chapter Name | RS Aggarwal Chapter 3 Linear Equations in Two Variables |
Book Name | RS Aggarwal Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
MCQ for Linear Equations in Two Variables Class 10 Maths
1. If 2x + 3y = 12 and 3x – 2y = 5 then
(a) x = 2, y = 3
(b) x = 2, y = - 3
(c) x = 3, y = 2
(d) x = 3, y = - 2
Solution
The given system of equations is
2x + 3y = 12 …(i)
3x – 2t = 5 …(ii)
Multiplying (i) by 2 and (ii) by 3 and then adding, we get
4x + 9x = 24 + 15
⇒ x = 39/13 = 3
Now, putting x = 3 in (i), we have
2 × 3 + 3y = 12
⇒ y = (12 – 6)/3 = 2
Thus, x = 3 and y = 2.
2. If x – y = 2 and 2/(x + y) = 1/5 then
(a) x = 4, y = 2
(b) x = 5, y = 3
(c) x = 6, y = 4
(d) x = 7, y = 5
Solution
(c) x = 6, y = 4
The given system of equations is
x – y = 2 ...(i)
x + y = 10 ...(ii)
Adding (i) and (ii), we get
2x = 12
⇒ x = 6
Now, putting x = 6 in (ii), we have
6 + y = 10
⇒ y = 10 – 6 = 4
Thus, x = 6 and y = 4.
3. If 2x/3 – y/2 + 1/6 = 0 and x/2 + 2y/3 = 3 then
(a) x = 2, y = 3
(b) x = - 2, y = 3
(c) x = 2, y = - 3
(d) x = - 2, y = - 3
Solution
The given system of equations is
2x/3 – y/3 = 1/6 ...(i)
x/2 + 2y/3 = 3 ...(ii)
Multiplying (i) and (ii) by 6, we get
4x – 3y = - 1 ...(iii)
3x + 4y = 18 ...(iv)
Multiplying (iii) by 4 and (iv) by 3 and adding, we get
16x + 9x = - 4 + 54
⇒ x = 50/25 = 2
Now, putting x = 2 in (iv), we have
3×2 + 4y = 18
⇒ y = (18 – 6)/4 = 3
Thus, x = 2 and y = 3.
4. If 1/x + 2/y = 4 and 3/y – 1/x = 11 then
(a) x = 2, y = 3
(b) x = - 2, y = 3
(c) x = -1/2
(d) x = -1/2, y = 1/3
Solution
The given system of equations is
1/x + 2/y = 4 ...(i)
3/y – 1/x = 11 ...(ii)
Adding (i) and (ii), we get
2/y + 3/y = 15
⇒ 5/y = 15
⇒ y = 5/15 = 1/3
Now, putting y = 1/3 in (i), we have
1/x + 2 × 3 = 4
⇒ 1/x = 4 – 6
⇒ x = - 1/2
Thus, x = 1/2 and y = 1/3.
5. If (2x + y + 2)/5 = (3x – y + 1)/3 = (3x + 2y + 1)/6 then
(a) x = 1, y = 1
(b) x = - 1, y = - 1
(c) x = 1, y = 2
(d) x = 2, y = 1
Solution
Consider,
(2x + y + 2)/5 = (3x – y + 1)/3 and (3x – y + 1)/3 = (3x + 2y + 1)/3,
Now simplifying these equations, we get
3(2x + y + 2) = 5(3x – y + 1)
⇒ 6x + 3y + 6 = 15x – 5y + 5
⇒ 9x – 8y = 1 ...(i)
And
6(3x – y + 1) = 3(3x + 2y + 1)
⇒ 18x - 6y + 6 = 9x + 6y + 3
⇒ 3x – 4y = - 1 ...(ii)
multiplying (ii) by 2 and subtracting it from (i)
9x – 6x = 1 + 2
⇒ x = 1
Now, putting x = 1 in (ii), we have
3×1 – 4y = - 1
⇒ y = (3 + 1)/4 = 1
Thus, x = 1, y = 1.
