extra questions of class 9 maths chapter 4 linear equation in two variable (excpt ncert)
Answers
Step-by-step explanation:
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
Write four solutions for each of the following equations:
(i) 2x + y = 7
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is:
2 units
0 unit
The taxi fare in a city is as follows: For the first kilometre, the fare is ₹ 8 and for the subsequent distance it is ₹5 per km. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information, and draw its graph.
Give the geometric representations of 2x+9 = 0 as an equation:
in one variable
in two variables
Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.
What is the number of solutions of the following pair of linear equations. x + 2y – 8 = 0 and 2x + 4y = 16?
a) 0
b) 1
c) 2
d) Infinite
What is the graph of the linear equation 2x +3y = 6 that cuts the y-axis at the point?
a) (2, 0)
b) (0, 3)
c) (3, 0)
d) (0, 2)
MCQ Questions - Linear equations in 2 variables
Q.1: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
x – y/5 – 10 = 0
Q.2. Write four solutions for each of the following equations:
2x + y = 7
Q.3: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Q.4: Draw the graph of each of the following linear equations in two variables:
y = 3x
Q.5: If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Q.6: Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.
Q.7: Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis?
Solutions:-
1) The equation x-y/5-10 = 0 can be written as:
(1)x + (-1/5) y + (-10) = 0
Now compare the above equation with ax + by + c = 0
Thus, we get;
a = 1
b = -⅕
c = -10
2) To find the four solutions of 2x + y = 7 we substitute different values for x and y
Let x = 0
Then,
2x + y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Then,
2x + y = 7
(2×1)+y = 7
2+y = 7
y = 7 – 2
y = 5
(1,5)
Let y = 1
Then,
2x + y = 7
2x+ 1 = 7
2x = 7 – 1
2x = 6
x = 3
(3,1)
Let x = 2
Then,
2x + y = 7
2(2)+y = 7
4+y = 7
y = 7 – 4
y = 3
(2,3)
The solutions are (0, 7), (1,5), (3,1), (2,3)
3) The given equation is
2x + 3y = k
According to the question, x = 2 and y = 1.
Now, Substituting the values of x and y in the equation 2x + 3y = k,
We get,
⇒(2 x 2)+ (3 × 1) = k
⇒4+3 = k
⇒7 = k
⇒k = 7
The value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k, is 7.
4) (For the 4th question graph is below)
To draw a graph of linear equations in two variables, let us find out the points to plot.
To find out the points, we have to find the values for which x and y satisfies the given equation.
Here,
y=3x
Substituting the values for x,
When x = 0,
y = 3x
y = 3(0)
⇒ y = 0
When x = 1,
y = 3x
y = 3(1)
⇒ y = 3
5) The given equation is
3y = ax + 7
According to the question, x = 3 and y = 4
Now, Substituting the values of x and y in the equation 3y = ax + 7,
We get,
(3×4) = (ax3) + 7
⇒ 12 = 3a+7
⇒ 3a = 12–7
⇒ 3a = 5
⇒ a = 5/3
The value of a, if the point (3, 4) lies on the graph of the equation 3y = ax + 7 is 5/3.
6) Given equation,
3x + 4y = 6.
We need at least 2 points on the graph to draw the graph of this equation,
Thus, the points the graph cuts
(i) x-axis
Since the point is on the x-axis, we have y = 0.
Substituting y = 0 in the equation, 3x + 4y = 6,
We get,
3x + 4×0 = 6
⇒ 3x = 6
⇒ x = 2
Hence, the point at which the graph cuts x-axis = (2, 0).
(ii) y-axis
Since the point is on the y-axis, we have, x = 0.
Substituting x = 0 in the equation, 3x + 4y = 6,
We get,
3×0 + 4y = 6
⇒ 4y = 6
⇒ y = 6/4
⇒ y = 3/2
⇒ y = 1.5
Hence, the point at which the graph cuts y-axis = (0, 1.5).
