GSEB Solutions for Class 9 Mathematics – Linear Equations in Two Variables

GSEB Solutions for Class 9 Mathematics – Linear Equations in Two Variables (English Medium)

Exercise 5:

Solution 1:

Solution 2(1):

Solution 2(2):

Solution 2(3):

Solution 2(4):

Solution 2(5):

Solution 2(6):

Solution 3(1):

Solution 3(2):

Solution 3(3):

Given,
a = 2k, b = 5k, c = 7k and k ≠ 0 and k ∈ R.
c + b – 5a = 7k + 5k – 5(2k) = 12k – 10k = 2k
c – b – a = 7k – 5k – 2k = 0
Now, (c + b – 5a, c – b – a)  º (2k, 0) is a solution of 2x + y – 8 = 0.
2(2k) + 0 – 8 = 0
∴4k = 8
∴k = 2

Solution 3(4):

Solution 3(5):

Given,
a = 2k, b = 5k, c = 7k and k ≠ 0 and k ∈ R.
2a + b – c – 2 = 2(2k) + 5k – (7k) – 2 = 2k – 2
3b + 2a – 3c + 2 = 3(5k) + 2(2k) – 3(7k) + 2 = -2k + 2 Now, (2a + b – c – 2, 3b + 2a – 3c + 2) º (2k – 2, -2k
+2) is the point of intersection of the co-ordinate axes i.e. the origin (0, 0).
(2k – 2, -2k +2) = (0, 0)
∴2k – 2 = 0 and -2k + 2 = 0
∴2k = 2 and 2k = 2
[∵ Both the equations are identical, we consider only one equation (2k = 2)]
∴ 2k = 2
∴ k = 1

Solution 4(1):

The given equation is x + y = 0.
∴ y = -x
For x = 0 ⇒ y = 0
For x = 2 ⇒ y = -2
For x = -2 ⇒ y = -(-2) = 2
Hence, the three solution of x + y = 0 can be given as (0, 0), (2, -2) and (-2, 2).
Now, plot the points (0, 0), (2, -2) and (-2, 2).
Draw the line passing through them to get the graph of x + y = 0.

The graph of x + y = 0 intersects both the axes at (0, 0).

Solution 4(2):

The given equation is x – y = 0.
∴ y = x
For x = 0 ⇒ y = 0
For x = 2 ⇒ y = 2
For x = -2 ⇒ y = -2
Hence, three solution of x – y = 0 can be given as (0, 0), (2, 2) and (-2, -2).
Now, plot the points (0, 0), (2, 2) and (-2, -2) and draw the line passing through them to get the graph of x – y = 0.

The graph of x – y = 0 intersects both the axes at (0, 0).

Solution 4(3):

The given equation is x + y = 2.
∴ y = 2 – x
For x = 0 ⇒ y = 2 – 0 = 2
For x = -1 ⇒ y = 2 – (-1) = 3
For x = 2 ⇒ y = 2 – 2 = 0
Hence, three solution of x + y = 2 can be given as (0, 2), (-1, 3) and (2, 0).
Now, plot the points (0, 2), (-1, 3) and (2, 0) and draw the line passing through them to get the graph of x + y = 2.

The graph of x + y = 2 intersects the X-axis at (2, 0) and the Y-axis at (0, 2).

Solution 4(4):

The given equation is x – y = 3.
∴ x – 3 = y
∴ y = x – 3
For x = 1 ⇒ y = 1 – 3 = -2
For x = 2 ⇒ y = 2 – 3 = -1
For x = 4 ⇒ y = 4 – 3 = 1
So, three solutions of x – y = 3 can be given as (1, -2), (2, -1) and (4, 1).
Now, plot the points (1, -2), (2, -1) and (4, 1).
Draw the line passing through them to get the graph of x – y = 3.

The graph of x – y = 3 intersects the X-axis at (3, 0) and the Y-axis at (0, -3).