6. If 3/(x + y) + 2/(x – y) = 2 and 9/(x + y) – 4/(x – y) = 1 then
(a) x = 1/2, y = 3/2
(b) x = 5/2, y = ½
(c) x = 3/2, y = ½
(d) x = 1/2, y = 5/2
Solution
(b) x = 5/2, y = 1/2
The given equations are
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), new system becomes
3u + 2v = 2 ...(iii)
9u – 4v = 1 ...(iv)
Now, multiplying (iii) by 2 and adding it with (iv), we get
6u + 9u = 4 + 1
⇒ u = 5/15 = 1/3
Again, multiplying (iii) by 2 and subtracting (iv) from, we get
6v + 4v = 6 – 1
⇒ v = 5/10 = 1/2
Therefore
x + y = 3 ...(v)
x – y = 2 ...(vi)
Adding (v) and (vi), we get
2x = 3 + 2
⇒ x = 5/2
Substituting x = 5/2, in (v), we have
5/2 + y = 3
⇒ y = 3 – 5/2 = 1/2
Thus, x = 5/2 and y = 1/2.
7. If 4x + 6y = 3xy and 8x + 9y = 5xy then
(a) x = 2, y = 3
(b) x = 1, y = 2
(c) x = 3, y = 4
(d) x = 1, y = - 1
Solution
The given equations are
4x + 6y = 3xy …(i)
8x + 9y = 5xy …(ii)
Dividing (i) and (ii) by xy, we get
6/x + 4/y = 3 …(iii)
9/x + 8/y = 5 …(iv)
Multiplying (iii) by 2 and subtracting (iv) from it, we get
12/x - 9/x = 6 – 5
⇒ 3/x = 1
⇒ x = 3
Substituting x = 3 in (iii), we get
6/3 + 4/y = 3
⇒ 4/y = 1
⇒ y = 4
Thus, x = 3 and y = 4.
8. If 29x + 37y = 103 and 37x + 29y = 95 then
(a) x = 1, y = 2
(b) x = 2, y = 1
(c) x = 3, y = 2
(d) x = 2, y = 3
Solution
The given system of equations is
29x + 37y = 103 ...(i)
37x + 29y = 95 ...(ii)
Adding (i) and (ii), we get
66x + 66y = 198
⇒ x + y = 3 ...(iii)
Subtracting (i) from (ii), we get
8x – 8y = - 8
⇒ x – y = - 1
Adding (iii) and (iv), we get
2x = 2
⇒ x = 1
Substituting x = 1 in (iii), we have
1 + y = 3
⇒ y = 2
Thus, x = 1 and y = 2
9. If 2(x+y) = 2(x–y) = √8 then the value of y is
(a) 1/2
(b) 3/2
(c) 0
(d) none of these
Solution
∵ 2x+y = 2x-y = √8
∴ x + y = x – y
⇒ y = 0
10. If 2/x + 3/y = 6 and 1/x + 1/2y = 2 then
(a) x = 1, y = 2/3
(b) x = 2/3, y = 1
(c) x = 1, y = 3/2
(d) x = 3/2, y = 1
Solution
(b) x = 2/3, y = 1
The given equations are
2/x + 3/y = 6 ...(i)
1/x + 1/2y = 2 ...(ii)
Multiplying (ii) by 2 and subtracting it from (ii), we get
3/y – 1/y = 6 – 4
⇒ 2/y = 2
⇒ y = 1
Substituting y = 1 in (ii), we get
1/x + 1/2 = 2
⇒ 1/x = 2 – 1/2
⇒ 1/x = 3/2
⇒ x = 2/3
11. The system of kx – y = 2 and 6x – 2y = 3 has a unique solution only when
(a) k = 0
(b) k ≠ 0
(c) k = 3
(d) k ≠ 3
Solution
The given equations are
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 given system to have a unique solution, we must have
a1/a2 ≠ b1/b2
⇒ k/6 ≠ -1/-2
⇒ k ≠ 3
11. The system x – 2y = 3 and 3x + ky = 1 have a unique solution only when?
(a) k = - 6
(b) k ≠ - 6
(c) k = 0
(d) k ≠ 0
Solution
The correct option is (b).
The given system of equations can be written as follows:
x – 2y – 3 = 0 and 3x + ky – 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = - 2, c1 = - 3, a2 = 3, b2 = k and c2 = - 1
∴ a1/a2 = 1/3, b1/b2 = -2/k and c1/c2 = -3/-1 = 3
These graph lines will intersect at a unique point when we have:
a1/a2 ≠ b1/b2
⇒ 1/3 ≠ -2/k
⇒ k ≠ - 6
Hence, k has all real values other than -6.
12. The system x + 2y = 3 and 5x + ky + 7 = 0 have no solution when?
(a) k = 10
(b) k ≠ 10
(c) k = -7/3
(d) k = - 21
Solution
The correct option is (a).
