Cross – multiplication Method of finding solution of a pair of Linear Equations:
For the pair of linear equations
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0,
x and y can be calculated as
EXERCISE 3.5
1.Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions. In case there is unique solution, find it by using cross – multiplication method.
( i ) x – 3 y – 3 = 0, 3 x – 9 y – 2 = 0
( ii ) 2 x + y = 5, 3 x + 2 y = 8
( iii ) 3 x – 5 y = 20, 6 x – 10 y = 40
( iv ) x – 3 y – 7 = 0, 3 x – 3 y – 15 = 0
Solution : ( i ) Given, x – 3 y – 3 = 0 and 3 x – 9 y – 2 = 0
a1 / a2 = 1 / 3 , b1 / b2 = – 3 /– 9 = 1 / 3, c1 / c2 = – 3 / – 2 = 3 / 2
( a1 / a2 ) = (b1 / b2 ) ≠ ( c1 / c2 )
Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.
( ii ) Solution : Given, 2 x + y = 5 and 3 x + 2 y = 8
a1 / a2 = 2 / 3 , b1 / b2 = 1 / 2 , c1 / c2 = – 5 / – 8
( a1 / a2) ≠ ( b1 / b2 )
Since they intersect at a unique point these equations will have a unique solution
By cross multiplication method:
x / ( b1 c2 – c1 b2 ) = y / ( c1 a2 – c2 a1 ) = 1 / ( a1b 2 – a2 b1 )
x / ( – 8 – (–10 ) ) = y / ( 15 + 16 ) = 1 / ( 4 – 3 )
x / 2 = y / 1 = 1
∴ x = 2 and y = 1
( iii ) Solution : Given, 3 x – 5 y = 20 and 6 x – 10 y = 40
( a1 / a2 ) = 3 / 6 = 1 / 2
( b1 / b2 ) = – 5 /– 10 = 1 / 2
( c1 / c2 ) = 20 / 40 = 1 / 2
a1 / a2 = b1 / b2 = c1 / c2
Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.
( iv ) Solution : Given, x – 3 y – 7 = 0 and 3 x – 3 y – 15 = 0
( a1 / a2 ) = 1 / 3
( b1 / b2 ) = − 3 / – 3 = 1
( c1 / c2 ) = – 7 / – 15
a1 / a2 ≠ b1 / b2
Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.
By cross multiplication method:
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.
2.( i ) For which values of a and b does the following pair of linear equations have an infinite number of solutions.
2 x + 3 y = 7
( a – b ) x + ( a + b ) y = 3 a + 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
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.
Solution : Let fixed monthly charge = Rs x and let charge of food for one day = Rs y
According to given conditions,
x + 20 y = 1000 … ( i ),
and x + 26 y = 1180 … ( ii )
Subtracting equation ( i ) from equation ( ii ), we get
6 y = 180 ⇒ y = 30
Putting value of y in ( i ), we get
x + 20 ( 30 ) = 1000
⇒ x = 1000 – 600 = 400
Therefore, fixed monthly charges = Rs 400 and, charges of food for one 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.
Solution : Let numerator = x and let denominator = y
According to given conditions
(x − 1)/y = 1/3 and x / ( y + 8) = 1/4
⇒ 3 x – 3 = y … ( i )
⇒ 4 x = y + 8 …( ii )
Subtracting equation ( i ) from ( ii ), we get
4 x – y −( 3 x − y ) = 8 – 3
⇒ x = 5
Putting value of x in ( i ), we get
3 ( 5 ) – y = 3 ⇒ 15 – y = 3
⇒ y = 12
∴ numerator = 5 and, denominator = 12
The fraction = 5 /12 .
( iii ) Yash scored 40 marks in a test, getting 3 marks for each right answers and losing 1 mark for each wrong answers. Had 4 marks been awarded for each correct answers and 2 marks been deducted for each incorrect answers, then Yash would have scored 50 marks. How many questions were there in the test.
Solution : Let number of correct answers = x
and let number of wrong answers = y
According to given conditions,
3 x – y = 40 … ( i )
And, 4 x − 2 y = 50 … ( ii )
From equation ( i ), y = 3 x − 40
Putting this in ( ii ), we get
4 x – 2 ( 3 x − 40 ) = 50 ⇒ 4 x − 6 x + 80 = 50
⇒ − 2 x = −30 ⇒ x = 15
Putting value of x in ( i ), we get
3 ( 15 ) – y = 40 ⇒ 45 – y = 40 ⇒ y = 45 – 40 ⇒ 5
Therefore, number of correct answers = x = 15 and number of wrong solution = y = 5
Total questions = x + y = 15 + 5 = 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.
Solution: Let speed of car which starts from part A = x km / hr
Let speed of car which starts from part B = y km / hr
According to given conditions,
( 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 : Let length of rectangle = x units and let breadth of rectangle = y units
Area = x y square units. According to given conditions,
x y – 9 = ( x − 5 ) ( y + 3 )
⇒ x y – 9 = x y + 3 x − 5 y – 15 ⇒ 3 x − 5 y = 6 … ( 1 )
And, x y + 67 = ( x + 3 ) ( y + 2 ) ⇒ x y + 67 = x y + 2 x + 3 y + 6
⇒ 2 x + 3 y = 61 … ( 2 )
From equation ( i ), 3 x = 6 + 5 y
⇒ x = (6 + 5 y) /3
Putting this in ( ii ), we get
2 + 3 y = 61 ⇒ 12 + 10 y + 9 y = 183
⇒ 19 y = 171 ⇒ y = 9 units
Putting value of y in ( ii ), we get
2 x + 3 ( 9 ) = 61 ⇒ 2 x = 61 – 27 = 34
⇒ x = 17 units
∴ length = 17 units and breadth = 9 units
Additional Questions:
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.
( a ) 3 x – 5 y = 20 : 3 x + 2 y = 40
( b ) 3 x – 7 y – 21 = 0: 6 x – 9 y –18 = 0
2.For which values of a and b does the following pair of linear equations have an infinite number of solutions?
3 x + 5 y = 10
( a – b ) x + ( a + b ) y = 3 a + b – 2
3. Solve the following pair of linear equations by the substitution and cross – multiplication methods :
6 x + 8 y = 12
4 x + 5 y = 16.