Solution 4(5):

Solution 4(6):

Solution 4(7):

Solution 4(8):

Solution 4(9):

Solution 4(10):

4y – 8 = 0
∴ 4y = 8
∴ y = 2
But, x is absent in the equation.
So, for any value of x, the value of y is 2 always.
Three solution of 4y – 8 = 0 can be given as (0, 2), (-2 , 2) and (2, 2).
Now, plot the points (0, 2), (-2 , 2) and (2, 2).
Draw the line passing through them to get the graph of 4y – 8 = 0.

The graph of 4y – 8 = 0 intersects the yY-axis at (0, 2), but it does not intersect the X-axis parallel to the X-axis.

Solution 5:

1. The given equation is 3x + 2 = -x + 10.
   ∴3x + x = 10 – 2
   ∴4x = 8
   ∴x = 2
   To represent the solution in the Cartesian plane, x = 2 is taken as the linear equation x + 0y = 2.
   But, y is absent in the equation.
   So, for any value of y, the value of x is always 2.
   Then three solutions of the equation can be given as (2, 0), (2, -5) and (2, 2)
2. The given equation is 4y – 3 = y + 6.
   ∴4y – y = 6 + 3
   ∴3y = 9
   ∴y = 3
   To represent the solution in the Cartesian plane, y = 3 is taken as the linear equation 0x + y = 3.
   But, x is absent in the equation.
   So, for any value of x, the value of y is always 3.
   Then three solutions of the equation can be given as (0, 3), (2, 3) and (-3, 3)
3. The given equation is 2x + 3 = x – 1.
   ∴ 2x – x =  -1 – 3
   ∴ x = -4
   To represent the solution in the Cartesian plane, x = -4 is taken as the linear equation x + 0y = -4.
   But, y is absent in the equation.
   So, for any value of y, the value of x is always -4.
   Then three solutions of the equation can be given as (-4, 3), (-4, 0) and (-4, -3).
4. The given equation is 3y + 3 = x – 1.
   ∴ 3y – 2y = -3 – 2
   ∴ y = – 5
   To represent the solution in the Cartesian plane, x = -5 is taken as the linear equation 0x + y = -5.
   But, y is absent in the equation.
   So, for any value of x, the value of y is always -5. Then three solutions of the equation can be given as (-4, -5), (0, -5) and (4, -5).

The geometric representation of all the above solutions. i.e.  x = 2, y = 3, x = -4 and y = -5 on the same number line is shown below:

For geometric representation of all the above solutions
in the same Cartesian plane, we plot the three ordered pairs (x, y) obtained for each of them on the same graph paper and draw the line each passing through those three points.

Solution 6:

1. x = 4
   Here, y is absent in the equation.
   So, for any value of y, the value of x is always 4.
   Three solutions of x = 4 can be given as (4, 2), (4, 0) and (4, -2).
2. y = 4
   Here, x is absent in the equation.
   So, for any value of x, the value of y is always 4.
   Three solutions of y = 4 can be given as (2, 4), (0, 4) and (-2, 4).
3. x = -4
   Here, y is absent in the equation.
   So, for any value of y, the value of x is always 4.
   Three solutions of x = -4 can be given as (-4, 2), (-4 , 0) and (-4, -2).
4. y = -4
   Here, x is absent in the equation.
   So, for any value of x, the value of y is -4 always.
   Three solutions of y = -4 can be given as (2, -4), (0, -4) and (-2, -4).
5. y = x
   Three solutions can be given as (2, 2), (0 , 0) and (-2, -2).
6. y = -x
   Three solutions can be given as (2, -2), (0, 0) and (-2, 2).

To draw the graph of all the above in R² on the same graph paper, plot the three ordered pairs (x, y) obtained for each of the lines on the same graph paper.
Draw a line each passing through those three points.

The points where these lines intersect each other are (0, 0), (4, 4), (-4, 4), (-4, -4) and (4, -4)

Solution 7:

Solution 8:

Solution 9(1):

C. I and III
In the equation y = x, both the x-coordinate and the y-coordinate have the same sign. So the points representing its solutions will lie in the Ist quadrant or the IIIrd quadrant or coincide with the origin.

Solution 9(2):

d. 1st, 2nd and 4th all
Line x + y = 2 passes through the points (2, 0) and (0, 2). Between these points the line passes through the 1st quadrant and beyond those points, the line passes through the 2nd quadrant on one side and the 4th quadrant on the other side.

