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Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables

Q1. Solve:

Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables
Hint: Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables  ...(1)
Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables    ...(2)

Sol. 

Subtracting (1) from (2), Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables = an – bm or Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables
Thus, Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables
Ans.Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables

Q2. Solve : 141x + 103y = 217;   103x + 141y = 27

Sol. 141x + 103y = 217 .......... (i)

103x + 141y = 27......... (ii)

(i) × 141 - (ii) × 103

(141)2x + (103)(141)y = 217  ×  141

(130)2x + (103)(141)y = 27  ×  103

 Subtracting both equations, we get 

x= (217 x 141)-(27 x 103)/ 1412- 1032

x=3

From equation (i)

141 x 3+ 103 x y=217

y= -2

Q3. Cost of a pen is two and half times the cost of a pencil. Express this situation as a linear equation in two variables.

Sol. Let cost of a pen be ₹ x and cost of a pencil be ₹ y. According to statement of the question, we have x = 5/2 y ⇒ 2x = 5y or 2x – 5y = 0

Q4. 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.

Sol.

We have the equation,

y = 9x – 7

For A (1, 2),

Substituting (x,y) = (1, 2),

We get,

2 = 9(1) – 7

2 = 9 – 7

2 = 2

For B (–1, –16),

Substituting (x,y) = (–1, –16),

We get,

–16 = 9(–1) – 7

-16 = – 9 – 7

-16 = – 16

For C (0, –7),

Substituting (x,y) = (0, –7),

We get,

– 7 = 9(0) – 7

-7 = 0 – 7

-7 = – 7

Hence, the 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 solutions of the linear equation y = 9x – 7

Therefore, the points A (1, 2), B (–1, –16), C (0, –7) lie on the graph of linear equation y = 9x – 7.

Q5. Solve : x + y = 18; y + z = 12;    z + x = 16

Sol. 

x+y = 18 ; y+z = 12 ;z+x = 16.
x = 18-y ∴ z+18-y = 16 ⇒ y-z = z & y + z = 12
adding -
2y=14⇒y = 7 & z = 5
∴ x = 18-7 = 11
∴ x = 11, y = 7 & z = 5

Q6. Write four solutions for the following equation:

πx + y = 9

Sol. To find the four solutions of πx + y = 9 we substitute different values for x and y

Let x = 0

Then,

πx + y = 9

(π × 0)+y = 9

y = 9

(0,9)

Let x = 1

Then,

πx + y = 9

(π×1)+y = 9

π+y = 9

y = 9-π

(1,9-π)

Let y = 0

Then,

πx + y = 9

πx +0 = 9

πx = 9

x =9/π

(9/π,0)

Let x = -1

Then, πx + y = 9

(π(-1))+y = 9

-π + y = 9

y = 9+π

(-1,9+π)

The solutions are (0,9), (1,9-π),(9/π,0),(-1,9+π)

The document Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 4 HOTS Questions - Linear Equations in Two Variables

1. What is a linear equation in two variables?
Ans. A linear equation in two variables is an equation that can be written in the form Ax + By = C, where A, B, and C are constants, and x and y are the variables.
2. How do you graph a linear equation in two variables?
Ans. To graph a linear equation in two variables, first solve for y to put the equation in slope-intercept form (y = mx + b). Then, plot the y-intercept (b) on the y-axis and use the slope (m) to find other points on the line.
3. What is the slope-intercept form of a linear equation?
Ans. The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept.
4. How can you determine if two linear equations in two variables are parallel?
Ans. Two linear equations in two variables are parallel if they have the same slope but different y-intercepts. This means that the lines will never intersect.
5. Can a linear equation in two variables have more than one solution?
Ans. Yes, a linear equation in two variables can have infinitely many solutions if the lines are the same (coincide) or if the lines are parallel and coincide.
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