**Linear Equations in Two Variables**

**Chapter 4**

**Exercise 4.3**

**EX 4.3 QUESTION 1.**

**Draw the graph of each of the following linear equations in two variables:**

**(i) x + y = 4**

**(ii) x – y = 2**

**(iii) y = 3x**

**(iv) 3 = 2x + y**

**Solution:**

(i) x + y = 4

⇒ y = 4 – x

putting x = 0, then y = 4 – 0 = 4

x = 1, then y =4 – 1 = 3

x = 2, then y = 4 – 2 = 2

(ii) x – y = 2

⇒ y = x – 2

putting x = 0, then y = 0 – 2 = -2

x = 1, then y = 1 – 2 = -1

x = 2, then y = 2 – 2 = 0

(iii) y = 3x

Putting x = 0,then y = 3(0) ⇒ y = 0

x = 1, then y = 3(1) = 3

x= -1, then y = 3(-1) = -3

(iv) 3 = 2x + y

⇒ y = 3 – 2x

Putting x = 0, then y = 3 – 2(0) = 3

x = 1,then y = 3 – 2(1) = 3 – 2 = 1

x = 2,then y = 3 – 2(2) = 3 – 4 = -1

**EX 4.3 QUESTION 2.**

**Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?**

**Solution:**

(2, 14) ⇒ x = 2 and y = 14

Equations of two lines passing through (2,14) are given by: x + y = 16, and 7x – y = 0

There is an infinite number of lines which passes through the point (2, 14) because an infinite number of lines can pass through a point.

**EX 4.3 QUESTION 3.**

**If the point (3, 4) lies on the graph of the equation 3y = ax+7, find the value of a.**

**Solution:**

3y = ax+7

we have( 3,4) which means x = 3 and y = 4

putting the values of x and y in the equation 3y = ax+7,

(3×4) = (a×3)+7

⟹ 12 = 3a+7

⟹ 3a = 12–7

⟹ 3a = 5

⟹ a = 5/3

Thus, the required value of a is 5/3.

**EX 4.3 QUESTION 4.**

**The taxi fare In a city Is as follows: For the first kilometre, the fare Is ₹8 and for the subsequent distance it is ₹5 per km. Taking the distance covered as x km and total fare as ₹y, write a linear equation for this Information, and draw Its graph.**

**Solution:**

Total distance covered = x km

Total taxi fare = ** **₹y

Fare for 1km = ** **₹ 8

Remaining distance = (x – 1) km

∴ Fare for (x – 1)km = ** **₹5 x(x – 1)

Total taxi fare = Rs. 8 + ** **₹ 5(x – 1)

acc. to question.

The total fare = Fare of first km+ fare after the first km

y = 8 + 5(x – 1) = y = 8 + 5x – 5

⇒ y = 5x + 3,

now, Solving the equation,

y = 5x + 3

putting x = 0, then y = 5(0) + 3 ⇒ y = 3

x = -1, then y = 5(-1) + 3 ⇒ y = -2

x = -2, then y = 5(-2) + 3 ⇒ y = -7

**EX 4.3 QUESTION 5.**

** From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.**

**For Fig. 4. 6**

**(i) y = x**

**(ii) x+y = 0**

**(iii) y = 2x**

**(iv) 2+3y = 7x**

**For Fig. 4. 7**

**(i) y = x+2**

**(ii) y = x–2**

**(iii) y = –x+2**

**(iv) x+2y = 6**

**Solution:**

For the For Fig.(4.6)

x + y = 0 is the correct linear equation. It satisfies the point [(-1, 1) = -1 + 1 = 0 , (0,0) and (1,-1) = 1 + (-1) = 0]

For the For Fig.(4.6)

y = -x + 2 is the correct linear equation. It satisfies the point

[(-1,3) 3 = -1(-1) + 2 = 3 = 3 , (0,2) ⇒ 2 = -(0) + 2 ⇒ 2 = 2 and (2, 0) ⟹ 0= 2+2]

**EX 4.3 QUESTION 6.**

**If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is**

**(i) 2 units**

**(ii) 0 unit**

**Solution:**

Constant force =5 units.

Let the distance travelled = x units

and work done = y units.

Work done = Force x Distance

⇒ y = 5 × x ⇒ y = 5x

When x = 0, then y = 5(0) = 0

x = 1, then y = 5(1) = 5

x = -1, then y = 5(-1) = -5

(i) Distance travelled =2 units means x = 2

∴ If x = 2, then y = 5(2) = 10

⇒ Work done = 10 units.

(ii) Distance travelled = 0 unit means x = 0

∴ If x = 0 ⇒ y = 5(0) – 0

⇒ Work done = 0 unit.

**EX 4.3 QUESTION 7.**

**Yamini and Fatima, two students of Class IX of a school, together contributed ₹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₹xand ₹y.) Draw the graph of the same.**

**Solution:**

Let Yamini contribute = ₹ x

and Fatima contribute = ₹ y

∴ x + y = 100 ⇒ y = 100 – x

when x = 0, y = 100 – 0 = 100

x = 50, y = 100 – 50 = 50

x = 100, y = 100 – 100 = 0

**EX 4.3 QUESTION 8.**

**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 = (**** )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.**

**Solution:**

(i) F = (9/5 )C + 32

When C = 0 , F = (9/5 ) x 0 + 32 = 32

When C = 15, F = (9/5)(-15) + 32= -27 + 32 = 5

When C = -10, F = (-10)+32 = -18 + 32 = 14

(ii) When C = 30,

F = (9/5)C +32

F = (9×30)/5+32

= (9×6)+32

= 54+32

= 86^{o}F

(iii) When F = 95,

95 = (9/5)C +32

(9/5)C = 95-32

(9/5)C =63

C = (63×5)/9

=35^{o}C

(iv) When C = 0,

F = (9/5)C +32

F = (9×0)/5 +32

=0+32

=32^{o}F

When F = 0,

0 = (9/5)C+32

(9/5)C = 0-32

(9/5)C = -32

C = (-32×5)/9

=17.7777

=17.8^{o}C

(v) When F = C,

C = (9/5)C+32

C – (9/5)C = 32

(5-9)C/5 =32

(-4/5)C = 32

(-4/5)C = (-32×5)/4

= – 40^{o}C