Solution 9(3):

c.II and IV
If x + y = 0 ⇒ x = -y .
Here, the values of x and y will have the same numbers with opposite signs. Thus, all those solutions will either be in the II^nd quadrant or the IV^thquadrant.

Solution 9(4):

c. c = 0
Origin i.e. (0, 0) is always solution of ax + by = 0 for any values of a and b. So the graph of ax + by = c passes through the origin if c = 0.

Solution 9(5):

d. infinitely many solutions
A linear equation in two variables always has infinitely many solutions.

Solution 9(6):

c. 45
Given, x = 2, y = 5 is a solution of 5x + 7y – k = 0.
∴ 5(2) + 7(5) – k = 0
∴ 10 + 35 = k
∴ 45 = k
∴ k = 45

Solution 9(7):

Solution 9(8):

Solution 9(9):

Solution 9(10):

c. lines through origin
The equation y = mx gives mx – y + 0 = 0 in standard form. Here, c = 0, so the equation y = mx represents lines through the origin for different values of m.

Solution 9(11):

c. parallel to X-axis
The graph of y = k (constant k ≠ 0) is a line parallel to the X-axis.

Solution 9(12):

b. parallel to Y-axis
The graph of the equation x = k (constant k ≠ 0) is a line parallel to the Y-axis.

Solution 9(13):

b. (-1, 3)
For (1, 2)
2x + 3y = 2(1) + 3(2) = 8 ≠ 7
For (-1, 3),
2x + 3y = 2(-1) + 3(3) = 7
For (-2, 5),
2x + 3y = 2(-2) + 3(5) = 11 ≠ 7
For (-2, 4),
2x + 3y = 2(-2) + 3(4) = 8 ≠ 7
Thus, one of the solutions of the linear equation 2x + 3y = 7 is (-1, 3).

Solution 9(14):

a. x – 3 = 0
Line x – 3 = 0 ⇒ x = 3
The graph of x = k (constant k ≠ 0) is a line parallel to the Y-axis.

Solution 9(15):

a. x + y = 0
In line x + y = 0 we have, a = 1, b = 1 and c = 0.
The graph of ax + by + c = 0 is a line passing through the origin if a ≠ 0, b ≠ 0 and c = 0.

Exercise 5.1:

Solution 1:

Let the cost of a notebook be Rs. x and the cost of a pen be Rs. y.
Now, cost of a notebook = Twice the cost of a pen
∴ x = 2 × y
∴ x = 2y
Thus, x = 2y is the representation of the given statement in the form of a linear equation in two variables.

Solution 2:

Exercise 5.2:

Solution 1(1):

Solution 1(2):

Solution 2:

Solution 3(1):

Solution 3(2):

2x = 4
∴ x = 2
But y is absent in the given equation.
Hence, for any value of x, the value of y is always 2.
Thus, (2, 1), (2, 2) and (2, 3) are three elements of the solution set of 2x = 4.

Solution 3(3):

Solution 3(4):

Solution 3(5):

Solution 3(6):

The equation x + y = 0 can be written as
y = -x
For x = -2, y = 2
So, (-2, 2) is a solution of x + y = 0.
For x = 0, y = 0
So, (0, 0) is a solution of x + y = 0.
For x = 2, y = -2
So (2, -2) is a solution of x + y = 0.
Thus (-2, 2), (0, 0) and (2, -2) are three elements of the solution set of x + y = 0.

Solution 3(7):

Solution 3(8):

Solution 4:

Option (3) is true.
3y = 2x + 7 is a linear equation in two variables and hence, it has infinitely many solutions.

Solution 5:

Solution 6(1):

x = 1, y = 2 is a solution of the equation 3x – 2y = k.
So, the equation is satisfied by replacing x by 1 and y by 3.
∴ 3(1) – 2(2) = k
∴ 3 – 4 = k
∴ -1 = k
∴ k = -1

Solution 6(2):

Solution 6(3):

Solution 6(4):

Exercise 5.3:

Solution 1(1):

Given equation is x + y = 6.
∴ y = 6 – x
For x = 0 ⇒ y = 6 – 0 = 6 i.e. y = 6
For x = 6 ⇒ y = 6 – 6 = 0 i.e. y = 0
For x = 3 ⇒ y = 6 – 3 = 3 i.e. y = 3
Hence, the three elements of the solution set of x + y = 6 can be given in the tabular form as below:

x	0	6	3
y	6	0	3

Now, plot the points (0, 6), (6, 0) and (3, 3) on the graph paper.
Draw the unique line passing through them which is the graph of equation x + y = 6.

