(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