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Important Questions for Class 9 Maths Chapter 4 FAQs
What are the conditions for a linear equation in two variables?
An equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero. For example, 10x+4y = 3 and -x+5y = 2 are linear equations in two variables.
What is the unique solution in linear equations in two variables?
Every linear equation in one variable has a unique solution. But a pair of linear equations have two solutions i.e. one for x and the other for y, satisfying both the equations. If a given set of linear equation intersects at a point, the solution will be unique for both the equations.
What is a pair of linear equations in two variables problem?
Linear equations in two variables are equations which can be expressed as ax + by + c = 0, where a, b and c are real numbers and both a, and b are not zero. The solution of such equations is a pair of values for x and y which makes both sides of the equation equal.
What is the general form of a linear equation in two variables?
Ax+By=C
Who discovered the linear equation in two variables?
Sir William Rowan Hamilton, an Irish mathematician, invented linear equations in the year 1843.
Important Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
Important Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables has been provided here. Students can refer to these questions before their examinations for better preparation.
Neha Tanna22 Oct, 2024
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Important Questions for Class 9 Maths Chapter 4:
Linear Equations in Two Variables introduces the concept of equations involving two variables and how they can be represented graphically. Important topics include the general form of linear equations (ax + by + c = 0), solutions of linear equations, plotting graphs, and finding the x- and y-intercepts.
Key questions often focus on solving equations, verifying solutions, and graphing lines on a coordinate plane. Students are also expected to understand the relationship between coefficients and the slope of the line and parallel and intersecting lines.
Important Questions for Class 9 Maths Chapter 4 Overview
Chapter 4 of Class 9 Maths, Linear Equations in Two Variables, is crucial for building a strong foundation in algebra and graphing. It introduces linear equations involving two variables, which form the basis for understanding higher-level concepts like coordinate geometry, simultaneous equations, and calculus in advanced classes. This chapter covers important concepts like the general form of linear equations, solutions of equations, plotting them on a graph, and interpreting their graphical representations.
Students learn how to find the x- and y-intercepts, and how the coefficients affect the slope and position of lines. Understanding this chapter is key for solving problems related to real-life situations, such as determining the relationship between quantities and predicting outcomes, making it a highly important chapter.
Important Questions for Class 9 Maths Chapter 4 PDF
Chapter 4 of Class 9 Maths, Linear Equations in Two Variables, is essential for understanding algebraic concepts and their graphical representations. This chapter forms the basis for future studies in geometry and algebra. Below, we have provided a PDF containing important questions that cover key concepts, helping students practice and enhance their problem-solving skills effectively.
Important Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
Below is the Important Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables -
Question 1: Define the following linear equations in the form ax + by + c = 0 and show the values of a, b and c in every individual case:
(i) x – y/5 – 10 = 0
(ii) -2x+3y = 6
(iii) y – 2 = 0
Answer 1:
(i) The equation x-y/5-10 = 0
(1)x + (-1/5) y + (-10) = 0
Directly compare the above equation with ax + by + c = 0
Therefore, we get;
a = 1
b = -⅕
c = -10
(ii) –2x + 3y = 6
Re-arranging the provided equation, we obtain,
–2x + 3y – 6 = 0
The required equation –2x + 3y – 6 = 0 can be written as,
(–2)x + 3y +(– 6) = 0
Directly comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0
We obtain a = –2
b = 3
c = -6
(iii) y – 2 = 0
y – 2 = 0
The required equation y – 2 = 0 can be written as,
0x + 1y + (–2) = 0
Directly comparing 0x + 1y + (–2) = 0 with ax + by + c = 0
We obtain a = 0
b = 1
c = –2
Question 2:
The price of a notebook is twice the cost of a pen. Note a linear equation in two variables to illustrate this statement.
(Taking the price of a notebook to be ₹ x and that of a pen to be ₹ y)
Answer 2:
Let the price of one notebook be = ₹ x
Let the price of one pen be = ₹ y
As per the question,
The price of one notebook is twice the cost of one pen.
i.e., the price of one notebook = 2×price of a pen
x = 2×y
x = 2y
x-2y = 0
x-2y = 0 is the required linear equation in two variables to illustrate the statement, ‘The price of one given notebook is twice the cost of a pen.
Question 3:
Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
Answer 3:
(i) 2x + 9 = 0
We have, 2x + 9 = 0
2x = – 9
x = -9/2
which is the required linear equation in one variable, that is, x only.