Solution 1(2):

Given equation is x – y = 2.
∴ y = x – 2
For x = -1 ⇒ y = -1 – 2 = -3 i.e. y = -3
For x = 3 ⇒ y = 3 – 2 = 1 i.e. y = 1
For x = 4 ⇒ y = 4 – 2 = 2 i.e. y = 2
Hence, three elements of the solution set of x – y = 2 can be given in the tabular form as below:

x	-1	3	4
y	-3	1	2

Now, plot the points (-1, -3), (3, 1) and (4, 2) on the graph paper.
Draw the unique line passing through them which is the graph of the equation x – y = 2.

Solution 1(3):

Solution 1(4):

Given equation is y = 3x.
For x = 0, y = 3(0) = 0
For x = 3, y = 3(3) = 9
For x = -1, y = 3(-1) = -3
Hence, three elements of the solution set of y = 3x can be given in the tabular form as below:

x	0	3	-1
y	0	9	-3

Now, plot the points (0, 0), (3, 9) and (-1, -3) on the graph paper.
Draw the unique line passing through them which is the graph of the equation y = 3x.

Solution 1(5):

Given equation is y = x + 1.
For x = 0, y = 0 + 1 = 1
For x = -2, y = -2 + 1 = -1
For x = 3, y = 3 + 1 = 4
Hence, three elements of the solution set of y = x + 1 can be given in the tabular form as below :

X	0	-2	3
y	1	-1	4

Now, plot the points (0, 1), (-2, -1) and (3, 4) on the graph paper.
Draw the unique line passing through them which is the graph of the equation y = x + 1.

Solution 1(6):

Given equation is 3x + y = 2
∴  y = 2 – 3x
For x = 0, y = 2 – 3(0) = 2
For x = 1, y = 2 – 3(1) = -1
For x = -1, y = 2 – 3(-1) = 5
Hence, three elements of the solution set of 3x + y = 2 can be given in the tabular form as below:

X	0	1	-1
Y	2	-1	5

Now, plot the points (0, 2), (1, -1) and (-1, 5) on the graph paper.
Draw the unique line passing through them which is the graph of the equation 3x + y = 2.

Solution 2:

Solution 3:

We know that infinitely many lines can pass through a given point.
So, we can give equations of many lines passing through the given point (2, 3).
Equation of four such lines with 3 elements of the solution set of each of line:
Line 1: x + y = 5

x	0	3	2
y	5	2	3

Now, we plot the points (0, 5), (3, 2) and (2, 3).
Draw the line passing through them.
Line 2: x – y = -1

x	0	-2	2
y	1	-1	3

Now, we plot the points (0, 1), (-2, -1) and (2, 3).
Draw the line passing through them.
Line 3: 3x + 2y = 12

x	0	4	2
y	6	0	3

Now, we plot the points (0, 6), (4, 0) and (2, 3).
Draw the line passing through them.
Line 4: 3x – 2y = 0

x	0	-2	2
y	0	-3	3

Now, we plot the points (0, 0), (-2, -3) and (2, 3).
Draw the line passing through them.

The table below gives the points of intersection of each line with the co-ordinate axes.

Line	Point of intersection with the X-axis	Point of intersection with the Y-axis
Line 1: x + y = 5 Line 2: x – y = -1 Line 3: 3x + 2y = 12 Line 4: 3x – 2y = 0	(5, 0) (-1, 0) (4, 0) (0, 0)	(0, 5) (0, 1) (0, 6) (0, 0)

 Solution 4(1):

Solution 4(2):

Solution 4(3):

Solution 4(4):

Solution 4(5):

Exercise 5.4:.

Solution 1(1):

The given equation is y = -4.
When y = -4 is taken as an equation in one variable, geometric representation is the point on the number line corresponding to -4 as shown below.

