(1) 2x + y = 7 ⇒ y = 7 - 2x Putting x = 0, we have, y = 7 - 2 x 0 = 7, therefore, (0, 7) is a solution of the equation. Putting x = 1, we have, y = 7 - 2 x 1 = 5, therefore, (1, 5) is a solution of the equation. Putting x = 2, we have, y = 7 - 2 x 2 = 3, therefore, (2, 3) is a solution of the equatiRead more
(1) 2x + y = 7 ⇒ y = 7 – 2x
Putting x = 0, we have, y = 7 – 2 x 0 = 7, therefore, (0, 7) is a solution of the equation.
Putting x = 1, we have, y = 7 – 2 x 1 = 5, therefore, (1, 5) is a solution of the equation.
Putting x = 2, we have, y = 7 – 2 x 2 = 3, therefore, (2, 3) is a solution of the equation
Putting x = 3, we have, y = 7 – 2 x 3 = 1, therefore, (3, 1) is a solution of the equation.
Hence, (0, 7), (1, 5), (2, 3) and (3, 1) are the four solutions of the equation 2x + y = 7.
(ii) πx + y = 9 ⇒ y = 9 – πx
Putting x = 0, we have, y = 9 – π × 0 = 9, therefore, (0, 9) is a solution of the equation.
Putting x = 1, we have, y = 9 – π × 1 = 9 – π, therefore, (1, 9 – π) is a solution of the equation.
Putting x = 2, we have, y = 9 – π × 2 = 9 – 27, therefore, (2, 9- 2π) is a solution of the equation.
Putting x = 3, we have, y = 9 – πx 3 = 9 – 3π, therefore, (3, 9 – 3π) is a solution of the equation.
Hence, (0, 9), (1, 9 – π), (2, 9 – 2π) and (3, 9 -3π) are the four solutions of the equation πx + y = 9.
(iii) x = 4y
Putting y = 0, we have, x = 4 × 0 = 0, therefore, (0, 0) is a solution of the equation.
Putting y = 1, we have, x = 4 × 1 = 4, therefore, (4, 1) is a solution of the equation.
Putting y = 2, we have, x = 4 × 2 = 8, therefore, (8, 2) is a solution of the equation.
Putting y = 3, we have, x = 4 × 3 = 12, therefore, (12, 3) is a solution of the equation.
Hence, (0, 0), (4, 1), (8, 2) and (12, 3) are the four solutions of the equation x = 4y.
Figure 1 x + y = 0 is the correct equation as it satisfies the points (-1, 1), (0, 0) and (1, -1). Figure 2 y = -x + 2 is the correct equation as it satisfies the points (-1, 3), (0, 2) and (2, 0).
Figure 1
x + y = 0 is the correct equation as it satisfies the points (-1, 1), (0, 0) and (1, -1).
Figure 2
y = -x + 2 is the correct equation as it satisfies the points (-1, 3), (0, 2) and (2, 0).
Let the contribution by Yamini = ₹x Let the contribution by Fatima = ₹y According to question, x + y = 100 ⇒ y = 100 – x For the graph: Putting x = 0, we have, y = 100 - 0 = 100 Putting x = 10, we have, y = 100 - 10 = 90 Putting x = 20, we have, y = 100 - 20 = 80 Hence, A(0, 100), B(10, 90) and C(20Read more
Let the contribution by Yamini = ₹x
Let the contribution by Fatima = ₹y
According to question, x + y = 100
⇒ y = 100 – x
For the graph:
Putting x = 0, we have, y = 100 – 0 = 100
Putting x = 10, we have, y = 100 – 10 = 90
Putting x = 20, we have, y = 100 – 20 = 80
Hence, A(0, 100), B(10, 90) and C(20, 80) are the solutions of equation.
Table lamp is 9 unit away form the sitting place and 4 units awaw form the left side. Hence, its coordinates ar (4,9). Here is the Video Explanation of this Question😁
Table lamp is 9 unit away form the sitting place and 4 units awaw form the left side. Hence, its coordinates ar (4,9).
Write four solutions for each of the following equations:
(1) 2x + y = 7 ⇒ y = 7 - 2x Putting x = 0, we have, y = 7 - 2 x 0 = 7, therefore, (0, 7) is a solution of the equation. Putting x = 1, we have, y = 7 - 2 x 1 = 5, therefore, (1, 5) is a solution of the equation. Putting x = 2, we have, y = 7 - 2 x 2 = 3, therefore, (2, 3) is a solution of the equatiRead more
(1) 2x + y = 7 ⇒ y = 7 – 2x
Putting x = 0, we have, y = 7 – 2 x 0 = 7, therefore, (0, 7) is a solution of the equation.
Putting x = 1, we have, y = 7 – 2 x 1 = 5, therefore, (1, 5) is a solution of the equation.
