Sunday, December 16, 2018

Pair of Linear Equations in two Variables Exercise 3.5

Pair of Linear Equations in two Variables Exercise 3.5

 

Page No: 62

1. Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0  ; 3x – 9y – 2 =0
(ii) 2x + y = 5 ; 3x +2y =8
(iii) 3x – 5y = 20 ; 6x – 10y =40
(iv) x – 3y – 7 = 0 ; 3x – 3y – 15= 0

Answer

(i) x – 3y – 3 = 0
3x – 9y – 2 =0
a1/a2 = 1/3
b1/b2 = -3/-9 = 1/3 and
c1/c2 = -3/-2 = 3/2
a1/a2 = b1/b c1/c2

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5
3x +2y = 8
a1/a2 = 2/3
b1/b2 = 1/2 and
c1/c2 = -5/-8 = 5/8
a1/a2 ≠ b1/b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,
x/b1c2-b2c= y/c1a2-c2a= 1/a1b2-a2b1
x/-8-(-10) = y/-15+16 = 1/4-3
x/2 = y/1 = 1
x/2 = 1, y/1 = 1
∴ = 2, = 1.

(iii) 3x – 5y = 20
6x – 10y = 40
a1/a2 = 3/6 = 1/2
b1/b2 = -5/-10 = 1/2 and
c1/c2 = -20/-40 = 1/2
a1/a2 = b1/b= c1/c2

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
a1/a2 = 1/3
b1/b2 = -3/-3 = 1 and
c1/c2 = -7/-15 = 7/15
a1/a2 ≠ b1/b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

x/45-(21) = y/-21-(-15) = 1/-3-(-9)
x/24 = y/-6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
= 4 and = -1
∴ = 4, = -1.

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y =7

(a – b)x + (a + b)y = 3a +b –2

Answer

2x + 3y -7 = 0

(a – b)x + (a + b)y - (3a +b –2) = 0

a1/a2 = 2/a-b = 1/2
b1/b2 = -7/a+b and
c1/c2 = -7/-(3a+b-2) = 7/(3a+b-2)
For infinitely many solutions,a1/a2 = b1/b2 = c1/c2

2/a-= 7/3a+b-26a + 2b - 4 = 7a - 7b
a - 9b = -4 ... (i)

2/a-= 3/a+b
2a + 2b = 3a - 3b
a - 5b = 0 ... (ii)

Subtracting equation (i) from (ii), we get
4b = 4
b = 1
Putting this value in equation (ii), we get
a - 5 × 1 = 0
a = 5
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

(ii) For which value of k will the following pair of linear equations have no solution?

3x + = 1

(2k –1)x + (k –1)y = 2k + 1    

Answer

3x + -1 = 0

(2k –1)x + (k –1)y - (2k + 1) = 0

a1/a2 = 3/2k-1
b1/b2 = 1/k-1 and
c1/c2 = -1/-2k-1 = 1/2k+1
For no solutions,
a1/a2 = b1/b c1/c2
3/2k-1 = 1/k-1 ≠ 1/2k+1

3/2k-1 = 1/k-1
3k - 3 = 2k - 1
k = 2
Hence, for k = 2, the given equation has no solution.

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x +5y = 9
3x +2y = 4

Answer

8x +5y = 9 ... (i)
3x +2y = 4 ... (ii)
From equation (ii), we get
x = 4-2y/3 ... (iii)
Putting this value in equation (i), we get
8(4-2y/3) + 5y = 9
32 - 16y +15y = 27
-y = -5
y = 5 ... (iv)
Putting this value in equation (ii), we get
3x + 10 = 4
x = -2
Hence, x = -2, = 5
By cross multiplication again, we get

8x + 5y -9 = 0

3x + 2y - 4 = 0

x/-20-(-18) = y/-27-(-32) = 1/16-15
x/-2 = y/5 = 1/1
x/-2 = 1 and y/5 = 1
x = -2 and y = 5

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Answer

Let be the fixed charge of the food and be the charge for food per day.

According to the question,

x + 20y = 1000 ... (i)

x + 26y = 1180 ... (ii)
Subtracting equation (i) from equation (ii), we get

6y = 180

y = 180/6 = 30
Putting this value in equation (i), we get
x + 20 × 30 = 1000
x = 1000 - 600
x = 400
Hence, fixed charge = Rs 400 and charge per day = Rs 30

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

Answer

Let the fraction be x/y
According to the question,

x-1/y = 1/3
⇒ 3x - y = 3... (i)
x/y+8 = 1/4
⇒ 4x - y = 8 ... (ii)
Subtracting equation (i) from equation (ii), we get

x = 5 ... (iii)

Putting this value in equation (i), we get

15 - = 3

= 12
Hence, the fraction is 5/12.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Answer

Let the number of right answers and wrong answers be and respectively.

According to the question,

3x - = 40 ... (i)

4x - 2y = 50

⇒ 2x - = 25 ... (ii)

Subtracting equation (ii) from equation (i), we get
= 15 ... (iii)
Putting this value in equation (ii), we get

30 - = 25

= 5

Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Answer

Let the speed of 1st car and 2nd car be u km/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (u - v) km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (v) km/h

According to the question,

5(v) = 100

⇒ u - v = 20 ... (i)

1(u + v) = 100 ... (ii)

Adding both the equations, we get

2u = 120

u = 60 km/h ... (iii)

Putting this value in equation (ii), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer

Let length and breadth of rectangle be unit and y unit respectively.
Area = xy
According to the question,
(x - 5) (y + 3) = xy - 9
⇒ 3x - 5y - 6 = 0 ... (i)
(x + 3) (y + 2) = xy + 67
⇒ 2x - 3y - 61 = 0 ... (ii)
By cross multiplication, we get
x/305-(-18) = y/-12-(-183) = 1/9-(-10)
x/323 = y/171 = 1/19
= 17, y = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

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