Let y = -4 be an equation in two variables where x is absent in the equation.
So, for any value of x, the value of y is always -4.
Therefore, (-3, -4), (0, -4) and (3, -4) are three elements of the solution set of y = -4.
Now, plot the points (-3, -4), (0, -4) and (3, -4) and draw the line passing through them.

Solution 1(2):

Solution 2(1):

The given equation is 3y = 6
∴ y = 2
But, x is absent in the equation.
Three elements of the solution set of 3y = 6 are (-2, 2), (0, 2) and (2, 2).
Now, plot the points (-2, 2), (0, 2) and (2, 2).
Draw the line passing through them which is the graph of the equation 3y = 6.

Solution 2(2):

The given equation is x = 4.
But, y is absent in the equation.Three elements of the solution set of x = 4 are (4, -3), (4, 0) and (4, 3).
Now, plot the points (4, -3), (4, 0) and (4, 3) and draw the line passing through them which is the graph of the equation x = 4.

Solution 2(3):

2y = 10
∴ y = 5
But, x is absent in the equation.
Three elements of the solution set of 2y =10 are (-2, 5), (0, 5) and (4, 5).
Now, plot the points (-2, 5), (0, 5) and (4, 5) and draw the line passing through them which is the graph of the equation 2y = 10.

Solution 2(4):

5x + 10 = 0
∴ 5x = -10
∴ x = -2
But, y is absent in the equation.
Three elements of the solution set of 5x + 10 = 0 are (-2, -3), (-2, 0) and (-2, 4).
Now, plot the points (-2, -3), (-2, 0) and (-2, 4) and draw the line passing through them which is the graph of the equation 5x + 10 = 0.

Solution 3(1):

The given equation is 2x + 1 = x – 2.
∴ 2x – x = – 2 – 1
∴ x = -3

• The solution of 2x + 1 = x – 2 is represented on the number line l as shown below:

• To represent the solution of 2x + 1 = x – 2 in the Cartesian plane, take the equation x = -3 as linear equation in two variables.
  But, y is absent in the equation.
  Three elements of the solution set of 2x + 1 = x -2 are (-3, -3), (-3, 0) and (-3, 4).
  Now, plot the points (-3, -3), (-3, 0) and (-3, 4) and draw the line passing through them which is the graph of the equation 2x + 1 = x – 2 in the Cartesian plane.

Solution 3(2):

The given equation is y – 1 = 2y – 5.
∴ y – 2y = -5 + 1
∴ -y = -4
∴ y = 4

• The solution of y – 1 = 2y – 5 is represented on the number line l as shown below :

• To represent the solution of y – 1 = 2y – 5 in the Cartesian plane, we take the equation y = 4 as linear equation in two variables.
  But, x is absent in the equation.
  Three elements of the solution set of y – 1 = 2y – 5 are (-1, 4), (0, 4) and (3, 4).
  Now, plot the points (-1, 4), (0, 4) and (3, 4).
  Draw the line passing through them which is the graph of the equation y – 1 = 2y – 5.

Solution 4:

The equation of the line is y = x + 1.
For x = 0 ⇒ y = 0 + 1 = 1
For x = -2 ⇒ y = -2 + 1 = -1
For x = 2 ⇒ y = 2 + 1 = 3
Hence, three elements of the solution set of y = x + 1 are (0, 1), (-2, -1) and (2, 3).
Now, plot the points (0, 1), (-2, -1) and (2, 3).
Draw the line passing through them which is the graph of the equation y = x + 1.
The equation of the line is x + y – 3 = 0.
∴ y = 3 – x
For x = -1, y = 3 – (-1) = 3 + 1 = -4
For x = 0, y = 3 – 0 = 3
For x = 1, y = 3 – 1 = 2
Hence, three elements of the solution set of x + y – 3 can be given as (-1, -4), (0, 3) and (1, 2).
Now, plot the points (-1, -4), (0, 3) and (1, 2).
Draw the line passing through them which is the graph of the equation x + y – 3 = 0.

The lines representing y = x + 1 and x + y – 3 = 0 intersect each other at point (1, 2).