Linear Equation in Two Variables · 10/9/2009 · (b)Linear Equations in Two Variables: - An...

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Linear Equation in Two Variables

• Linear Equations• Solution Of A Linear Equation• Graph Of A Linear Equation In Two Variables• Equations Of Lines Parallel To X-axis And Y- axis

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(a) Linear Equation in One Variable: - An equation like ax + b ≠ 0, i.e. 2x + 1 = 0, is having one variable x. This is known as linear equation in one variable. Such equations have a unique solution.

The solution of a linear equation is not affected: -

(i) When same number is added (or subtracted) from both sides of the equation.

(ii) When we multiply (or divide) both sides of the equation by the same non-zero number.

(b) Linear Equations in Two Variables: -An equation like ax + by + c = 0 {where a, b and c are real numbers and at least one of a, b is non-zero}, i.e. 2x + 3y + 1 = 5.This equation is having two variables x and y, that is why such equations are known as linear equations in two variables.




EXERCISE 1

Q1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent the above statement.

Sol. Let the cost of a notebook be x and that of a pen be y.According to question,The cost of a notebook is twice the cost of a pen∴ x = 2y⇒ x - 2y = 0.

Q2. 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 - 5y

- 10 = 0 (iii) -2x + 3y = 6 (iv) x = 3y

(v) 2x = -5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x

Sol. (i) 2x + 3y = 9.35 ⇒ 2x + 3y - 9.35 = 0

Comparing the equation with ax + by + c = 0, we have

a = 2, b = 3, c = 9.35

(ii) x - 5y

- 10 = 0


a = 1, b = -51

, c = -10

(iii) -2x + 3y = 6 ⇒ -2x + 3y – 6 = 0


a = -2, b = 3, c = -6

(iv) x = 3y ⇒ x - 3y = 0


a = 1, b = -3, c = 0

(v) 2x = -5y ⇒ 2 x + 5y = 0





a = 2, b = 5, c = 0

(vi) 3x + 2 = 0 ⇒ 3 x + 0y + 2 = 0


a = 3, b = 0, c = 2

(vii) y – 2 = 0 ⇒ 0 x + 1y - 2 = 0


a = 0, b = 1, c = -2

(viii) 5 = 2x ⇒ -2x - 0y + 5 = 0


a = -2, b = 0, c = 5




EXERCISE 2

Q1. Which one of the following statements is true and why ?

y = 3x + 5 has

(i) a unique solution(ii) only two solutions(iii) infinitely many solutions.

Sol. Since y = 3x + 5 is a linear equation in two variables. We know that a linear equation in two variables has infinite solutions.

Hence option (iii) is true.

Q2. Write four solutions for each of the following equations: (i) 2x + y = 7 (ii) π x + y = 9 (iii) x = 4y.

Sol. (i) 2x + y = 7⇒ y = 7 – 2x …………equation (1)

When x = 0, Putting the value in equation (1)y = 7 – 2 × 0, ⇒ y = 7 – 0, ⇒ y = 7.

Putting the value x = 1 in equation (1)y = 7 – 2 × 1, ⇒ y = 7 – 2, ⇒ y = 5.



Hence, four solutions for equation 2x + y = 7 are (0,7), (1,5), (2,3), (3,1).

(ii) π x + y = 9 ⇒ y = 9 – π x …………equation (1)

When x = 0, Putting the value in equation (1)y = 9 – π × 0, ⇒ y = 9 – 0, ⇒ y = 9.

Putting the value x = 1 in equation (1)y = 9 – π × 1, ⇒ y = 9 –π .

Putting the value x = -1 in equation (1)y = 9 – π × -1, ⇒ y = 9 +π . {since -1 × -1 = +1}

Putting the value x = 2 in equation (1)




y = 9 – π × 2, ⇒ y = 9 – 2π .

Hence, four solutions for equation π x + y = 9 are (0, 9), (1, 9 –π ), (-1, 9 +π ), (2, 9 – 2π ).

