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# Short Answer Type Questions- Linear Equations in Two Variables Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Full Circle

## Class 9 : Short Answer Type Questions- Linear Equations in Two Variables Class 9 Notes | EduRev

The document Short Answer Type Questions- Linear Equations in Two Variables Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. Is (3, 2) a solution of x + y = 6?
Solution: (3, 2) means x = 3 and y = 2
∴ Substituting x = 3 and y = 2 in x + y = 6,
we have 3 + 2 = 6
⇒ 5 = 6 which is not correct Since
L.H.S. ≠ R.H.S.
∴ (3, 2) is not a solution of x + y = 6.

Question 2. Is  a solution of 2x + 3y = 12?
Solution: The given equation is 2x + 3y = 12  ...(1)

Here Solution =

⇒ x = 2 and y = (8/3)
Substituting x = 2 and y =(8/3)  in (1), we get

⇒ 4 + 8 = 12
⇒ 12 = 12
∵ L.H.S. = R.H.S

is a solution of 2x + 3y = 12.

Question 3. (i) Write the equation of x-axis.
(ii) Write the equation of y-axis.
Solution: (i) Since, the y-coordinate is zero at every point on the x-axis.
∴ The equation of x-axis is y = 0.
(ii) Since, the x-coordinate is zero at every point on the y-axis.
∴ The equation of y-axis is x = 0.

Question 4. Express in the form of ax + by + c = 0 and write the value of a, b and c.
Solution: We have
...(1)

Comparing (1) with ax + by + c = 0, we have
a = –2, b = (3/2) and c = –4.

Question 5. Express 2x = 5 in the form ax + by + c = 0 and find the value of a, b and c.
Solution: 2x = 5 can be written as 2x – 5 = 0
⇒ 2x + (0)y – 5 = 0
⇒ 2x + (0)y + (–5) = 0                                ...(1)
Comparing (1) with ax + by + c = 0, we get
a = 2,  b = 0 and  c = –5.

Question 6. Write two solutions of 3x + y = 8.
Solution: We have 3x + y = 8
For x = 0, we have 3(0) + y = 8
⇒ 0 x y= 8 ⇒ y = 8

.e. (0, 8) is a solution.
For x = 1, we have 3(1) + y = 8
⇒ 3 + y = 8
⇒ y = 8 – 3 = 5

i.e. (1, 5) is another solution.

Question 7. If x = –1 and y = 2 is a solution of kx + 3y = 7, find the value k.
Solution: We have kx + 3y = 7                          ...(1)
∴ Putting x = –1 and y = 2 in (1), we get
k(–1) + 3(2) = 7
⇒ –k + 6 = 7
⇒ –k = 7 – 6 = 1
⇒ k= –1
Thus, the required value of k = –1.

Question 8. Show that x = 2 and y = 1 satisfy the linear equation 2x + 3y = 7.
Solution: We have 2x + 3y = 7                       ...(1)
Since, x = 2 and y = 1 satisfy the equation (1).
∴ Substituting x = 2 and y = 1 in (1), we get
L.H.S. = 2(2) + 3(1) = 4 + 3 = 7
= R.H.S.
Since, L.H.S. = R.H.S.
∴ x = 2 and y = 1 satisfy the given equation.

Question 9. Write four solutions of 2x + 3y = 8.
Solution: We have 2x + 3y = 8   ...(1)
Let us assume x = 0.
∴ Substituting x = 0 in (1), we get
2(0) + 3y = 8
⇒ 0 + 3y = 8
⇒ y= (8/3)

is a solution of (1)

Again assume y = 0.
∴ From (1), we have
2x + 3(0) = 8
⇒ 2x = 8
⇒   x= (8/2)  = 4

∴ (4, 0) is a solution of (1).
Again assume x = 1.
∴ From (1), we have
2(1) + 3y = 8
⇒ 3y = 8 – 2 = 6
⇒ y = (6/3) = 2
∴ (1, 2) is a solution of (1).
Again assume x = 2.
∴ From (1), we have
2(2) + 3y = 8
⇒ 4 + 3y = 8
⇒ 3y = 8 – 4 = 4
⇒ y = (4/3)

is a solution of (1).

∴ The required four solutions are:,  (4, 0), (1, 2) and

Question 10. Draw the graph of the equation 2x – 3y = 12. At what points, the graph of the equation cuts the x-axis and the y-axis?
Solution: We have 2x – 3y = 12

∴ We get the following table

Plotting the ordered pairs (0, –4), (3, –2) and (6, 0) and joining them, we get a straight line PQ.
Thus, PQ is the graph of 2x – 3y = 12.

From the graph, we see that, it (in line PQ) cuts the x-axis at the point (6, 0) and the y-axis at the point (0, –4).

Question 11. Draw the graph of 9x – 5y + 160 = 0. From the graph find the value of y when x = 5.
Solution: We have: 9x – 5y + 160 = 0

When x = 0,
When x = –10,
When x = –15,
We get the table:

Plotting (0, 32), (–10, +14) and (–15, 5) on the graph paper and joining them, we get a straight line.
From the graph, we find for x = 5, the value of y = 41.

Question 12. From the following graph, find four solutions of the equation representing the line AB.

Solution: From the graph, we find that the points (2, 0), (0, 3), (4, –3) and (–2, 6) lie on AB.

are four solutions of the equation representing the straight line AB.

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