Linear Equations in Two Variables: Solutions for Practice Set 1.2 of Maharashtra State Board for Maths 1 Algebra 10th Standard.  

Practice Set 1.2

Q.1) Complete the following table to draw graph of the equations

      a)  x+y = 3
x 3
y 5 3
(x,y) (3,0) (0,3)

    b) x-y = 4
x -1 0
y 0 -4
(x,y) (0,-4)

Solution:-

a) x + y = 3

put x = 3
3 + y = 3
y = 3 - 3
y = 0

put y = 5
x + 5 = 3
x = 3 - 5
x = -2

put y = 3
x + y = 3
x + 3 = 3
x = 3 - 3
x = 0

x 3 -2 0
y 0 5 3
(x,y) (3,0) (0,3)

b) x - y = 4

put y = 0
x - 0 = 4
x = 4 + 0
x = 4

put x = -1
(-1) - y = 4
- y = 4 + 1
- y = 5
y = - 5

x 4 -1 0
y 0 -5 -4
(x,y) (4 , 0) (-1 , -5) (0 , -4)


Q.2) Solve the following simultaneous equations graphically. 

x + y = 6;  x - y = 4

Solution:-

The given simultaneous equation are

1) x + y = 6 

∴ y = 6 - x

put x = 0

y = 6 - 0

∴ y = 6

put y = 2

y = 6 - 2

∴ y = 4

put y = 6

y = 6 - 6

y = 0

x 0 2 6
y 6 4 0
(x,y) (0,6) (2,4) (6,0)

2) x - y = 4
∴ y = x - 4

put x = 0
∴ y = 0 - 4
∴ y = -4

put x = 4
∴ y = 4 - 4
∴ y = 0

put x = 5
∴ y = 5- 4
∴ y = 1

x 0 4 5
y -4 0 1
(x,y) (0,-4) (4,0) (5,1)


Point of intersection of two lines is (5,1).
∴ x = 5 and y = 1 is the solution of the simultaneous equations x + y = 6 and x - y = 4. 

Q.3) Solve the following simultaneous equations graphically.
x + y = 5 ; x - y = 3

Solution:-

The given simultaneous equation are

1) x + y = 5

In the equation x + y = 5

putting x = 0

0 + y = 5

∴ y = 5

putting y = 0

x + 0 = 5

∴ x = 5

putting y = 3

x + 3 = 5

∴ x = 5 - 3

∴ x = 2

x 0 5 2
y 5 0 3
(x,y) (0,5) (5,0) (2,3)

In the equation x - y = 3
putting x = 0
0 - y = 3
∴ y = -3

putting x = 3
3 - y = 3
∴  y = 0

putting x = 5
5 - y = 3
∴ y = 2

x 3 0 5
y 0 -3 2
(x,y) (3,0) (0,-3) (5,2)


The point of intersection of the two lines is (4,1).

Q.4) Solve the following simultaneous equations graphically.
x + y = 0 ; 2x - y = 9

Solution:-

In the equation x + y = 0
putting x = 1
1 + y = 0
∴ y = -1

putting x = 2
2 + y = 0
∴ y = -2

putting x = 3
3 + y = 0
∴ y = -3

putting x = 5
5 + y = 0
∴ y = -5

x 1 2 3 5
y -1 -2 -3 -5
(x,y) (1,-1) (2,-2) (3,-3) (5,-5)

In the equation 2x - y = 9
putting x = 0
2(0) - y = 9
0 - y = 9
∴ y = -9

putting y = -5
2x - (-5) = 9
2x + 5 = 9
2x = 9 - 5
2x = 4
x=4/2
x = 2

putting y = -3
2x - (-3) = 9
2x + 3 = 9
2x = 9 - 3
2x = 6
x = 6/2
x = 3

putting y = -1
2x - (-1) = 9
2x + 1 = 9
2x = 9 - 1
2x = 8
x = 8/2
x = 4

x 0 2 3 4
y -9 -5 -3 -1
(x,y) (0,-9) (2,-5) (3,-3) (4,-1)

Point of intersection of two lines is (3,-3).