Therefore, x= -9/2 is a unique solution on the number line as shown below:
(ii) 2x +9=0
We can write 2x + 9 = 0 in the two variables as 2x + 0, y + 9 = 0
or x = −9−0.y/2
∴ When y = 1, x = −9−0.(1)/2 = -9/2
y=2 , x = −9−0.(2)/2 = -9/2
y = 3, x = −9−0.(3)/2= -9/2
Therefore, we obtain the following table:
X
-9/2
-9/2
-9/2
Y
1
2
3
Now, plotting the ordered pairs (−9/2,3), (−9/2,3) and (−9/2,3) on graph paper and connecting them, we get a line PQ as the solution of 2x + 9 = 0.
Question 4:
Note four solutions individually for the following equations:
(i) 2x + y = 7
Answer 4:
For the four answers of 2x + y = 7, we replace different values for x and y
Let x = 0
Then,
2x + y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Now,
2x + y = 7
(2×1)+y = 7
2+y = 7
y = 7 – 2
y = 5
(1,5)
Let y = 1
Now,
2x + y = 7
2x+ 1 = 7
2x = 7 – 1
2x = 6
x = 3
(3,1)
Let x = 2
Now,
2x + y = 7
2(2)+y = 7
4+y = 7
y = 7 – 4
y = 3
(2,3)
The answers are (0, 7), (1,5), (3,1), (2,3)
Question 5: The linear equation 2x -5y = 7 has
(A) A unique solution
(B) Two solutions
(C) Infinitely many solutions
Answer 5:
(C) Infinitely many solutions
Solution:
Linear equation: The equation of two variables which gives a straight line graph is called a linear equation.
Here the linear equation is 2x – 5y = 7
Let y = 0, then the value of x is:
2x – 5(0)=7
2x =7
x = 7/2
Now, let y = 1, then the value of x is:
2x – 5 (1) =7
2x -5 =7
2x = 7 + 5
2x =12
x = 12/2
x = 6
Here for different values of y, we are getting different values of x
Therefore, the equation has infinitely many solutions
Question 6:
Represent the following linear equations in the form ax + by + c = 0 and show the required values of a, b and c in every case:
Answer 6:
(i) x –(y/5)–10 = 0
The required equation x –(y/5)-10 = 0 can be written as,
1x+(-1/5)y +(–10) = 0
Comparing the given equation x+(-1/5)y+(–10) = 0 with ax+by+c = 0
We obtain,
a = 1
b = -(1/5)
c = -10
(ii) –2x+3y = 6
–2x+3y = 6
Rearranging the equation, we obtain,
–2x+3y–6 = 0
The required equation –2x+3y–6 = 0 can be written as,
(–2)x+3y+(– 6) = 0
Comparing the given equation (–2)x+3y+(–6) = 0 with ax+by+c = 0
We obtain a = –2
b = 3
c =-6
(iii) x = 3y
x = 3y
Rearranging the equation, we obtain,
x-3y = 0
The required equation x-3y=0 can be written as,
1x+(-3)y+(0)c = 0
Comparing the given equation 1x+(-3)y+(0)c = 0 with ax+by+c = 0
We obtain a = 1
b = -3
c =0
(iv) 2x = –5y
2x = –5y
Rearranging the equation, we obtain,
2x+5y = 0
The required equation 2x+5y = 0 can be written as,
2x+5y+0 = 0
Comparing the given equation 2x+5y+0= 0 with ax+by+c = 0
We obtain a = 2
b = 5
c = 0
(v) 3x+2 = 0
3x+2 = 0
The required equation 3x+2 = 0 can be written as,
3x+0y+2 = 0
Comparing the given equation 3x+0+2= 0 with ax+by+c = 0
We obtain a = 3
b = 0
c = 2
(vi) y–2 = 0
y–2 = 0
The required equation y–2 = 0 can be written as,
0x+1y+(–2) = 0
Comparing the given equation 0x+1y+(–2) = 0 with ax+by+c = 0
We obtain a = 0
b = 1
c = –2
(vii) 5 = 2x
5 = 2x
Rearranging the equation, we obtain,
2x = 5
i.e., 2x–5 = 0
The required equation 2x–5 = 0 can be written as,
2x+0y–5 = 0
Comparing the given equation 2x+0y–5 = 0 with ax+by+c = 0
We obtain a = 2
b = 0
c = -5
Question 7:
Note four solutions individually for the following equations:
πx + y = 9
Answer 7:
For the four answers of πx + y = 9, we replace other values for x and y
Let x = 0
Now,
πx + y = 9
(π × 0)+y = 9
y = 9
(0,9)
Let x = 1
Now,
πx + y = 9
(π×1)+y = 9
π+y = 9
y = 9-π
(1,9-π)
Let y = 0
Now,
πx + y = 9
πx +0 = 9
πx = 9
x =9/π
(9/π,0)
Let x = -1
Now,
Put x=2, we have
πx + y = 9
π(2) + y = 9
y = 9 – 2π
The answers are (0,9), (1,9-π),(9/π,0),(2,9 – 2π)
Question 8:
Find out the value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k.
