CBSE - Linear Equations in Two Variables

The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be x and that of a pen to be y).

Answer

Let the cost of pen be y and the cost of notebook be x.
A/q,
Cost  of a notebook = twice the pen = 2y.
∴2y = x
⇒ x - 2y = 0
This is a linear equation in two variables to represent this statement.

Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35               (ii) x - y/5 - 10 = 0                  (iii) -2x + 3y = 6                (iv) x = 3y
(v) 2x = -5y                        (vi) 3x + 2 = 0                        (vii) y - 2 = 0                     (viii) 5 = 2x

Answer

(i) 2x + 3y = 9.35
⇒ 2x + 3y - 9.35 = 0
On comparing this equation with ax + by + c = 0, we get
a = 2x, b = 3 and c = -9.35

(ii) x - y/5 - 10 = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -1/5 and c = -10


(iii) -2x + 3y = 6
⇒ -2x + 3y - 6 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 3 and c = -6

(iv) x = 3y
⇒ x - 3y = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -3 and c = 0

(v) 2x = -5y
⇒ 2x + 5y = 0
On comparing this equation with ax + by + c = 0, we get
a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0
⇒ 3x + 0y + 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 3, b = 0 and c = 2

(vii) y - 2 = 0
⇒ 0x + y - 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 0, b = 1 and c = -2

(viii) 5 = 2x
⇒ -2x + 0y + 5 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 0 and c = 5

Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,               (ii) only two solutions,             (iii) infinitely many solutions

Answer

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

Write four solutions for each of the following equations:
    (i) 2x + y = 7             (ii) πx + y = 9                (iii) x = 4y

Answer

(i) 2x + y = 7
⇒ y = 7 - 2x
→ Put x = 0,
y = 7 - 2 × 0 ⇒ y = 7
(0, 7) is the solution.
→ Now, put x = 1
y = 7 - 2 × 1 ⇒ y = 5
(1, 5) is the solution. 
→ Now, put x = 2
y = 7 - 2 × 2 ⇒ y = 3
(2, 3) is the solution. 
→ Now, put x = -1
y = 7 - 2 × -1 ⇒ y = 9
(-1, 9) is the solution.
The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9
⇒ y = 9 - πx 
→ Put x = 0,
y = 9 - π×0 ⇒ y = 9
(0, 9) is the solution.
→ Now, put x = 1
y = 9 - π×1 ⇒ y = 9-π
(1, 9-π) is the solution. 
→ Now, put x = 2
y = 9 - π×2 ⇒ y = 9-2π
(2, 9-2π) is the solution. 
→ Now, put x = -1
y = 9 - π× -1 ⇒ y = 9+π
(-1, 9+π) is the solution.
The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y
→ Put x = 0,
0 = 4y ⇒ y = 0
(0, 0) is the solution.
→ Now, put x = 1
1 = 4y ⇒ y = 1/4
(1, 1/4) is the solution. 
→ Now, put x = 4
4 = 4y ⇒ y = 1
(4, 1) is the solution. 
→ Now, put x = 8
8 = 4y ⇒ y = 2
(8, 2) is the solution.
The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

Check which of the following are solutions of the equation x - 2y = 4 and which are not:
    (i) (0, 2)              (ii) (2, 0)             (iii) (4, 0)            (iv) (√2, 4√2)              (v) (1, 1)

Answer

(i) Put x = 0 and y = 2 in the equation x - 2y = 4.
0 - 2×2 = 4
⇒ -4 ≠ 4
∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x - 2y = 4.
2 - 2×0 = 4
⇒ 2 ≠ 4
∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x - 2y = 4.
4 - 2×0 = 4
⇒ 4 = 4
∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x - 2y = 4.
√2 - 2×4√2 = 4 ⇒ √2 - 8√2 = 4 ⇒ √2(1 - 8) = 4
⇒ -7√2  ≠ 4
∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x - 2y = 4.
1 - 2×1 = 4
⇒ -1 ≠ 4
∴ (1, 1) is not a solution of the given equation.

Given equation = 2x + 3y = k
x = 2, y = 1 is the solution of the given equation.
A/q,
Putting the value of x and y in the equation, we get
2×2 + 3×1 = k
⇒ k = 4 + 3
⇒ k = 7

(i) x + y = 4
Put x = 0 then y = 4
Put x = 4 then y = 0

(ii) x - y = 2
Put x = 0 then y = -2
Put x = 2 then y = 0

(iii) y = 3x
Put x = 0 then y = 0
Put x = 1 then y = 3

(iv) 3 = 2x + y
Put x = 0 then y = 3
Put x = 1 then y = 1

Here, x = 2 and y =14.
Thus, x + y = 1 
also, y = 7x ⇒ y - 7x = 0
∴ The equations of two lines passing through (2, 14) are
x + y = 1 and y - 7x = 0.
There will be infinite such lines because infinite number of lines can pass through a given point.

The point (3, 4) lies on the graph of the equation. 
∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get
3×4 = a×3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 - 7
⇒ a = 5/3

Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x - 5
⇒ y = 5x + 3
 

From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
   For Fig. 4. 6                           For Fig. 4.7
   (i) y = x                                  (i) y = x + 2
   (ii) x + y = 0                           (ii) y = x – 2
   (iii) y = 2x                              (iii) y = –x + 2
   (iv) 2 + 3y = 7x                     (iv) x + 2y = 6

Answer

In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).
∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.

In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).
∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points

Let the distance traveled by the body be x and y be the work done by the force.
y ∝ x (Given)
⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)
A/q,
(i) When x = 2 units then y = 10 units
(ii) When x = 0 unit then y = 0 unit
 

Answer

Let the contribution amount by Yamini be x and contribution amount by Fatima be y.
A/q,
x + y = 100
When x = 0 then y = 100
When x = 50 then y = 50
When x = 100 then y = 0
 

In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

F = (9/5)C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Answer

(i) F = (9/5)C + 32

When C = 0 then F = 32

also, when C = -10 then F = 14

(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30 + 32
⇒ F = 54 + 32
⇒ F = 86

(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C + 32
⇒ (9/5)C = 95 - 32
⇒ C = 63 × 5/9
⇒ C = 35

(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9

Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0 + 32
⇒ F = 32

(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F - 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.

Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
y = 3

(ii) in two variables, it is represented as a line parallel to X-axis.

0x + y = 3

Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
x = -9/2

(ii) in two variables, it is represented as a line parallel to Y-axis.

2x + 0y + 9 = 0