The given system of equations can be written as follows:
x + 2y – 3 = 0 and 5x + ky + 7 = 0
The given equations care of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = - 3, a2 = 5, b2 = k and c2 = 7
∴ a1/a2 = 1/5, b1/b2 = 2/k and c1/c2 = -3/7
For the system of equations to have no solution, we must have:
a1/a2 = b1/b2 ≠ c1/c2
∴ 1/5 = 2/k ≠ -3/7
⇒ k = 10
14. If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is
(a) -5/4
(b) 2/5
(c) 3/2
(d) 15/4
Solution
The given system of equations can be written as follows:
3x + 2ky – 2 = 0 and 2x + 5y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b1y + c2 = 0
Here, a1 = 3, b1 = 2k, c1 = -2, a2 = 2, b2 = 5 and c2 = 1
∴ a1/a2 = 3/2, b1/b2 = 2k/5 and c1/c2 = -2/1
For parallel lines, we have:
a1/a2 = b1/b2 ≠ c1/c2
∴ 3/2 = 2k/5 ≠ -2/1
⇒ k = 15/4
15. For what value of k do the equations kx – 2y = 2 and 3x + y = 5 represent two lines intersecting at a unique point ?
(a) k = 3
(b) k = -3
(c) k = 6
(d) all real values except -6
Solution
(a) all real values except -6
The given system of equations can be written as follows:
kx – 2y – 3 = 0 and 3x + y – 5 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = -2, c1 = - 3 and a2 = 3, b2 = 1 and c2 = - 5
∴ a1/a2 = k/3, b1/b2 = -2/1 and c1/c2 = -3/-5 = 3/5
thus, for these graph lines to intersect at a unique point, we must have:
a1/a2 ≠ b1/b2
⇒ k/3 ≠ -2/1
⇒ k ≠ -6
Hence, the graph lines will intersect at all real values of k except -6.
16. The pair of equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Solution
The given system of equations can be written as:
x + 2y + 5 = 0 and -3x – 6y + 1 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = 5, a2 = - 3, b2 = -6 and c2 = 1
∴ a1/a2 = 1/-3, b1/b2 = 2/6 = 1/-3 and c1/c2 = 5/1
∴ a1/a2 = b1/b2 ≠ c1/c2
Hence, the given system has no solution.
17. The pair of equations 2x + 3y = 5 and 4x + 6y = 15 has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Solution
(d) no solution
The given system of equations can be written as:
2x + 3y – 5 = 0 and 4x + 6y – 15 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = -5, a2 = 4, b2 = 6 and c2 = - 15
∴ a1/a2 = 2/4 = 1/2, b1/b2 = 3/6 = 1/2 and c1/c2 = -5/-15 = 1/3
∴ a1/a2 = b1/b2 ≠ c1/c2
Hence, the given system has no solution.
18. If a pair of linear equations is consistent, then their graph lines will be
(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident
Solution
(d) intersecting or coincident
If a pair of linear equations is consistent, then the two graph lines either intersect at a point or coincidence.
19. If a pair of linear equations is inconsistent, then their graph lines will be
(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident
Solution
(a) parallel
If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since, there is no common solution, their graph lines do not intersect. Hence, they are parallel.
20. In a ∆ABC, ∠C = 3∠B = 2(∠A + ∠B), then ∠B = ?
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Solution
Let ∠A = x° and ∠B = y°
∴ ∠A = 3∠B = (3y)°
Now, ∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ...(i)
Also, ∠C = 2(∠A + ∠B)
⇒ 3y = 2(x + y)
⇒ 2x – y = 0 ...(ii)
On multiplying (ii) by 4, we get:
8x – 4y = 0 ...(iii)
On adding (i) and (iii) we get:
9x = 180
⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180
⇒ 4y = (180 – 20) = 160
⇒ y = 40
∴ x = 20 and y = 40
∴ ∠B = y° = 40°
21. In a cyclic quadrilateral ABCD, it is being given that ∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?
Solution
(b) 80°
The correct option is (b).
In a cyclic quadrilateral ABCD.