(iii) x = 4y …………equation (1)

When y = 0, Putting the value in equation (1)x = 4 × 0, ⇒ x = 0.

Putting the value y = 1 in equation (1)x = 4 × 1, ⇒ y = 4.

Putting the value y = -1 in equation (1)x = 4 × -1, ⇒ y = -4.

Putting the value y = 2 in equation (1)

x = 4 × 2, ⇒ y = 8.

Hence, four solutions for equation x = 4y are (0, 0), (1, 4), (-4,-1), (8, 2).

Q3. 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)

Sol. (i) The given equation is x – 2y = 4L.H.S = x – 2y

Putting the value x = 0 and y = 2 to verify the solution (0, 2)

= 0 – 2 × 2 = 0 – 4 = -4 ≠ R.H.S.

∴ (0, 2) is not a solution of the equation x – 2y = 4.

(ii) The given equation is x – 2y = 4 L.H.S = x – 2y


= 2 – 2 × 0 = 2 – 0 = 2 ≠ R.H.S.


(iii) The given equation is x – 2y = 4 L.H.S = x – 2y


= 4 – 2 × 0 = 4 – 0 = 4 = R.H.S.

∴ (4, 0) is a solution of the equation x – 2y = 4.




(iv) The given equation is x – 2y = 4 L.H.S = x – 2y

Putting the value x = 0 and y = 2 to verify the solution ( 2 , 4 2 )

= 2 – 2 × 4 2 = 2 – 8 2 = 7 2 ≠ R.H.S.

∴ ( 2 , 4 2 ) is not a solution of the equation x – 2y = 4.

(v) The given equation is x – 2y = 4 L.H.S = x – 2y


= 1 – 2 × 1 = 1 – 2 = -1 ≠ R.H.S.


Q4. Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Sol. The given equation is 2x + 3y = k Putting the given value x = 2 and y = 1 in the equation

⇒ 2 × 2 + 3 × 1 = k

⇒ 4 + 3 = k

⇒ 7 = k

Hence, k = 7

EXERCISE 3




Q1. Draw the graph of each of the following linear equation in two variables:

(i) x + y = 4 (ii) x – y = 2 (iii) y = 3x (iv) 3 = 2x + y

Sol. (i) The given equation isx + y = 4

⇒ y = 4 – x …………….equation (1)

Now , putting the value x = 0 in equation (1)

⇒ y = 4 – 0 = 4. So the solution is (0, 4)





So, the table of the different solutions of the equation is

0 1 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

(ii) The given equation isx - y = 2

⇒ x = 2 + y …………….equation (1)


x 0 1 2y 4 3 2



Now , putting the value y = 0 in equation (1)

⇒ x = 2 + 0 = 2. So the solution is (2, 0)






2 3 40

0.5

1

1.5

2

2.5

x-y=2

y

x

y

(iii) The given equation is y = 3x

⇒ y = 3x …………….equation (1)


⇒ y = 3 × 0 = 0. So the solution is (0, 0)




x 2 3 4y 0 1 2






0 1 20

1

2

3

4

5

6

7

y=3x

y

x

y

(iii) The given equation is3 = 2 x + y

⇒ 2 x + y = 3

⇒ y = 3 – 2x …………….equation (1)

Now, putting the value x = 0 in equation (1)

⇒ y = 3 – 2 × 0



⇒ y = 3 – 2 × 1



x 0 1 2y 0 3 6




⇒ y = 3 – 2 × 2

⇒ y = 3 – 4 = -1. So the solution is (2, -1)


0 1 2-1.5-1

-0.50

0.51

1.52

2.53

3.5

3=2x+y

y

x

y

Q2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Sol. Since the given solution is (2, 14)

Therefore, x = 2 and y = 14

∴ One equation is x + y = 2 + 14 = 16

⇒ x + y = 16

Second equation is x – y = 2 – 14 = -12

⇒ x - y = -12

Third equation is y = 7x


x 0 1 2y 3 1 -1



⇒ 0 = 7x – y

⇒ 7x - y = 0In fact we can find infinite equations because through one point infinite lines pass.

