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RS Aggarwal Solutions: Exercise 3D - Linear Equations in two variables Notes | EduRev

Class 10 : RS Aggarwal Solutions: Exercise 3D - Linear Equations in two variables Notes | EduRev

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Exercise 3D
1. Show that the following system of equations has a unique solution:
3x + 5y = 12,
5x + 3y = 4.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
Page 2

Exercise 3D
1. Show that the following system of equations has a unique solution:
3x + 5y = 12,
5x + 3y = 4.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:

3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36
On subtracting (iii) from (iv), we get:
16x = -16
x = -1
On substituting x = -1 in (i), we get:
3(-1) + 5y = 12
5y = (12 + 3) = 15
y = 3
Hence, x = -1 and y = 3 is the required solution.
2. Show that the following system of equations has a unique solution:
2x - 3y = 17,
4x + y = 13.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
2x - 3y -
4x + y -
The given equations are of the form
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13
Now,
= = and = = -3
Since, , therefore the system of equations has unique solution.
Using cross multiplication method, we have
Page 3

Exercise 3D
1. Show that the following system of equations has a unique solution:
3x + 5y = 12,
5x + 3y = 4.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:

3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36
On subtracting (iii) from (iv), we get:
16x = -16
x = -1
On substituting x = -1 in (i), we get:
3(-1) + 5y = 12
5y = (12 + 3) = 15
y = 3
Hence, x = -1 and y = 3 is the required solution.
2. Show that the following system of equations has a unique solution:
2x - 3y = 17,
4x + y = 13.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
2x - 3y -
4x + y -
The given equations are of the form
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13
Now,
= = and = = -3
Since, , therefore the system of equations has unique solution.
Using cross multiplication method, we have

= =
= =
= =
= =
x =  , y =
x = 4, y = -3
Hence, x = 4 and y = -3.
3. Show that the following system of equations has a unique solution:
+ = 3, x 2y = 2.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
+ = 3
= 3
2x + 3y = 18
2x + 3y 1
and
x 2y = 2
x 2y
These equations are of the forms:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
2x + 3y
x 2y
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y
3x - 6y
On adding (v) from (vi), we get:
7x = 42
x = 6
On substituting x = 6 in (iii), we get:
Page 4

Exercise 3D
1. Show that the following system of equations has a unique solution:
3x + 5y = 12,
5x + 3y = 4.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:

3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36
On subtracting (iii) from (iv), we get:
16x = -16
x = -1
On substituting x = -1 in (i), we get:
3(-1) + 5y = 12
5y = (12 + 3) = 15
y = 3
Hence, x = -1 and y = 3 is the required solution.
2. Show that the following system of equations has a unique solution:
2x - 3y = 17,
4x + y = 13.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
2x - 3y -
4x + y -
The given equations are of the form
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13
Now,
= = and = = -3
Since, , therefore the system of equations has unique solution.
Using cross multiplication method, we have

= =
= =
= =
= =
x =  , y =
x = 4, y = -3
Hence, x = 4 and y = -3.
3. Show that the following system of equations has a unique solution:
+ = 3, x 2y = 2.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
+ = 3
= 3
2x + 3y = 18
2x + 3y 1
and
x 2y = 2
x 2y
These equations are of the forms:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
2x + 3y
x 2y
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y
3x - 6y
On adding (v) from (vi), we get:
7x = 42
x = 6
On substituting x = 6 in (iii), we get:

