RD SHARMA Solutions for Class 9 Maths Chapter 7 - Linear Equations in Two Variables
Chapter 7 - Linear Equations in Two Variables Exercise Ex. 7.1
Chapter 7 - Linear Equations in Two Variables Exercise Ex. 7.2
If x = 1 and y = 6 is solution of the equation 8x - ay + a2= 0, find the value of a.
Write two solutions of the form x = 0, y = a and x = b, y = 0 for the follwoing equation: 5x - 2y = 10
Write two solutions of the form x = 0, y = a and x = b, y = 0 for the following equation: -4x + 3y = 12
Chapter 7 - Linear Equations in Two Variables Exercise Ex. 7.3
Plot the points (3,5) and (-1,3) on a graph paper and verify that the straight line passing through these points also passes through the point (1,4).
The given points on the graph:
It is dear from the graph, the straight line passing through these points also passes through the point (1,4).
From the choices given below, choose the equation whose graph is given in fig.,
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x
From the choices given below, choose the equation whose graph is given in fig.,
(i) y = x + 2
(ii) y = x - 2
(iii) y = -x + 2
(iv) x + 2y = 6
Draw the graph of the equation 2x + 3y = 12. Find the graph, find the coordinates of the point.
(i) whose y-coordinate is 3.
(ii) whose x-coordinate is -3






The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten's digit of the number are x and y respectively, then write the linear equation representing the above statement.
Draw the graph of y = |x|.
We have,
y = |X| ...(i)
Putting x = 0, we get y = 0
Putting x = 2, we get y = 2
Putting x = -2, we get y = 2
Thus, we have the following table for the points on graph of |x|.
x | 0 | 2 | -2 |
y | 0 | 2 | 2 |
The graph of the equation y = |x|:
Draw the graph of y = |x| + 2.
We have,
y = |x| + 2 ...(i)
Putting x = 0, we get y = 2
Putting x = 1, we get y = 3
Putting x = -1, we get y = 3
Thus, we have the following table for the points on graph of |x| + 2:
x | 0 | 1 | -1 |
y | 2 | 3 | 3 |
The graph of the equation y = |x| + 2:
Ravish tells his daughter Aarushi, "Seven years ago, I was seven times as old as you were then. Also, three years form now, I shall be three times as old as you will be". If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.
Chapter 7 - Linear Equations in Two Variables Exercise Ex. 7.4
y + 3 = 0
y = -3
Point A represents -3 on number line.
On Cartesian plane, equation represents all points on x axis for which y = -3
y = 3
Point A represents 3 on number line.
On Cartesian plane, equation represents all points on x axis for which y = 3
Give the geometrical representation of 2x + 13 = 0 as an equation in
One variable
Give the geometrical representation of 2x + 13 = 0 as an equation in
Two variables
Solve the equation 3x + 2 = x - 8, and represent the solution on (i) the number line.
Solve the equation 3x + 2 = x - 8, and represent the solution on (ii) the Cartesian plane.
On Cartesian plane, equation represents all points on y axis for which x = -5
Write the equation of the line that is parallel to x-axis and passing through the point
(i) (0,3)
(ii) (0,-4)
(iii) (2,-5)
(iv) (3,4)
(i) The equation of the line that is parallel to x-axis and passing through the point (0,3) is y = 3
(ii) The equation of the line that is parallel to x-axis and passing through the point (0,-4) is y = -4
(iii) The equation of the line that is parallel to x-axis and passing through the point (2,-5) is y = -5
(iv) The equation of the line that is parallel to x-axis and passing through the point (3, 4) is y = 4
Chapter 7 - Linear Equations in Two Variables Exercise 7.33
If (4, 19) is a solution of the equation y = ax + 3, then a =
(a) 3
(b) 4
(c) 5
(d) 6
y = ax + 3
If (4, 19) is its solution, then it must satisfy the equation.
Thus, we have
19 = a × 4 + 3
i.e. 4a = 16
i.e. a = 4
Hence, correct option is (b).
If (a, 4) lies on the graph of 3x + y = 10, then the value of a is
(a) 3
(b) 1
(c) 2
(d) 4
3x + y = 10
If (a, 4) lies on its graph, then it must satisfy the equation.
Thus, we have
3(a) + 4 = 10
i.e. 3a = 6
i.e. a = 2
Hence, correct option is (c).
The graph of the linear equation 2x - y = 4 cuts x-axis at
(a) (2, 0)
(b) (-2, 0)
(c) (0, -4)
(d) (0, 4)
On x-axis, the y-co-ordinate is always 0.
So, 2x - y = 4 will cut the x-axis where y = 0
i.e. 2x = 4
i.e. x = 2
Thus, 2x - y = 4 will cut the x-axis at (2, 0).
Hence, correct option is (a).
How many linear equations are satisfied by x = 2 and y = -3?
(a) Only one
(b) Two
(c) Three
(d) Infinitely many
From Point (2, -3) there are infinitely many lines passing in every-direction.
So (2, -3) is satisfied with infinite linear equations.
Hence, correct option is (d).
The equation x - 2 = 0 on number line is represented by
(a) a line
(b) a point
(c) infinitely many lines
(d) two lines
Given equation is x – 2 = 0.
If this is treated as an equation in one variable x only, then it has the unique solution x = 2, which is a point on the number line.
However, when treated as an equation in two variables, it can be expressed as x - 2 = 0.
So as, an equation in two variables, x – 2 = 0 is represented by a single line parallel to y-axis at a distance of 2 units.
Hence, correct option is (a).
x = 2, y = -1 is a solution of the linear equation
(a) x + 2y = 0
(b) x + 2y = 4
(c) 2x + y = 0
(d) 2x + y = 5
Substituting x = 2 and y = -1 in the following equations:
L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 = R.H.S.
L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 ≠ 4 ≠ R.H.S.
L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 0 ≠ R.H.S.
L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 5 ≠ R.H.S.
Hence, correct option is (a).
If (2k - 1, k) is a solution of the equation 10x - 9y = 12, then k =
(a) 1
(b) 2
(c) 3
(d) 4
If (2k - 1, k) is solution of equation 10x - 9y = 12, then it must satisfy this equation.
Thus, we have
10(2k - 1) - 9k = 12
20k - 10 - 9k = 12
11k = 22
k = 2
Hence, correct option is (b).
The distance between the graph of the equations x = -3 and x = 2 is
(a) 1
(b) 2
(c) 3
(d) 5
The distance between the graph of the equations x = -3 and x = 2
= 2 - (-3)
= 2 + 3
= 5
Hence, correct option is (d).
The distance between the graphs of the equations y = -1 and y = 3 is
(a) 2
(b) 4
(c) 3
(d) 1
The distance between given two graphs
= 3 - (-1)
= 3 + 1
= 4
Hence, correct option is (b).
If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length
(a) 4 units
(b) 3 units
(c) 5 units
(d) none of these
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