Putting x = 2, we have, y = 7 – 2 x 2 = 3, therefore, (2, 3) is a solution of the equation
Putting x = 3, we have, y = 7 – 2 x 3 = 1, therefore, (3, 1) is a solution of the equation.
Hence, (0, 7), (1, 5), (2, 3) and (3, 1) are the four solutions of the equation 2x + y = 7.
(ii) πx + y = 9 ⇒ y = 9 – πx
Putting x = 0, we have, y = 9 – π × 0 = 9, therefore, (0, 9) is a solution of the equation.
Putting x = 1, we have, y = 9 – π × 1 = 9 – π, therefore, (1, 9 – π) is a solution of the equation.
Putting x = 2, we have, y = 9 – π × 2 = 9 – 27, therefore, (2, 9- 2π) is a solution of the equation.
Putting x = 3, we have, y = 9 – πx 3 = 9 – 3π, therefore, (3, 9 – 3π) is a solution of the equation.
Hence, (0, 9), (1, 9 – π), (2, 9 – 2π) and (3, 9 -3π) are the four solutions of the equation πx + y = 9.
(iii) x = 4y
See lessPutting y = 0, we have, x = 4 × 0 = 0, therefore, (0, 0) is a solution of the equation.
Putting y = 1, we have, x = 4 × 1 = 4, therefore, (4, 1) is a solution of the equation.
Putting y = 2, we have, x = 4 × 2 = 8, therefore, (8, 2) is a solution of the equation.
Putting y = 3, we have, x = 4 × 3 = 12, therefore, (12, 3) is a solution of the equation.
Hence, (0, 0), (4, 1), (8, 2) and (12, 3) are the four solutions of the equation x = 4y.
From the choices given below, choose the equation whose graphs are given in Figures.
Figure 1 x + y = 0 is the correct equation as it satisfies the points (-1, 1), (0, 0) and (1, -1). Figure 2 y = -x + 2 is the correct equation as it satisfies the points (-1, 3), (0, 2) and (2, 0).
Figure 1
See lessx + y = 0 is the correct equation as it satisfies the points (-1, 1), (0, 0) and (1, -1).
Figure 2
y = -x + 2 is the correct equation as it satisfies the points (-1, 3), (0, 2) and (2, 0).
Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.
Let the contribution by Yamini = ₹x Let the contribution by Fatima = ₹y According to question, x + y = 100 ⇒ y = 100 – x For the graph: Putting x = 0, we have, y = 100 - 0 = 100 Putting x = 10, we have, y = 100 - 10 = 90 Putting x = 20, we have, y = 100 - 20 = 80 Hence, A(0, 100), B(10, 90) and C(20Read more
Let the contribution by Yamini = ₹x
See lessLet the contribution by Fatima = ₹y
According to question, x + y = 100
⇒ y = 100 – x
For the graph:
Putting x = 0, we have, y = 100 – 0 = 100
Putting x = 10, we have, y = 100 – 10 = 90
Putting x = 20, we have, y = 100 – 20 = 80
Hence, A(0, 100), B(10, 90) and C(20, 80) are the solutions of equation.
In which quadrant or on which axis do each of the points (– 2, 4), (3, – 1), (– 1, 0), (1, 2) and (– 3, – 5) lie? Verify your answer by locating them on the Cartesian plane.
(-2, 4): Second Quadrant (3,-1): Fourth Quadrant (-1,0): x-axis (1,2): First Quadrant (-3,-5): Third Quadrant Here is the video explanation😊✌
(-2, 4): Second Quadrant
(3,-1): Fourth Quadrant
(-1,0): x-axis
(1,2): First Quadrant
(-3,-5): Third Quadrant
Here is the video explanation😊✌
See lessHow will you describe the position of a table lamp on your study table to another person?
Table lamp is 9 unit away form the sitting place and 4 units awaw form the left side. Hence, its coordinates ar (4,9). Here is the Video Explanation of this Question😁
Table lamp is 9 unit away form the sitting place and 4 units awaw form the left side. Hence, its coordinates ar (4,9).
Here is the Video Explanation of this Question😁
See less(Street Plan): A city has two main roads which cross each other at the center of the city. These two roads are along the North-South direction and East-West direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are about 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North – South direction and another in the East – West direction. Each cross street is referred to in the following manner: If the 2nd street running in the North – South direction and 5th in the East – West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find: (i) how many cross – streets can be referred to as (4, 3). (ii) how many cross – streets can be referred to as (3, 4).
(i) Only one cross-street referred to as (4,3). (ii) Only one cross-street referred to as (3, 4). You can see here for better explanation✌😁
(i) Only one cross-street referred to as (4,3).
(ii) Only one cross-street referred to as (3, 4).
You can see here for better explanation✌😁
See lessWhat is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
(i) x-axis and y-axis
(i) x-axis and y-axis
See lessWhat is the name of each part of the plane formed by these two lines?
Quadrant
Quadrant
See lessWrite the name of the point where these two lines intersect.
Origin
Origin
See less