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
Solution:
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0
a1/a2 = 1/3 ; b1/b2 = -3/-9 = 1/3 ; c1/c2 = -3/-2 = 3/2
a1/a2 = b1/b2 ≠ 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 ; c1/c2 = -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-b2c1) = y/(c1a2-c2a1) = 1/(a1b2-a2b1)
x/[-8-(-10)] = y/[-15+16] = 1/(4-3)
x/2 = y/1 = 1
x/2 = 1 ; y/1 = 1
x = 2 ; y = 1
Therefore, x = 2, y = 1
(iii) 3x – 5y = 20
6x – 10y = 40
a1/a2 = 3/6 = ½ ; b1/b2 = -5/-10 = 1/2 ; c1/c2 = -20/-40 = 1/2
a1/a2 = b1/b2 = 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 ; 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
x = 4 and y = -1
Thus, x = 4 and y = -1
2.
(i) For which values of a and b will the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a –b)x + (a + b)y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x+ y = 1
(2k-1)x + (k-1)y = 2k +1
Solution:
(i) 2x + 3y -7 = 0
(a-b)x + (a+b)y – (3x +b -2) = 0
a1/a2 = 2/(a-b) ; b1/b2 = 3/(a+b) ; c1/c2 = -7/-(3a+b-2) = 7/(3a+b-2) ;
For infinitely many solutions,
a1/a2 = b1/b2 = c1/c2
2/(a-b) = 7/(3a+b-2)
6a+2b-4=7a-7b
a-9b = -4 ————(1)
2/(a-b) = 3/(a+b)
2a + 2b = 3a – 3b
a – 5b = 0 ————(2)
Subtracting (1) from (2), we obtain
4b = 4
b = 1
Substituting this in equation(2), we obtain
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) 3x + y – 1 = 0
(2k-1)x +(k-1)y – 2k -1 = 0
a1/a2 = 3/(2k-1) ; b1/b2 = 1/(k-1) ; c1/c2 = -1/-2k-1 = 1/2k+1 ;
For no solution;
a1/a2 = b1/b2 ≠ 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
- Solve the following pair of linear equations by the substitution and cross multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
8x + 5y = 9 ———-(1)
3x + 2y = 4 ———–(2)
From equation (ii), we get
x = (4-2y)/3 ————(3)
Substituting this value in equation (1), we get
8([4-2y)/3) + 5y = 9
32 – 16y + 15y = 27
-y = -5
y = 5
Substituting the value of y in equation (2)
3x + 10 = 4
x = -2
Hence x = -2 , y = 5
Again by cross multiplication method, 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
- 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.
(ii). A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes ¼ when 8 is added to its denominator. Find the fraction.
(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?
(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?
(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.
Solution:
(i) Let x be the fixed charge of the food and y be the charge for food per day.
Given
x + 20y = 1000 ————-(1)
x + 26 y = 1180 ———–(2)
Subtracting equation (1) from equation (2), we get
6y = 180
y = 30
Substituting this value in equation (1),
x + 20 x 30 = 1000
x = 1000 – 600
x = 400
Hence, fixed charge = Rs 400
And charge per day = Rs 30
(ii)Let the fraction be x/y.
Given
x-1/y = 1/3 i.e., 3x – y = 3 ———–(1)
x/(y+8) = ¼ i.e., 4x – y = 8 ————(2)
Subtracting equation (1) from equation (2),
x = 5 ———(3)
Putting this value in equation (1), we obtain
15 – y = 3
y = 12
Hence the fraction is 5/12.
(iii)Let the number of right answers and wrong answers be x and y respectively.
Given,
3x – y = 40 ————-(1)
4x – 2y = 50
i.e., 2x – y = 25 ————-(2)
Subtracting equation (2) from equation (1),
x = 15 ———–(3)
Substituting the value of x in equation (2),
30 – y = 25
y = 5
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20
(iv)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 = (u + v) km/h
Given
5 ( u – v) = 100
i.e., u – v = 20 ————-(1)
1(u+v) = 100 ————-(2)
Adding both the equations,
2u = 120
u = 60 km/h ———(3)
Substituting the value of u in equation (2),
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h
(v) Let length and breadth of rectangle be x unit and y unit respectively.
Area = xy
Given,
(x – 5)(y+3) = xy – 9
i.e., 3x – 5y – 6 = 0 ———-(1)
(x + 3)(y + 2) = xy + 67
i.e., 2x + 3y – 61 = 0 ———-(2)
By cross-multiplication method, we obtain
x/[305-(-18)] = y/[-12-(-183)] = 1/[9-(10)]
x/323 = y/171 = 1/19
x = 17, y = 9
Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.
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