Q.5) Solve the following simultaneous equations graphically.
3x - y = 2 ; 2x - y = 3

Solution:-

In the equation 3x - y = 2
putting x = 0
3(0) - y = 2
0 - y = 2
y = -2

putting x = 1
3(1) - y = 2
3 - y = 2
y = 1

putting x = 2
3(2) - y = 2
6 - y = 2
y = 4
x 0 1 2
y -2 1 4
(x,y) (0,-2) (1,1) (2,4)

In the equation 2x - y = 3
putting y = -3
2x - (-3) = 3
2x + 3 = 3
2x = 3 - 3
2x = 0
x = 0

putting y = -1
2x - (-1) = 3
2x + 1 = 3
2x = 3 - 1
2x = 2
x = 2/2
x = 1

putting y = 1
2x - (1) = 3
2x = 3 + 1
2x = 4
x = 4/2
x = 2

x 0 1 2
y -3 -1 1
(x,y) (0,-3) (1,-1) (2,1)

The point of intersection of the two lines is (-1,-5).

Q.6) Solve the following simultaneous equations graphically.
3x - 4y = -7 ; 5x - 2y = 0

Solution:-

In the equation 3x - 4y = -7
putting y = 2.5 
3x - 4(2.5) = -7
3x - 10 = -7
3x = -7 + 10
3x = 3
x = 3/3
x = 1

putting x = 0
3(0) - 4y = -7
0 - 4y = -7
-4y = -7
y = -7/-4
y = 1.75

putting y = 0
 3x - 4(0) = -7
3x - 0 = -7
3x = -7 + 0
3x = -7
x = -7/3
x = -2.3

x 1 0 -2.3
y 2.5 1.75 0
(x,y) (1,2.5) (0,1.75) (-2.3,0)

In the equation 5x - 2y = 0
putting x = 0
5(0) - 2y = 0
0 - 2y = 0
-2y = 0
y = 0/-2
y = 0

putting x = 2
5(2) - 2y = 0
10 - 2y = 0
-2y = 0 - 10
-2y = -10
y = -10/-2
y = 5

putting x = 4
5(4) - 2y = 0
20 - 2y = 0
-2y = 0 -20 
-2y = -20 
y = -20/-2
y = 10

x 0 2 4
y 0 5 10
(x,y) (0,0) (2,5) (4,10)


The point of intersection of two lines is (1,2.5).

Q.6) Solve the following simultaneous equations graphically.
2x - 3y = 4 ; 3y - x = 4

Solution:-

In the equation 2x - 3y = 4
putting x = 2
2(2) - 3y = 4
4 - 3y = 4
-3y = 4 - 4
-3y = 0
y = 0/-3
y = 0

putting x = 3.5
2(3.5) - 3y = 4
7 - 3y = 4
-3y = 4 -7
-3y = -3
y = -3/-3
y = 1

putting x = 1
2(1) - 3y = 4
2 - 3y = 4
-3y = 4 -2
-3y = 2
y = 2 /-3
y = -0.6

x 2 3.5 1
y 0 1 -0.6
(x,y) (2,0) (3.5,1) (1,-0.6)

In the equation  3y - x = 4
putting x = -4
3y- (-4) = 4
3y + 4 = 4
3y = 4 - 4
3y = 0
y = 0/3
y = 0

putting x = -1
3y - (-1) = 4
3y + 1 = 4
3y = 4 - 1
3y = 3
y = 3/3
y = 1

putting x = 2
3y - 2 = 4
3y = 4 + 2
3y = 6
y = 6/3
y = 2 

x

-4

2

-1

y

0

 2 1
(x,y) (0,-4) (2,2) (-1,1)



Point of intersection of the two lines is (8,4).