∠A = (x + y + 10)°
∠B = (y + 20)°
∠C = (x + y – 30)°
∠D = (x + y)°
We have:
∠A + ∠C = 180° and ∠B + ∠D = 180° [Since ABCD is a cyclic quadrilateral]
Now, ∠A + ∠C = (x + y + 10)° + (x + y – 30)° = 180°
⇒ 2x + 2y – 20 = 180
⇒ x + y – 10 = 90
⇒ x + y = 160 ...(i)
Also, ∠B + ∠D = (y + 20)° + (x + y)° = 180°
⇒ x + 2y + 20 = 180
⇒ x + 2y = 160
On subtracting (i) from (ii), we get:
y = (160 – 100) = 60
On substituting y = 60 in (i), we get:
x + 60 = 100
⇒ x = (100 – 60) = 40
∴ ∠B = (y + 20)° = (60 + 20)° = 80°
22. 5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is
(a) 45 years
(b) 50 years
(c) 47 years
(d) 40 years
Solution
(d) 40 years
Let the man’s present age be x years.
Let his son’s present age be y years.
Five years later:
(x + 5) = 3(y + 5)
⇒ x + 5 = 3y + 15
⇒ x – 3y = 10 ...(i)
Five tears ago:
(x – 5) = 7(y – 5)
⇒ x – 5 = 7y – 35
⇒ x – 7y = - 30 ...(ii)
On subtracting (i) from (ii), we get:
x – 3 × 10 = 10
⇒ x – 30 = 10
⇒ x = (10 + 30) = 40 years
Hence, the man’s present age is 40 years.
23.
| Assertion (A) | Reason (R) |
The system of equations x + y – 8 = 0 and x – y – 2 = 0 has a unique solutions. | The system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution when a1/a2 = b1/b2. |
The correct answer is
(a)
(b)
(c)
(d)
Solution
Option (c) is the correct answer.
Clearly, Reason (R) is false.
On solving x + y = 8 and x – y = 2, we get:
x = 5 and y = 3
Thus, the given system has a unique solution. So, assertion (A) is true.
∴ Assertion (A) is true and Reason (R) is false.
24. The graphs of the equations 6x – 2y + 9 = 0 and 3x – y + 12 = 0 are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other
Solution
(b) parallel
The given equations are as follows:
6x – 2y + 9 = 0 and 3x – y + 12 = 0
They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 6, b1 = - 2, c1 = 9 and a2 = 3, b2 = - 1 and c2 = 12
∴ a1/a2 = 6/3 = 2/1, b1/b2 = -2/-1 = 2/1 and c1/c2 = 9/12 = 3/4
∴ a1/a2 = b1/b2 ≠ c1/c2
The given system has no solution.
Hence, the lines are parallel.
25. The graphs of the equations 2x + 3y – 2 = 0 and x – 2y – 8 = 0 are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other
Solution
The given equations are as follows:
2x + 3y – 2 = 0 and x – 2y – 8 = 0
They are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 3, c1 = - 2 and a2 = 1, b2 = - 2 and c2 = -8
∴ a1/a2 = 2/1, b1/b2 = 3/-2 and c1/c2 = -2/-8 = ¼
∴ a1/a2 ≠ b1/b2
The given system has a unique solution.
Hence, the lines intersect exactly at one point.
26. The graphs of the equations 5x – 15y = 8 and 3x – 9y = 24/5 are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other
Solution
(a) coincident
The correct option is (a).
The given system of equations can be written as follows:
5x – 15y – 8 = 0 and 3x – 9y – 24/5 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 5, b1 = - 15, c1 = - 8 and a2 = 3, b2 = - 9 and c2 = -24/5
∴ a1/a2 = 5/3, b1/b2 = -15/-9 = 5/3 and c1/c2 = -8× 5/-24 = 5/3
∴ a1/a2 = b1/b2 = c1/c2
The given system of equations will have an infinite number of solutions.
Hence, the lines are coincident.
27. The sum of the digits of a two digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is
(a) 96
(b) 69
(c) 87
(d) 78
Solution
(a) 96
Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
According to the question, we have:
x + y = 15 ...(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) = (10x + y) + 9
⇒ 10y + x – 10x – y = 9
⇒ 9y – 9x = 9
⇒ y – x = 1 ...(ii)
On adding (i) and (ii), we get:
2y = 16
⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15
⇒ x = (15 – 8) = 7
Number = (10x + y) = 10×7 + 8 = 70 + 8 = 78
Hence, the required number is 78.
Exercise – Formative Assessment
1. The graphic representation of the equations = x + 2y = 3 and 2x + 4y + 7 = 0 gives a pair of
(a) parallel lines
(b) intersecting lines
(c) coincident lines
(d) none of these
Solution
The given system of equations can be written as follows:
x + 2y – 3 = 0 and 2x + 4y + 7 = 0
The given equations are of the following form:
a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = - 3 and a2 = 2, b2 = 4 and c2 = 7
∴ a1/a2 = 1/2, b1/b2 = 2/4 = 1/2 and c1/c2 = -3/7
∴ a1/a2 = b1/b2 ≠ c1/c2
So, the given system has no solution.
