# RBSE Class 9 Maths Important Questions Chapter 4 Linear Equations in Two Variables

Rajasthan Board RBSE Class 9 Maths Important Questions Chapter 4 Linear Equations in Two Variables Important Questions and Answers.

Rajasthan Board NCERT New Syllabus RBSE Solutions for Class 9 Guide Pdf free download in Hindi Medium and English Medium are part of RBSE Solutions. Here we have given RBSE Class 9th Books Solutions.

## RBSE Class 9 Maths Chapter 4 Important Questions Linear Equations in Two Variables

I. Multiple Choice Questions :
Choose the correct answer from the given options.

Question 1.
The number of lines passing through a point (2, 3) are :
(a) only one
(b) two
(c) infinite
(d) None of these
(c) infinite

Question 2.
Which of the following are solutions of the equation x + 2y = 7?
(a) x = 3, y = 5
(b) x = 3, y = - 5
(c) x = 3, y = 2
(d) x = 0, y = 7
(c) x = 3, y = 2

Question 3.
The graph of equation 2x + 3y = 12 represents a :
(a) point
(b) straight line
(c) triangle
(d) square
(b) straight line

Question 4.
The graph of equation 3x + 4y = 12 intersects -e-axis at :
(a) (4, 0)
(b) (0, 3)
(c) (4, 3)
(d) (0, 0)
(a) (4, 0)

Question 5.
The graph of y = b is a straight line :
(a) parallel to x-axis
(b) parallel to y-axis
(c) passes through origin
(d) coincident on x-axis
(a) parallel to x-axis

Question 6.
x = 5, y = 2 is a solution of the linear equation :
(a) x + 2y = 7
(b) 5x + 2y = 7
(c) x + y = 7
(d) 5x + y = 7
(c) x + y = 7

Question 7.
Any point on the fine y = x is of the form :
(a) (a, a)
(b) (0, a)
(c) (a, 0)
(d) (a, - a)
(a) (a, a)

Question 8.
The number of linear equations in x and y which can be satisfied by x = 1 and y = 2 (are) :
(a) only one
(b) two
(c) three
(d) infinitely many
(d) infinitely many

Question 9.
The graph of the linear equation y = x passes through the point :
(a) $$\left(\frac{3}{2},-\frac{3}{2}\right)$$
(b) (0, $$\frac{3}{2}$$)
(c) (1, 1)
(d) $$\left(-\frac{1}{2}, \frac{1}{2}\right)$$
(c) (1, 1)

Question 10.
The equation 2x + 5y = 7 has unique solution of x, y are :
(a) natural numbers
(b) positive real numbers
(c) real numbers
(d) rational numbers
(a) natural numbers

Question 11.
The equation y + 3 = 0 represents a line :
(a) parallel of y-axis
(b) parallel to x-axis
(c) passing through origin
(d) passing through the point (- 3, 0)
(b) parallel to x-axis

Question 12.
Which of the following equations represents a line parallel to x-axis?
(a) 3x + 2 = 0
(b) 3y + 2 = 0
(c) 3x + 2y = 0
(d) 3x - 2y = 0
(b) 3y + 2 = 0

Question 13.
The equation x = 2y represents a line :
(a) parallel to x-axis
(b) parallel to y-axis
(c) passing through origin
(d) passing through the point (1, 2)
(c) passing through origin

Question 14.
Which of the following equations has a graph shown in the adjoining figure?

(a) y = x
(b) x + y = 0
(c) y = 2x
(d) 2 + 3y = 7x
(b) x + y = 0

Question 15.
Choose the equation whose graph is represented below:

(a) 2x + 3y = 7
(b) 2x + 3y = 9
(c) 2x + 3y = 6
(d) x + y = 2
(c) 2x + 3y = 6

II. Fill in the Blanks :

Question 1.
If the linear equation 3x - ay = 6 has one solution as (4, 3), then a = ______________.
2

Question 2.
If the point (3, 4) lies on the graph of 3y = ax + 7, then the value of a = ______________
$$\frac{5}{3}$$

Question 3.
The solution of the linear equation x + 2y = 8, which represents a point on x-axis ______________
(8, 0)

Question 4.
The equation of the x-axis is given by ______________
y = 0

Question 5.
The linear equation for the statement “Age of ‘x’ exceeds age of y by 7 years” = ______________
x - y - 7 = 0

III. True/ False:
State whether the following statements are True or False.

Question 1.
ax + by + c = 0, where a, b and c are real numbers, is a linear equation in two variables.
False

Question 2.
The graph of the equation y = 3x + 5 passes through the origin.
False

Question 3.
The line parallel to the x-axis at distance 3 units above x-axis is given by the line y = 3.
False

Question 4.
The graph of every linear equation in two variables need not be a straight line.
True

Question 5.
Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.
False

Question 6.
The graph of the equation y = 1 is a line parallel to y-axis.
False

Question 7.
A linear equation in two variables has no more than two solutions.
False

Question 8.
The graph of the line y = -2 is parallel to x-axis at a distance of 2 units below the origin.
True

IV. Match the Columns :

Question 1.
Match the column I with the column II.