Q3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a?

Sol. The given equation is 3y = ax + 7 ……………….equation (1)

According to problem, point (3, 4) lie on it.So, putting the value x = 3 and y = 4 in equation (1)

⇒ 3 × 4 = a × 3 + 7

⇒ 12 = 3a + 7

⇒ 12 – 7 = 3a

⇒ 5 = 3a

⇒ 35

= a

⇒ a = 35

Now putting the value of a in equation (1)

3y = 35

x + 7

Now for x = 3 and y = 4

L.H.S = 3y = 3 × 4 = 12

R.H.S = 35

x + 7 = 35

× 3 + 7 = 5 + 7 = 12

Hence, L.H.S = R.H.S

Or, 3y = 35

x + 7.

Q4. The taxi fare in a city is as follows: For the first kilometer, the fare is Rs. 8 and for the subsequent distance it is Rs. 5 per kilometer. Taking the distance covered as x km and total fares as Rs. y, write a linear equation for this information, and draw its graph.




Sol. Given,Taxi fare for first kilometer = Rs. 8Taxi fare for subsequent distance = Rs. 5

Total distance covered = x Total fare = y

Since the fare for first kilometer = Rs. 8

According to problem,

Fare for (x – 1) kilometer = 5(x-1)

So, the total fare = 5(x-1) + 8

∴ y = 5(x-1) + 8

⇒ y = 5x – 5 + 8

⇒ y = 5x + 3

Hence, y = 5x + 3 is the required linear equation.

Now the equation is

y = 5x + 3 …………….equation (1)

Now, putting the value x = 0 in equation (1)

⇒ y = 5× 0 + 3

⇒ y = 0 + 3 = 3 So the solution is (0, 3)


⇒ y = 5× 1 + 3

⇒ y = 5 + 3 = 8. So the solution is (1, 8)


⇒ y = 5× 2 + 3

⇒ y = 10 + 3 = 13. So the solution is (2, 13)





0 1 20

2

4

6

8

10

12

14

y=5x+3

y

x

y

Q5. From the choices given below, choose the equation whose graphs are given in Fig. 1 and Fig. 2.

(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


x 0 1 2y 3 8 13



-1 0 1-1.5

-1

-0.5

0

0.5

1

1.5

y

x

y

-1 0 20

0.5

1

1.5

2

2.5

3

3.5

y

x

y

Sol. From the given Fig. 1, the solutions of the equation are (-1, 1), (0, 0) and (1, -1)

Therefore the equation which satisfies these solutions is the correct equation.

Equation (ii) x + y = 0 , satisfies these solutions.

Proof:

Putting the value x = -1 and y = 1 in the equation x + y = 0




L.H.S = x + y = -1 + 1 = 0 = R.H.S

Putting the value x = 0 and y = 0

L.H.S = x + y = 0 + 0 = 0 = R.H.S

Putting the value x = 1 and y = -1

L.H.S = x + y = 1 + (-1) = 1 – 1 = 0 = R.H.S

Hence, option (ii) x + y = 0 is correct.

From the given Fig. 2, the solutions of the equation are (-1, 3), (0, 2) and (2, 0)

Therefore the equation which satisfies these solutions is the correct equation.

Equation (i) y = -x + 2 , satisfies these solutions.

Proof:

Putting the value x = -1 and y = 3 in the equation y = -x + 2 ⇒ x + y = 2

L.H.S = x + y = -1 + 3 = 2 = R.H.S


L.H.S = x + y = 0 + 2 = 2 = R.H.S


L.H.S = x + y = 2 + 0 = 2 = R.H.S

Hence, option (i) y = -x + 2 is correct.

Q6. 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 is 5 units. Read from the graph the work done when the distance traveled by the body is

(i) 2 units (ii) 0 units

Sol. Let x units be the distance traveled by the body and y units be the work done by constant force.