Hence, the lines are parallel.
2. If 2x – 3y = 7 and (a + b)x – (a + b – 3)y = (4a + b) have an infinite number of solutions, then
(a) a = 5, b = 1
(b) a = -5, b = 1
(c) a = 5, b = - 1
(d) a = -5, b = - 1
Solution
(d) a = -5, b = -1
The given system of equations can be written as follows:
2x – 3y – 7 = 0 and (a + b)x – (a + b – 3)y – (4a + b) = 0
The given equations are of the following form:
a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = -3, c1 = - 7 and a2 = (a + b), b2 = -(a + b – 3) and c2 = -(4a + b)
∴ a1/a2 = 2/(a + b), b1/b2 = -3/-(a + b – 3) = 3/(a + b – 3) and c1/c2 = -7/-(4a + b) = 7/(4a + b)
For an infinite number of solutions, we must have:
a1/a2 = b1/b2 = c1/c2
∴ 2/(a + b) = 3/(a + b – 3) = 7/(4a + b)
Now, we have:
2/(a + b) = 3/(a + b – 3)
⇒ 2a + 2b – 6 = 3a + 3b
⇒ a + b + 6 = 0 ...(i)
Again, we have:
3/(a + b – 3) = 7/(4a + b)
⇒ 12a + 3b = 7a + 7b – 21
⇒ 5a – 4b + 21 = 0 ...(ii)
On multiplying (i) by 4, we get:
4a + 4b + 24 = 0 ...(iii)
On adding (ii) and (iii), we get:
9a = -45
⇒ a = -5
On substituting a = -5 in (i), we get:
-5 + b + 6 = 0
⇒ b = - 1
∴ a = -5 and b = - 1
3. The pair of equations 2x + y = 5, 3x + 2y = 8 has
(a) a unique solution
(b) two solutions
(c) no solution
(d) infinitely many solutions
Solution
The given system of equations can be written as follows:
2x + y – 5 = 0 and 3x + 2y – 8 = 0
The given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 2, b1 = 1, c1 = -5 and a2 = 3, b2 = 2 and c2 = -8
∴ a1/a2 = 2/3, b1/b2 = 1/2 and c1/c2 = -5/-8 = 5/8
∴ a1/a2 ≠ b1/b2
The given system has a unique solution.
Hence, the lines intersect at one point.
4. If x = -y and y > 0, which of the following is wrong?
(a) x2y > 0
(b) x + y = 0
(c) xy < 0
(d) 1/x – 1/y = 0
Solution
(d) 1/x - 1/y = 0
Given:
x = -y and y > 0
Now, we have:
(i) x2y
On substituting x = -y, we get:
(-y)2y = y3 > 0 (∵ y > 0)
This is true.
(ii) x + y
On substituting x = -y, we get:
(-y ) + y = 0
This is also true.
(iii) xy
On substituting x = -y, we get:
(-y)y = -y2 (∵ y > 0)
This is again true.
(iv) 1/x – 1/y = 0
⇒ (y – x)/xy = 0
On substituting x = -y, we get:
y – (-y)/(-y)y = 0 ⇒ 2y/-y2 = 0
⇒ 2y = 0
⇒ y = 0
5. Show that the system of equations -x + 2y + 2 = 0 and 1/2.x - 1/4.y – 1 = 0 has a unique solution.
Solution
The given system of equations:
-x + 2y + 2 = 0 and 1/2.x – 1/4.y – 1 = 0
The given equations are of the following form:
a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = -1, b1 = 2, c1 = 2 and a2 = 1/2, b2 = -1/4 and c2 = - 1
∴ a1/a2 =-1/(1/2) = -2, b1/b2 = 2/(-1/4) = -8 and c1/c2 = 2/-1 = -2
∴ a1/a2 ≠ b1/b2
The given system has a unique solution.
Hence, the lines intersect at one point.
6. For what values of k is the system of equations kx + 3y = (k – 2), 12x + ky = k inconsistent ?
Solution
The given system of equations can be written as follows:
kx + 3y – (k – 2) = 0 and 12x + ky – k = 0
The given equations are of the following form:
a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = k, b1 = 3, c1 = -(k – 2) and a2 = 12, b2 = k and c2 = -k
∴ a1/a2 = k/12, b1/b2 = 3/k and c1/c2 = -(k – 2)/-k = (k – 2)/k
For inconsistency, we must have:
a1/a2 = b1/b2 ≠ c1/c2
⇒ k/12 = 3/k ≠ (k – 2)k
⇒ k2 = (3 × 12) = 36
Hence, the pair of equations is inconsistent if k = ±6.