 Column I Column II 1. If the point (1,2) lies on the line 2x + ay = 8 then a = 1 2. The number of solutions of the equation x + 5y = 11 where x, y are natural number is (are) 2 3. The number of lines passing through (3, 4) and parallel to x-axis is (are) 3

 Column I Column II 1. If the point (1,2) lies on the line 2x + ay = 8 then a = 3 2. The number of solutions of the equation x + 5y = 11 where x, y are natural number is (are) 2 3. The number of lines passing through (3, 4) and parallel to x-axis is (are) 1

V. Very Short Answer Type Questions :

Question 1.
Find the value ‘m’ if (- m, 3) is a solution of equation 4x + 9y - 3 = 0.
If (- m, 3) is a solution of the equation
4x + 9y - 3 = 0, then 4(- m) + 9(3) -3 = 0
⇒ - 4m + 27 - 3 = 0
⇒ - 4m + 24 = 0
⇒ 4m = 24
⇒ m = $$\frac{24}{4}$$ = 6

Question 2.
Give the equations of two lines passing through (3, 14). How many more such lines are there, and why?
The equations of two lines passing through (3, 14) can be taken as
x + y = 17 and 7x - y = 7
There are infinitely many such fines because through a point an infinite number of fines can be drawn.

Question 3.
If the point (3, 4) lies on the graph of the equation = ax + 7, find the value of a.
8y = ax + 7, then
8(4) = a(3) + 7
⇒ 32 = 3a + 7
⇒ 32 - 7 = 3a
⇒ 25 = 3a
⇒ 3a = 25
⇒ a = $$\frac{25}{3}$$

Question 4.
Express the linear equation 7 = 2x in the form ax + by + c = 6 and also write the values of a, b and c.
7 = 2x
⇒ 2x - 7 = 0
⇒ 2x + 0y-7 = 0 .
Comparing with ax + by + c = 0, we get :a = 2, 6 = 0, c = - 7

Question 5.
Write two solutions of the linear equation x + 2y = 1.
x + 2y = 1
⇒ 2y = 1 - x
⇒ y = $$\frac{1-x}{2}$$
When x = 1 , y = $$\frac{1-1}{2}$$ = 0
When x = 3, y = $$\frac{1-3}{2}=\frac{-2}{2}$$ = -1
Two solution are (1, 0); (3, -1)

Question 6.
The cost of a pen is ₹ 5 less than half the cost of a notebook. Write this statement as a linear equation in two variables.
Let the cost of notebook be x and cost of pen be y.
∴ According to the given condition
y = $$\frac{1}{2}$$x - 5
2y = x - 10
⇒ x - 2y - 10 = 0 is the linear equation in two variables.

VI. Short Answer Type Questions :

Question 1.
Determine the point on the graph of the linear equation 2x + 5y = 19, whose ordinate is lg times its abscissa.
Let x and y be the abscissa and ordinate, respectively of the point lie on given line 2x + 5y = 19. ...............(i)
Then, by given condition,
Ordinate (y) = 1$$\frac{1}{2}$$x (Abscissa) ...............(ii)
⇒ y = $$\frac{3}{2}$$x

On putting y = $$\frac{3}{2}$$x in equation (i), we get:
2x + 5 ($$\frac{3}{2}$$) = 19
⇒ 4x + 15x = 38
⇒ 19x = 38
⇒ x = 2

On substituting the value of * in equation (ii), we get:
y = $$\frac{3}{2}$$ × 2 = 3
Hence, the required point is (2, 3).

Question 2.
Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5?
Given that, y varies directly as x.
i.e. y ∝ x ⇒ y = kx .............(i)
Where, k is an arbitrary constant.
Given, y = 12 and x = 4
Then, from equation (i),
12 = 4k
⇒ k = 3
On putting the value of k in equation (i), we get:
y = 3x .............(ii)
Which is the required linear equation.
When x = 5, then from equation (ii),
y = 3 × 5 = 15

Question 3.
Compare the; equation $$\frac{x}{3}+\frac{3}{2}$$y + 4 = 2y - 3 and lx + my - n = 0 and write the value of l, m and n.
$$\frac{x}{3}+\frac{3}{2}$$y + 4 = 2y - 3
$$\frac{x}{3}+\frac{3}{2}$$y + 7 = 0
Comparing with lx + my - n = 0, we get: l = $$\frac{1}{3}$$x,m = - $$\frac{1}{2}$$,n = -7