According to problem, y = 5x …………….equation (1)











0 1 20

2

4

6

8

10

12

y=5x

y

x

y

(i) When x = 2 units (distance)

Putting the value x in equation (1)

y = 5x

⇒ y = 5 × 2 = 10.

Hence, Work done = 10.

(ii) When x = 0 units (distance)

Putting the value x in equation (1)


x 0 1 2y 0 5 10



y = 5x

⇒ y = 5 × 0 = 0.

Hence, Work done = 0.

Q7. 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 this data satisfies. (You may take their contributions as Rs. x and Rs. y). Draw the graph of the same.

Sol. Let the contribution of Yamini be x and that of Fatima be y.

According to problem, x + y = 100 …………..(1)


⇒ 0 + y = 100

⇒ y = 100 . So the solution is (0, 100)


⇒ 50 + y = 100, ⇒ y = 100 – 50



⇒ 100 + y = 100, ⇒ y = 100 – 100




x 0 50 100y 100 50 0



0 50 1000

20

40

60

80

100

120

x+y=100

y

x

y

Q8. In countries like the 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 = (59

) 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.

Sol. Given equation is F = (59

) C + 32

(i) We have to take Celsius along x-axis and Fahrenheit along y-axis

Let C be x and F be ySo the equation will be

y = (59

) x + 32 ………………equation (1)





⇒ y = (59

) x 0 + 32 = 0 + 32 = 32. So the solution is (0, 32)


⇒ y = (59

) x 5 + 32 = 9 + 32 = 41. So the solution is (5, 41)

Putting the value x = -5 in equation (1)

⇒ y = (59

) x (-5) + 32 = -9 + 32 = 23. So the solution is (-5,

23)


0 5 -50

5

10

15

20

25

30

35

40

45

y=(9/5)x+32

y

x

y

(ii) When C = 30°,

F = 59

× 30 + 32 = 9 × 6 + 32 = 54 + 32 = 86°

(iii) When F = 95°,


x 0 5 -5y 32 41 23



95 = 59

× C + 32

⇒ 95 – 32 = 59

C ⇒ 63 = 59

C

⇒ 59

C = 63 ⇒ C = 9563×

= 7 × 5 = 35°.

(iv) When C = 0,

F = 59

× 0 + 32 = 0 + 32 = 32°.

(v) When x° F = x° C

F = (59

) C + 32

⇒ x = (59

) x + 32

⇒ x - 59

x = 32 ⇒ 595 xx −

= 32

⇒ 54x−

= 32 ⇒ - 4x = 32 × 5

⇒ x = 4532

−×

= -8 × 5 = - 40°C.




EXERCISE 4

Q1. Give the geometric representations of y = 3 as an equation.

(i) in one variable (ii) in two variables.

Sol. (i) y = 3

(ii) y = 3 ⇒ 0x + y = 3 ⇒ 0x + y – 3 = 0 which is in fact y = 3

It is a line parallel to x-axis at a positive distance of 3 from it. We have two solution for it. i.e. (0, 3), (1, 3).

(0, 3) (1, 3)

Q2. Give the geometric representation of 2x + 9 = 0 as and equation,


0 1 2 3

A B3

2

1

1 2 3-1-2-1

-2

XX’

Y

Y’



(i) in one variable (ii) in two variables.

Sol. (i) 2x + 9 = 0

⇒ 2x = -9 ⇒ x = - 29

= - 4.5

(iii) Given equation is

2x + 9 = 0

⇒ 2x + 0y + 9 = 0 {we know that it is actually 2x + 9 = 0}

⇒ x = - 29

= - 4.5

It is line parallel to y-axis at a negative distance we have the two points lying it, the points are A(-4.5, 0), B(-4.5, 2).

(-4.5, 2)

(-4.5, 0)


0 21 3 X-1-2-3-4-5X’

-4.5

A

B

X

-2

-1

0

1

2

Y

Y’

-1-2-3-4-5X’


Linear Equation in Two Variables · 10/9/2009 · (b)Linear Equations in Two Variables: - An...

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