RD Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Solutions
Question - 1 : - Solve the following systems of equations graphically :x + y = 3
2x + 5y = 12 (C.B.S.E. 1997)
Answer - 1 : -
x + y = 3
=> x = 3 – y
Substituting some different values of y, we get the corresponding values of x as shown below
Now plot the points on the graph and join them 2x + 5y = 12
2x = 12 – 5y
⇒ x =
Substituting some different values of y, we get the corresponding values of x as shown below
Now plot the points on the graph and join them we see that these two lines intersect each other at (1, 2)
x = 1, y = 2
x – 2y = 5
2x + 3y = 10 (C.B.S.E. 1997)
Answer - 2 : -
x – 2y = 5 => x = 5 + 2y
Substituting some different values of y, we get the corresponding values of x as shown below
Now plot the points are the graph and join them
2x + 3y = 10 => 2x = 10 – 3y
⇒ x =
Substituting some different values of y We get the corresponding values of x as shown below :
Now plot the points on the graph and join them we see that these two lines intersect each other at (5, 0)
x = 5, y = 0
3x + y + 1 = 0
2x – 3y + 8 = 0 (C.B.S.E. 1996)
Answer - 3 : -
3x + y + 1 = 0
y = -3x – 1
Substituting the values of x, we get the corresponding values of y, as shown below
Now plot the points on the graph and join them
2x – 3y + 8 = 0
⇒ 2x = 3y – 8
⇒ x =
Substituting some different values of y, we get the corresponding values of x as shown below
Now plot the points on the graph and join then we see that these two lines intersect, each other at (-1, -2)
x = -1, y = 2
2x + y – 3 = 0
2x – 3y – 7 = 0 (C.B.S.E. 1996)
Answer - 4 : -
2x + y – 3 = 0 => y = -2x + 3
Substituting some different values of x, we get the corresponding values of y as shown below:
Now plot the points and join them 2x – 3y – 7 = 0
2x = 3y +7
x =
Substituting some different values of y, we get corresponding values of x as shown below:
Now plot the points on the graph and join them we see that these two lines intersect each other at (2, -1)
Question - 5 : - x + y = 6
x – y = 2 (C.B.S.E. 1994)
Answer - 5 : -
x + y = 6 => x = 6 – y
Substituting some different values of y, we get the corresponding values of x as shown under
Now plot the points on the graph and join them
x – y = 2 ⇒ x = 2 + y
Substituting some different values of y, we get the corresponding values of x as shown below:
Now plot the points on the graph and join them
We see that there two lines intersect each other at (4, 2)
x = 4, y = 2
Question - 6 : - x – 2y = 6
3x – 6y = 0 (C.B.S.E. 1995)
Answer - 6 : -
x – 2y = 6
x = 6 + 2 y
Substituting some different values ofy, we get the corresponding values of x as shown below:
Now plot the points and join them
3x – 6y = 0 ⇒ 3x = 6y ⇒ x = 2y
Substituting some different value of y, we get corresponding the values of x as shown below:
plot the points on the graph and join them We see that these two lines intersect each other at no point
The lines are parallel
There is no solution
Question - 7 : - x + y = 4
2x – 3y = 3 (C.B.S.E. 1995)
Answer - 7 : -
x + y = 4 => y = 4 – x
Substituting some different values of y, we get the corresponding values of x as shown below:
Now plot the points and join them 2x – 3y = 3
⇒ 2x = 3 + 3y
⇒ x =
Substituting some different values of y, we get the corresponding values of x as shown below:
Now plot the points on the graph and join them we see that these two lines intersect each other at (3, 1)
x = 3, y = 1
Question - 8 : - 2x + 3y= 4
x – y + 3 = 0 (C.B.S.E. 1995)
Answer - 8 : -
2x + 3y = 4
=> 2x = 4 – 3y
=> x =
Substituting some different values of y, we get corresponding values of x as shown below:
Now plot the points are the graph and join them
x – y + 3 = 0
x = y – 3
Substituting some different values of y, we get corresponding values of x as shown below:
Now plot the points on the graph and join them
We see that these two lines intersect each other at (-1, 2)
x = -1, y = 2
Question - 9 : - 2x – 3y + 13 =0
3x – 2y + 12 = 0 (C.B.S.E. 2001C)
Answer - 9 : -
2x – 3y + 13 = 0
2x = 3y – 13
=> x =
Substituting some different values of y, we get corresponding values of x as shown below
Plot the points on the graph and join them 3x – 2y + 12 = 0
3x = 2y – 12
x =
Substituting some different values of y, we get corresponding values of x as shown below
Plot the points and join them We see that these two lines intersect each other at (-2, 3)
x = -2, y = 3
Question - 10 : - 2x + 3y + 5 = 0
3x – 2y – 12 = 0 (C.B.S.E. 2001 C)
Answer - 10 : -
2x + 3y + 5 = 0
2x = – 3y – 5
x =
Substituting some different value of y, we get corresponding values of x as shown below
Now plot the points on the graph and join them
3x – 2y – 12 = 0
3x = 2y +12
x =
Substituting some different value of y, we get corresponding values of x as shown below:
Now plot the points on the graph and join them we see that these lines intersect each other at (2, -3)
x = 2, y = -3