7. Show that the equations 9x – 10y = 21, 3x/2 – 5y/3 = 7/2 have infinitely many solutions.
Solution
The given system of equations can be written as follows:
9x – 10y – 21 = 0 and (3x/2.)x – 5y/3 – 7/2 = 0
The given equations are of the following form:
a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 9, b1 = - 10, c1 = - 21 and a2 = 3/2, b2 = -5/3 and c2 = -7/2
∴ a1/a2 = 9/(3/2) = 6, b1/b2 = -10/(-5/3) = 6 and c1/c2 = -21 × 2/-7 = 6
∴ a1/a2 = b1/b2 = c1/c2
This shows that the given system of equations has an infinite number of solutions.
8. Solve the system of equations: x – 2y = 0, 3x + 4y = 20
Solution
The given equations are as follows:
x – 2y = 0 ...(i)
3x + 4y = 20 ...(ii)
on multiplying (i) by 2, we get:
2x – 4y = 0 ...(iii)
On adding (ii) and (iii), we get:
5x = 20
⇒ x = 4
On substituting x = 4 in (i), we get:
4 – 2y = 0
⇒ 4 = 2y
⇒ y = 2
Hence, the required solution is x = 4 and y = 2.
9. Show that the paths represented by the equations x – 3y = 2 and -2x + 6y = 5 are parallel.
Solution
The given system of equations can be written as follows:
x – 3y – 2 = 0 and -2x + 6y – 5 = 0
the given equations are of the following form:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 1, b1 = -3, c1 = - 2 and a2 = -2, b2 = 6 and c2 = - 5
∴ a1/a2 = 1/-2 = -1/2, b1/b2 = -3/6 = -1/2 and c1/c2 = -2/-5 = 2/5
∴ a1/a2 = b1/b2 ≠ c1/c2
Thus, the given system of equations has no solution.
Hence, the paths represented by the equations are parallel.
10. The difference between two numbers is 26 and one number is three times the other. find the numbers.
Solution
Let the larger number be x and the smaller number be y.
Then, we have:
x – y = 26 ...(i)
x = 3y ...(ii)
On substituting x = 3y in (i), we get:
3y – y = 26
⇒ 2y = 26
⇒ y = 13
On substituting y = 13 in (i), we get:
x – 13 = 26
⇒ x = 26 + 13 = 39
Hence, the required numbers are 39 and 13.
11. Solve: 23x + 29y = 98, 29x + 23y = 110
Solution
The given equation are as follows:
23y + 29y = 98 ...(i)
29x + 23y = 110 ...(ii)
On adding (i) and (ii), we get:
52x + 52y = 208
⇒ x + y = 4 ...(iii)
On subtracting (i) from (ii), we get:
6x – 6y = 12
⇒ x – y = 2 ...(iv)
On adding (iii) and (iv), we get:
2x = 6 ⇒ x = 3
On substituting x = 3 in (iii), we get:
3 + y = 4
⇒ y = 4 – 3 = 1
Hence, the required solution is x = 3 and y = 1.
12. Solve: 6x + 3y = 7xy and 3x + 9y = 11xy
Solution
The given equations are as follows:
6x + 3y = 7xy ...(i)
3x + 9y = 11xy ...(ii)
For equation (i), we have:
(6x +3y)/xy = 7
⇒ 6x/xy + 3y/xy = 7
⇒ 6/y + 3/x = 7 ...(iii)
For equation (ii), we have:
(3x + 9y)/xy = 11
⇒ 3x/xy + 9x/xy = 11
⇒ 3/y + 9/x = 11...(iv)
On substituting 3/y = v and 1/x = u in (iii) and (iv), we get:
6v + 3u = 7 ...(v)
3v + 9u = 11 ...(vi)
On multiplying (v) by 3, we get:
18v + 9u = 21 ...(vii)
On substituting y = 3/2 in (iii), we get:
6/(3/2) + 3/x = 7
⇒ 4 + 3/x = 7
⇒ 3/x = 3
⇒ 3x = 3
⇒ x = 1
Hence, the required solution is x = 1 and y = 3/2.
13. Find the value of k for which the system of equations 3x + y =1 and kx + 2y = 5 has
(i) a unique solution,
(ii) no solution.