Answer 8:
The provided equation is
2x + 3y = k
As per the given question, x = 2 and y = 1.
Then, Replacing 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 required value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k, is 7.
Question 9:
Establish that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the required linear equation y = 9x – 7.
Answer 9:
We include the equation,
y = 9x – 7
For A (1, 2),
Replacing (x,y) = (1, 2),
We obtain,
2 = 9(1) – 7
2 = 9 – 7
2 = 2
For B (–1, –16),
Replacing (x,y) = (–1, –16),
We get,
–16 = 9(–1) – 7
-16 = – 9 – 7
-16 = – 16
For C (0, –7),
Replacing (x,y) = (0, –7),
We obtain,
– 7 = 9(0) – 7
-7 = 0 – 7
-7 = – 7
Therefore, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Therefore, A (1, 2), B (–1, –16) and C (0, –7) are answers to the linear equation y = 9x – 7
Thus, points A (1, 2), B (–1, –16), and C (0, –7) lie on the graph of the linear equation y = 9x – 7.
Question 10: Note the linear equation such that every point on its graph has a coordinate 3 times its abscissa.
Answer 10:
As per the question,
A given linear equation such that every point on its graph has a coordinate(y) which is 3 times its
abscissa(x).
So we obtain
⇒ y = 3x.
Therefore, y = 3x is the required linear equation.
Question 11:
Illustrate the graph of the given linear equation 3x + 4y = 6. At what points does the graph cut the X and Y-axis?
Answer 11:
Given the equation,
3x + 4y = 6.
We need at least 2 points on the graph to illustrate the graph of this equation,
Therefore, the points the graph cuts
(i) x-axis
The given point is on the x-axis. We have y = 0.
Replacing y = 0 in the equation, 3x + 4y = 6,
We get,
3x + 4×0 = 6
⇒ 3x = 6
⇒ x = 2
Therefore, the point at which the graph cuts the x-axis = (2, 0).
(ii) y-axis
Since the point is on the y-axis, we have x = 0.
Replacing x = 0 in the equation, 3x + 4y = 6,
We obtain,
3×0 + 4y = 6
⇒ 4y = 6
⇒ y = 6/4
⇒ y = 3/2
⇒ y = 1.5
Thus, the point at which the graph cuts the x-axis = (0, 1.5).
By plotting the points (0, 1.5) and (2, 0) on the graph.
Question 12: Show that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the given graph of the linear equation y = 9x – 7.
Answer 12:
We have the given equation,
y = 9x – 7
For A (1, 2),
Substitute the values of (x,y) = (1, 2),
We obtain,
2 = 9 (1) – 7 = 9 – 7 = 2
For B (–1, –16),
Substitute the values of (x,y) = (–1, –16),
We obtain,
–16 = 9(–1) – 7 = – 9 – 7 = – 16
For C (0, –7),
Substitute the values of (x,y) = (0, –7),
We obtain,
– 7 = 9(0) – 7 = 0 – 7 = – 7
Thus, we locate that points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Thus, A (1, 2), B (–1, –16), and C (0, –7) are required solutions of the linear equation y = 9x – 7
Hence, the given points A (1, 2), B (–1, –16) and C (0, –7) lie on the graph of the required linear equation y = 9x – 7.
Benefits of Solving Important Questions for Class 9 Maths Chapter 4
Below we have provided some of the benefits of solving Important Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables -
Strengthens Conceptual Understanding:
Reinforces key concepts like linear equations, graphing, and slope.
Enhances Problem-Solving Skills:
Builds proficiency in solving different types of linear equation problems.
Improves Graphing Techniques:
Boosts the ability to plot and interpret linear equations on a coordinate plane.
Exam-Oriented Practice:
Familiarizes students with the format and difficulty of exam questions.
Boosts Confidence:
Regular practice reduces errors and increases confidence in tackling similar problems.
Prepares for Higher Studies:
Lays a solid foundation for more advanced algebra and geometry concepts.
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