2(6) + 3y = 18
3y = (18 - 12) = 6
y = 2
Hence, x = 6 and y = 2 is the required solution.
4. Find the value of k for which the system of equations has a unique solution:
2x + 3y = 5,
kx - 6y = 8.
Sol:
The given system of equations are
2x + 3y 5 = 0
kx - 6y - 8 = 0
This system is of the form:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= 3, c1 = -5 and a2 = k, b2 = -6, c2 = -8
Now, for the given system of equations to have a unique solution, we must have:
-4
Hence, -4
5. Find the value of k for which the system of equations has a unique solution:
x ky = 2,
3x + 2y + 5=0.
Sol:
The given system of equations are
x - ky 2 = 0
3x + 2y + 5 = 0
This system of equations is of the form:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 1, b1= -k, c1 = -2 and a2 = 3, b2 = 2, c2 = 5
Now, for the given system of equations to have a unique solution, we must have:
-
Hence, - .
Page 5

Exercise 3D
1. Show that the following system of equations has a unique solution:
3x + 5y = 12,
5x + 3y = 4.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:

3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36
On subtracting (iii) from (iv), we get:
16x = -16
x = -1
On substituting x = -1 in (i), we get:
3(-1) + 5y = 12
5y = (12 + 3) = 15
y = 3
Hence, x = -1 and y = 3 is the required solution.
2. Show that the following system of equations has a unique solution:
2x - 3y = 17,
4x + y = 13.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
2x - 3y -
4x + y -
The given equations are of the form
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13
Now,
= = and = = -3
Since, , therefore the system of equations has unique solution.
Using cross multiplication method, we have

= =
= =
= =
= =
x =  , y =
x = 4, y = -3
Hence, x = 4 and y = -3.
3. Show that the following system of equations has a unique solution:
+ = 3, x 2y = 2.
Also, find the solution of the given system of equations.
Sol:
The given system of equations is:
+ = 3
= 3
2x + 3y = 18
2x + 3y 1
and
x 2y = 2
x 2y
These equations are of the forms:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2
For a unique solution, we must have:
, i.e.,
Hence, the given system of equations has a unique solution.
Again, the given equations are:
2x + 3y
x 2y
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y
3x - 6y
On adding (v) from (vi), we get:
7x = 42
x = 6
On substituting x = 6 in (iii), we get:

2(6) + 3y = 18
3y = (18 - 12) = 6
y = 2
Hence, x = 6 and y = 2 is the required solution.
4. Find the value of k for which the system of equations has a unique solution:
2x + 3y = 5,
kx - 6y = 8.
Sol:
The given system of equations are
2x + 3y 5 = 0
kx - 6y - 8 = 0
This system is of the form:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 2, b1= 3, c1 = -5 and a2 = k, b2 = -6, c2 = -8
Now, for the given system of equations to have a unique solution, we must have:
-4
Hence, -4
5. Find the value of k for which the system of equations has a unique solution:
x ky = 2,
3x + 2y + 5=0.
Sol:
The given system of equations are
x - ky 2 = 0
3x + 2y + 5 = 0
This system of equations is of the form:
a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 1, b1= -k, c1 = -2 and a2 = 3, b2 = 2, c2 = 5
Now, for the given system of equations to have a unique solution, we must have:
-
Hence, - .

6. Find the value of k for which the system of equations has a unique solution:
5x 7y = 5,
2x + ky = 1.
Sol:
The given system of equations are
5x - 7y
2x + ky -
This system is of the form:
a1x+b1y+c1 = 0
a2x+b2y+c2 = 0
where, a1 = 5, b1= -7, c1 = -5 and a2 = 2, b2 = k, c2 = -1
Now, for the given system of equations to have a unique solution, we must have:
-
Hence, - .
7. Find the value of k for which the system of equations has a unique solution:
4x + ky + 8=0,
x + y + 1 = 0.
Sol:
The given system of equations are
4x + ky + 8 = 0
x + y + 1 = 0
This system is of the form:
a1x+b1y+c1 = 0
a2x+b2y+c2 = 0
where, a1 = 4, b1= k, c1 = 8 and a2 = 1, b2 = 1, c2 = 1
For the given system of equations to have a unique solution, we must have:
Hence,
8. Find the value of k for which the system of equations has a unique solution:
4x - 5y = k,
2x - 3y = 12.
Sol:
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