Solution
The given system of equations can be written as follows:
3x + y = 1
⇒ 3x + y – 1 = 0 ...(i)
kx + 2y = 5
⇒ kx + 2y – 5 = 0 ...(ii)
These equations are of the following form:
a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0
Here, a1 = 3, b1 = 1, c1 = -1 and a2 = k, b2 = 2 and c2 = -5
(i) For a unique solution, we must have:
a1/a2 ≠ b1/b2 i.e., 3/k ≠ 1/2
⇒ k = 6
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) In order that the given equations have no solution, we must have:
a1/a2 = b1/b2 ≠ c1/c2
3/k = 1/2 ≠ -1/-5
⇒ 3/k = 1/2 and 3/k = -1/-5
⇒ 3/k = 1/2 and 3/k = -1/-5
⇒ k = 6, k ≠ 15
Thus, for k = 6, the given system of equations will have no solution.
14. In a ∆ABC, ∠C = 3∠B = 2(∠A + ∠B), find the measure of each one of ∠A, ∠B and ∠C.
Solution
Let ∠A = x° and ∠B = y°
Then, ∠C = 3∠B = 3y°
Now, we have:
∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180°
⇒ x + 4y = 180 ...(i)
Also, ∠C = 2(∠A + ∠B)
⇒ 3y = 2(x + y)
⇒ 2x – y = 0 ...(ii)
On multiplying (ii) by 4, we get:
8x – 4y = 0 ...(iii)
On adding (i) and (iii), we get:
20 + 4y = 180
⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180°
⇒ 4y = (180 – 20) = 160
⇒ y = 40
∴ x = 20 and y = 40
∴ ∠A = 20°, ∠B = 40°, ∠C = (3 × 40°) = 120°
15. 5 pencils and 7 pens together cost Rs 195 while 7 pencils and 5 pens together cost Rs 153. Find the cost of each one on the pencil and pen.
Solution
Let the cost of each pencil be Rs. x and that of each pen be Rs y.
Then, we have:
5x + 7y = 195 ...(i)
7x + 5y = 153 ...(ii)
Adding (i) and (ii), we get:
12x + 12y = 348
⇒ 12(x + y) = 348
⇒ x + y = 29 ...(iii)
Subtracting (i) from (ii), we get:
2x – 2y = -42
⇒ 2(x – y) = -42
⇒ x – y = -21 ...(iv)
On adding (iii) and (iv), we get:
4 + y = 29
⇒ y = (29 – 4) = 25
Hence, the cost of each pencil is Rs 4 and the cost of each pen is Rs 25.
16. Solve the following system of equations graphically:
2x – 3y = 1, 4x – 3y + 1 = 0
Solution
On a graph paper, draw a horizontal X’OX and a vertical line YOY’ as the x-axis and y-axis, respectively.
Graph of 2x – 3y = 1
2x – 3y = 1
⇒ 3y = (2x – 1)
∴ y = (2x – 1)/3 ...(i)
Putting x = -1, we get:
y = -1
Putting x = 2, we get:
y = 1
Putting x = 5, we get:
y = 3
Thus, we have the following table for the equation 2x – 3y = 1.
x | -1 | 2 | 5 |
y | -1 | 1 | 3 |
Now, plots the point A(-1, -1), B(2, 1) and C(5, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extent it on both the sides.
Thus. the line AC is the graph of 2x – 3y = 1.
Graph of 4x – 3y + 1 = 0
4x – 3y + 1 = 0
⇒ 3y = (4x + 1)
∴ y = (4x +1)/3 ...(ii)
Putting x = - 1, we get:
y = - 1
Putting x = 2, we get:
y = 3
Putting x = 5, we get:
y = 7
Thus, we have the following table for the equation 4x – 3y + 1 = 0
x | -1 | 2 | 5 |
y | -1 | 3 | 7 |
Now, plot the point P(2, 3) and Q(5, 7). The point A(-1, -1) has already been plotted. Join PA and QP to get the graph line AQ. Extend it on both sides.
Thus, the line AQ is the graph of the equations 4x – 3y + 1 = 0
The two lines intersect at A(-1, -1).
Thus, x = -1 and y = -1 is the solution of the given system of equations.
17. Find the angles of a cyclic quadrilateral ABCD in which ∠A = (4x + 20)°, ∠B = (3x – 5)°, ∠C = 4y° and ∠D = (7y + 5)°.