Question 4.
Find at least 3 solutions for the following linear equation in two variables: 2x + 5y = 13
2x + 5y = 13
⇒ 5y = 13 - 2x
⇒ y = $$\frac{13-2 x}{5}$$
Put x = 0, then y = $$\frac{13-2(0)}{5}=\frac{13}{5}$$
Put x = 1, then y = $$\frac{13-2(1)}{5}=\frac{11}{5}$$
Put x = 2, then y = $$\frac{13-2(2)}{5}=\frac{9}{5}$$
Put x = 3, then y = $$\frac{13-2(3)}{5}=\frac{7}{5}$$
∴ (0, $$\frac{13}{5}$$), (1, $$\frac{11}{5}$$), (2, $$\frac{9}{5}$$) and (3, $$\frac{7}{5}$$) are the solutions of the equation 2x + 5y = 13.

Question 5.
How many solutions of the equation 2x + 1 = x - 3 are there on the :
(i) number line?
(ii) cartesian plane?
2x + 1 = x - 3
2x - x = - 1 - 3
x = - 4
(i) On number line, only one solution is possible i.e. x = - 4

(ii) On cartesian plane.
We have solutions as :
x + 0.y + 4 = 0 ..........(i)

At y = 1, y = 2, and y = 3, are x = -4, x = -4, x = 4. [from(i)]

∴ In cartesian plane infinitely many solutions are possible for x + O.y + 4 = 0. The graph is a line parallel to y-axis at a distance of - 4 units to the left of origin.

Question 6.
Solve for x : 3x - 12 + $$\frac{3}{7}$$x = 2(x - 1)
What type of graph is it in two dimensions?
3x - 12 + $$\frac{3}{7}$$x = 2(x - 1) 94
$$\frac{24}{7}$$x - 12 = 2x - 2
$$\frac{24}{7}$$x - 2x = 12 - 2
$$\frac{10 x}{7}$$ = 10
⇒ x = 7
The graph of this equation is a line parallel to y-axis at a distance of 7 units to the right of origin O.

VII. Long Answer Type Questions :

Question 1.
Give the geometrical representation of 2y + 7 = 0 as equation in :
(i) one variable,
(ii) two variables.
The given equation is
2y + 7 = 0

(i) In one variable
2y + 7 = 0
⇒ 2y = - 7
⇒ y = -$$\frac{7}{2}$$
The representation of y = $$\frac{7}{2}$$ on the number line is as shown fig. :

(ii) In two variables
2y + 7 = 0
0.x+2.y+7 = 0
It is a linear equation in two variables x and y. This is represented by a line. All the values of x are possible because O.x is always O. However, y must satisfy the relation 2y + 7 = 0,

i.e. y = -$$\frac{7}{2}$$. Hence, two solutions of the given equation are x = 0, y = -$$\frac{7}{2}$$ and x = 2, y = -$$\frac{7}{2}$$.

Thus, the graph AB is a line parallel to the x-axis at a distance of units below it.

Question 2.
Force applied on a body is directly proportional to the acceleration produced in the body. Write an equation to express the situation and plot the graph of the equation taking constant to he 5 units.
Let the force applied on a body be y units and the acceleration produced in the body be a units.

Then,
y ∝ x
⇒ y = kx
Here, k = 5 units (given)
∴ y = 5x ...............(1)

(1) is an equation to express the situation.
Table of solutions

We plot the points (0, 0) and (1, 5) on a graph paper and join the same by a ruler to get the line which is the graph of the equation y = 5x.

Question 3.
Draw the graph of the linear equation y = mx + c for m = 2 and c = 1. Read from the graph, the value of y when x = $$\frac{3}{2}$$.
Given equation is y = mx + c.
For m = 2 and c = 1, the linear equation is y = 2x + 1 ................(i)

When x = 0, then y = 2 × 0 + 1 = 1
When x = 1, then y = 2 × 1 + 1 = 3
When x = 2, then y = 2 × 2 + 1 = 5

Thus, we have the following table :

So, plot the points A(0,1), 5(1, 3) and C(2, 5) on the graph paper and join them by a line.
Thus, the line AC is the required graph of given linear equation.

Now, draw a line parallel to Y-axis at a g distance x = $$\frac{3}{2}$$ from O on X-axis. The line intersects the graph AC at point D. Now, draw a perpendicular from D on Y-axis, which intersects at y = 4.
Hence, it is clear from the graph that when x = $$\frac{3}{2}$$ then y = 4.

Last Updated on April 29, 2022, 2:33 p.m.
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Published April 29, 2022