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

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

``` Page 1

Exercise 3C
1. Solve the system of equations by using the method of  cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x =  = 3, y =  = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of  cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:
Page 2

Exercise 3C
1. Solve the system of equations by using the method of  cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x =  = 3, y =  = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of  cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:

3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of  cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of  cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:
Page 3

Exercise 3C
1. Solve the system of equations by using the method of  cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x =  = 3, y =  = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of  cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:

3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of  cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of  cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:

Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = -20
Hence, x = 5 and y = -20 is the required solution.
5. Solve the system of equations by using the method of  cross multiplication:
2x + 5y 1 = 0, 2x + 3y 3 = 0
Sol:
The given equations may be written as:
2x + 5y
2x + 3y
Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -1
Hence, x = 3 and y = -1 is the required solution.
6. Solve the system of equations by using the method of  cross multiplication:
2x + y 35 = 0,
3x + 4y 65 = 0
Sol:
The given equations may be written as:
2x + y
Page 4

Exercise 3C
1. Solve the system of equations by using the method of  cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x =  = 3, y =  = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of  cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:

3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of  cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of  cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:

Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = -20
Hence, x = 5 and y = -20 is the required solution.
5. Solve the system of equations by using the method of  cross multiplication:
2x + 5y 1 = 0, 2x + 3y 3 = 0
Sol:
The given equations may be written as:
2x + 5y
2x + 3y
Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -1
Hence, x = 3 and y = -1 is the required solution.
6. Solve the system of equations by using the method of  cross multiplication:
2x + y 35 = 0,
3x + 4y 65 = 0
Sol:
The given equations may be written as:
2x + y

3x + 4y
Here a1 = 2, b1 = 1, c1 = -35, a2 = 3, b2 = 4 and c2 = -65
By cross multiplication, we have:
= =
= =
= =
x = = 15, y = = 5
Hence, x = 15 and y = 5 is the required solution.
7. Solve the system of equations by using the method of  cross multiplication:
7x - 2y 3 = 0,
11x - y 8 = 0.
Sol:
The given equations may be written as:
7x - 2y
11x - y
Here a1 = 7, b1 = -2, c1 = -3, a2 = 11, b2 = - and c2 = -8
By cross multiplication, we have:
= =
= =
= =
x = = 1, y = = 2
Hence, x = 1 and y = 2 is the required solution.
8. Solve the system of equations by using the method of  cross multiplication:
+ 4 = 0, - = 0
Page 5

Exercise 3C
1. Solve the system of equations by using the method of  cross multiplication:
x + 2y + 1 = 0,
2x 3y 12 = 0.
Sol:
The given equations are:
2x 3y
Here a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = -3 and c2 = -12
By cross multiplication, we have:
= =
= =
= =
x =  = 3, y =  = -2
Hence, x = 3 and y = -2 is the required solution.
2. Solve the system of equations by using the method of  cross multiplication:
3x - 2y + 3 = 0,
4x + 3y 47 = 0
Sol:
The given equations are:

3x - 2y
4x + 3y
Here a1 = 3, b1 = -2, c1 = 3, a2 = 4, b2 = 3 and c2 = -47
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = 9
Hence, x = 5 and y = 9 is the required solution.
3. Solve the system of equations by using the method of  cross multiplication:
6x - 5y - 16 = 0,
7x - 13y + 10 = 0
Sol:
The given equations are:
6x - 5y -
7x -
Here a1 = 6, b1 = -5, c1 = -16, a2 = 7, b2 = -13 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 6, y = = 4
Hence, x = 6 and y = 4 is the required solution.
4. Solve the system of equations by using the method of  cross multiplication:
3x + 2y + 25 = 0, 2x + y + 10 = 0
Sol:
The given equations are:

Here a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1 and c2 = 10
By cross multiplication, we have:
= =
= =
= =
x = = 5, y = = -20
Hence, x = 5 and y = -20 is the required solution.
5. Solve the system of equations by using the method of  cross multiplication:
2x + 5y 1 = 0, 2x + 3y 3 = 0
Sol:
The given equations may be written as:
2x + 5y
2x + 3y
Here a1 = 2, b1 = 5, c1 = -1, a2 = 2, b2 = 3 and c2 = -3
By cross multiplication, we have:
= =
= =
= =
x = = 3, y = = -1
Hence, x = 3 and y = -1 is the required solution.
6. Solve the system of equations by using the method of  cross multiplication:
2x + y 35 = 0,
3x + 4y 65 = 0
Sol:
The given equations may be written as:
2x + y

3x + 4y
Here a1 = 2, b1 = 1, c1 = -35, a2 = 3, b2 = 4 and c2 = -65
By cross multiplication, we have:
= =
= =
= =
x = = 15, y = = 5
Hence, x = 15 and y = 5 is the required solution.
7. Solve the system of equations by using the method of  cross multiplication:
7x - 2y 3 = 0,
11x - y 8 = 0.
Sol:
The given equations may be written as:
7x - 2y
11x - y
Here a1 = 7, b1 = -2, c1 = -3, a2 = 11, b2 = - and c2 = -8
By cross multiplication, we have:
= =
= =
= =
x = = 1, y = = 2
Hence, x = 1 and y = 2 is the required solution.
8. Solve the system of equations by using the method of  cross multiplication:
+ 4 = 0, - = 0

Sol:
The given equations may be written as:
+
-
Here a1 =  , b1 =  , c1 = -4, a2 = , b2 = - and c2 = -
By cross multiplication, we have:
=  =
= =
= =
x = = 18, y = = 15
Hence, x = 18 and y = 15 is the required solution.
9. Solve the system of equations by using the method of  cross multiplication:
+ = 7, + = 17
Sol:
Taking = u and = v, the given equations become:
u + v = 7
2u + 3v = 17
The given equations may be written as:
u + v
2u + 3v
Here, a1 = 1, b1 = 1, c1 = -7, a2 = 2, b2 = 3 and c2 = -17
By cross multiplication, we have:
= =
= =
= =
u = = 4, v = = 3
= 4, = 3
x = , y =
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## Mathematics (Maths) Class 10

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