Solution
Given:
In a cyclic quadrilateral ABCD, we have:
∠A = (4x + 20)°
∠B = (3x – 5)°
∠C = 4y°
∠D = (7y + 5)°
∠A + ∠C = 180° and ∠B + ∠D = 180° [Since ABCD is a cyclic quadrilateral]
Now, ∠A + ∠C = (4x + 20)° + (4y)° = 180°
⇒ 4x + 4y + 20 = 180°
⇒ 4x + 4y = 180 – 20 = 160
⇒ x + y = 40 ...(i)
Also, ∠B + ∠D = (3x – 5)° + (7y + 5)° = 180°
⇒ 3x + 7y ...(ii)
On multiplying (i) by 3, we get:
3x + 3y = 120 ...(iii)
On subtracting (iii) from (ii), we get:
4y = 60
⇒ y = 15
On substituting y = 15 in (i), we get:
x + 15 = 40 ⇒ x = (40 – 15) = 25
Therefore, we have:
∠A = (4x + 20)° = (4×25 + 20)° = 120°
∠B = (3x – 5)° = (3×25 – 5)° = 70°
∠C = 4y° = (4×15)° = 60°
∠D = (7y + 5)° = (7×15 + 5)° = (105 + 5)°
= 110°
18. Solve for x and y: 35/(x + y) + 14/(x – y) = 19, 14/(x + y) + 35/(x – y) = 37
Solution
35/(x + y) + 14/(x – y) = 19 and 14/(x + y) + 35/(x – y) = 37
Taking 1/(x + y) = u and 1/(x – y)= v.
35u + 14v – 19 = 0 ...(i)
14u + 35v – 37 = 0 ...(ii)
Here, a1 = 35, b1 = 14, c1 = - 19 and a2 = 14, b2 = 35 and c2 = -37
By cross multiplication, we have:
∴ u/[14×(37) – 35×(-19)] = v/[(-19)×14 – (-37)×(35)] = 1/[35×35 – 14×14]
⇒ u/(-518 + 665) = v/(-266 + 1295) = 1/(1225 – 196)
⇒ u/147 = v/1029 = 1/1029
⇒ u = 147/1029 = 1/7, v = 1029/1029 = 1
⇒ 1/(x + y) = 1/7, 1/(x – y) = 1
∴ (x + y) = 7 ...(iii)
And, (x – y) = 1 ...(iv)
Again, the equations (iii) and (iv) can be written as follows:
x + y – 7 = 0 ...(v)
x – y – 1 = 0 ...(vi)
Here, a1 = 1, b1 = 1, c1 = 7 and a2 = 1, b2 = -1 and c2 = -1
By cross multiplication, we have:
∴ x/[1×(-1) – (-1)×(-7) = y/[(-7)× 1 – (-1)×1] = 1/[1×(-1) – 1×1]
⇒ x /(-1 – 7) = y/(-7 + x) = 1/(-1 -1)
⇒ x/-8 = y/-6 = 1/-2
⇒ x = -8/-2 = 4, y = -6/-2 = 3
Hence, x = 4 and y = 3 is the required solution.
19. If 1 is added to both of the numerator and denominator of a fraction, it becomes 4/5. If however, 5 is subtracted from both numerator and denominator, the fraction becomes 1/2. Find the fraction.
Solution
Let the required fraction be x/y.
Then, we have:
(x + 1)/(y + 1) = 4/5
⇒ 5(x + 1)= 4(y + 1)
⇒ 5x + 5 = 4y + 4
⇒ 5x – 4y = -1 ...(i)
Again, we have:
(x – 5)/(y – 5) = 1/2
⇒ 2(x – 5) = 1/(y – 5)
⇒ 2x – 10 = y – 5
⇒ 2x – y = 5 ...(ii)
On multiplying (i) from (iii), we get:
3x = (20 – (-1) = 20 + 1 = 21
⇒ 3x = 21
⇒ x = 7
On substituting x = 7 in (i), we get
5×7 – 4y = -1
⇒ 35 – 4y = -1
⇒ 4y = 36
⇒ y = 9
∴ x = 7 and y = 9
Hence, the required fraction is 7/9.
20. Solve: ax/b - by/a = (a + b), ax – by = 2ab
Solution
The given equations may be written as follows:
ax/b – by/a – (a + b) = 0 ...(i)
ax – by – 2ab = 0 ...(ii)
Here, a1 = a/b, b1 = -b/a, c1 = -(a + b) and a2 = a, b2 = -b and c2 = -2ab
By cross multiplication, we have:
