Equations reducible to a pair of Linear Equations in 2 variables:
Some equations may be in a form which can be reduced to a linear equation through substitution.
2 / x + 3 / y = 4
5 / x − 4 / y = 9
In this case, we may make the substitution
1 / x = u 1 / y = v
The pair of equations reduces to
2 u + 3 v = 4
5 u – 4 v = 9
The above pair of equations will be solved first.
After solving, substitute again to get the values of x and y.
EXERCISE 3.6
1.Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/2 x + 1/3 y = 2
and 1/3 x + 1/2 y = 13/6
Solution: Let us assume 1 / x = m and 1 / y = n , then the equation will change as follows.
m / 2 + n / 3 = 2
⇒ 3 m + 2 n – 12 = 0…………………….( 1 )
m / 3 + n / 2 = 13 / 6
⇒ 2 m + 3 n – 13 = 0……………………….( 2 )
Now, using cross multiplication method, we get,
m / [– 26 –(–36 ) ] = n / [– 24 – (– 39 ) ] = 1 / ( 9 – 4 )
m / 10 = n / 15 = 1 / 5
m / 10 = 1 / 5 and n / 15 = 1 / 5
So, m = 2 and n = 3 ( But 1 /x = m and 1/y = n)
1 / x = 2 and 1 / y = 3
x = 1 / 2 and y = 1 / 3
( ii ) 2 / √x + 3 / √y = 2
4 / √x + 9 / √y = – 1
Solution: Substituting 1 / √x = m and 1 / √y = n in the given equations, we get
2 m + 3 n = 2 …………… ( i )
4 m – 9 n = – 1 ……………… ( ii )
Multiplying equation ( i ) by 3, we get
6 m + 9 n = 6 ………………….. ( iii )
Adding equation ( ii ) and ( iii ), we get
10 m = 5
m = 1 / 2 …………………… ( iv )
Putting the value of ‘m’ in equation ( i ), we get
2 × 1 / 2 + 3 n = 2
3 n = 1 ⇒ n = 1 / 3
m = 1 / √x
½ = 1 / √x ⇒ x = 4
n = 1 / √y ⇒ 1 / 3 = 1 / √y ⇒ y = 9
Hence, x = 4 and y = 9
( iii ) 4 / x + 3 y = 14
3 / x – 4 y = 23
Solution: Putting in the given equation we get,
So, 4 m + 3 y = 14 ⇒ 4 m + 3 y – 14 = 0 ……………..…..( 1 )
3 m – 4 y = 23 ⇒ 3 m – 4 y – 23 = 0 ……………………….( 2 )
By cross multiplication method, we get,
m / ( – 69 – 56 ) = y / ( – 42 – ( – 92 ) ) = 1 / ( – 16 – 9 )
– m / 125 = y / 50 = – 1 / 25
– m / 125 = – 1 / 25 and y / 50 = – 1 / 25
m = 5 and b = – 2
m = 1 / x = 5
∴ x = 1 / 5 and y =– 2
( iv ) 5 / ( x – 1 ) + 1 / ( y – 2 ) = 2
6 / ( x – 1 ) – 3 / ( y – 2 ) = 1
Solution: Substituting 1 / ( x – 1 ) = m and 1 / ( y – 2 ) = n in the given equations, we get,
5 m + n = 2 …………………………( i )
6 m – 3 n = 1 ……………………….( ii )
Multiplying equation ( i ) by 3, we get
15 m + 3 n = 6 …………………….( iii )
Adding ( ii ) and ( iii ) we get
21 m = 7 ⇒ m = 1 / 3
Putting this value in equation ( i ), we get
5 × 1 / 3 + n = 2
n = 2 – 5 / 3 = 1 / 3 and m = 1 / ( x – 1 )
⇒ 1 / 3 = 1 / ( x – 1 ) ⇒ x = 4
n = 1 / ( y – 2 ) ⇒ 1 / 3 = 1 / ( y – 2 ) ⇒ y = 5
Hence, x = 4 and y = 5
( v ) ( 7 x – 2 y ) / x y = 5
( 8 x + 7 y ) / x y = 15
Solution: ( 7 x – 2 y ) / x y = 5
7 / y – 2 / x = 5…………………………..( i )
( 8 x + 7 y ) / x y = 15
8 / y + 7 / x = 15…………………………( ii )
Substituting 1 / x = m in the given equation we get,
– 2 m + 7 n = 5 ⇒ – 2 + 7 n – 5 = 0 …….( iii )
7 m + 8 n = 15 ⇒ 7 m + 8 n – 15 = 0 ……( iv )
By cross multiplication method, we get,
m / (–105 – ( – 40 ) ) = n / (– 35 – 30 ) = 1 / (– 16 – 49 )
m / (–65 ) = n / (– 65 ) = 1 / (– 65 )
m / 65 = 1 / – 65 ⇒ m = 1
n / (– 65 ) = 1 / (– 65 ) ⇒ n = 1
m = 1 and n = 1
But m = 1 / x = 1 and n = 1 / x = 1
Therefore, x = 1 and y = 1
( vi ) 6 x + 3 y = 6 x y
2 x + 4 y = 5 x y
Solution: x + 3 y = 6 x y
6 / y + 3 / x = 6
Let 1 / x = m and 1 / y = n
⇒ 6 n + 3 m = 6
⇒3 m + 6 n – 6 = 0………………( i )
2 x + 4 y = 5 x y
⇒ 2 / y + 4 / x = 5
⇒ 2 n + 4 m = 5
⇒ 4 m + 2 n – 5 = 0 ……………( ii )
3 m + 6 n – 6 = 0
4 m + 2 n – 5 = 0
By cross multiplication method,we get
m / (– 30 – ( –12 ) ) = n / ( – 24 – ( –15 ) ) = 1 / ( 6 – 24 )
m / – 18 = n / – 9 = 1 / − 18
m / – 18 = 1 / – 18 ⇒ m = 1
n / – 9 = 1 / – 18 ⇒ n = 1 / 2
m = 1 and n = 1 / 2
m = 1 / x = 1 and n = 1 /y = 1 / 2
x = 1 and y = 2
Hence, x = 1 and y = 2
( vii ) 10 / ( x + y ) + 2 / ( x – y ) = 4
15 / ( x + y ) – 5 / ( x – y ) = – 2
Solution: Substituting 1 / x + y = m and 1 / x – y = n in the given equations, we get,
10 m + 2 n = 4 ⇒ 10 m + 2 n – 4 = 0 ………………..…..( i )
15 m – 5 n = – 2 ⇒ 15 m – 5 n + 2 = 0
m / ( 4 – 20 ) = n / [– 60 –( 20 ) ] = 1 / (– 50 −30 )
m /– 16 = n / – 80 = 1 / – 80
m /– 16 = 1 / – 80 and n /– 80 = 1 / –80
m = 1 / 5 and n = 1
m = 1 / ( x + y ) = 1 / 5
x + y = 5 …………………………………………( iii )
n = 1 / ( x – y ) = 1
x – y = 1……………………………………………( iv )
Adding equation ( iii ) and ( iv ), we get
2 x = 6 ⇒ x = 3 …….( v )
Putting the value of x = 3 in equation ( 3 ), we get
y = 2
Hence, x = 3 and y = 2
( viii ) 1 / ( 3 x + y ) + 1 / ( 3 x – y ) = 3 / 4
1 / 2 ( 3 x + y ) – 1 / 2 ( 3 x – y ) = – 1 / 8
Solution: Substituting 1 / ( 3 x + y ) = m and 1 / ( 3 x – y ) = n in the given equations, we get,
m + n = 3 / 4 …………………………….…… ( 1 )
m / 2 – n / 2 = – 1 / 8
m – n = – 1 / 4 …………………………..…( 2 )
Adding ( 1 ) and ( 2 ), we get
2 m = 3 / 4 – 1 / 4 ⇒ 2 m = 1 / 2
Putting in ( 2 ), we get
1 / 4 – n = – 1 / 4
n = 1 / 4 + 1 / 4 = 1 / 2
m = 1 / ( 3 x + y ) = 1 / 4
3 x + y = 4 …………………………………( 3 )
n = 1 / ( 3 x – y ) = 1 / 2
3 x – y = 2 ………………………………( 4 )
Adding equations ( 3 ) and ( 4 ), we get
6 x = 6
x = 1
Putting in ( 3 ), we get
3 ( 1 ) + y = 4
y = 1
Hence, x = 1 and y = 1
Formulate these problems as a part of equations, and hence find their solutions.
( i ) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Solution:
Let speed of rowing in still water = x km / h
Let speed of current = y km / h
So, speed of rowing downstream = ( x + y ) km / h
And, speed of rowing upstream = ( x − y ) km / h
According to given conditions,
x + y = 20 / 2 and 4 / x − y = 2
⇒ 2 x + 2 y = 20 and 2 x − 2 y = 4
⇒ x + y = 10 …….… ( 1 )
and x – y = 2 ………… ( 2 )
Adding ( 1 ) and ( 2 ), we get
2 x = 12 ⇒ x = 6
Putting x = 6 in ( 1 ), we get
6 + y = 10
⇒ y = 10 – 6 = 4
Therefore, speed of rowing in still water = 6 km / h
Speed of current = 4 km / h
( ii ) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Solution: Let us consider,
Number of days taken by women to finish the work = x
Number of days taken by men to finish the work = y
Work done by women in one day = 1 / x
Work done by women in one day = 1 / y
As per the question given,
4 ( 2 / x + 5 / y ) = 1
( 2 / x + 5 / y ) = 1 / 4
And, 3 ( 3 / x + 6 / y ) = 1
( 3 / x + 6 / y ) = 1 / 3
Now, put 1 / x = m and 1 / y = n, we get,
2 m + 5 n = 1 / 4 ⇒ 8 m + 20 n = 1 …………………( 1 )
3 m + 6 n = 1 / 3 ⇒ 9 m + 18 n = 1 ………………( 2 )
By cross multiplication method, we get ,
m / ( 20 – 18 ) = n / ( 9 – 8 ) = 1 / ( 180 – 144 )
m / 2 = n / 1 = 1 / 36 ⇒ m / 2 = 1 / 36
m = 1 / 18 ⇒ m = 1 / x = 1 / 18 ⇒ x = 18
n = 1 / y = 1 / 36 ⇒ y = 36
∴ Number of days taken by women to finish the work = 18
Number of days taken by men to finish the work = 36.
( iii ) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution: ( iii ) Let us consider,
Speed of the train = x km / h
Speed of the bus = y km / h
According to the given question,
60 / x + 240 / y = 4 …………………( 1 )
100 / x + 200 / y = 25 / 6 …………….( 2 )
Put 1 / x = m and 1 / y = n, in the above two equations;
60 m + 240 n = 4……………………..( 3 )
100 m + 200 n = 25 / 6
600 m + 1200 n = 25 ………………….( 4 )
Multiply equation. 3 by 10, to get,
600 m + 2400 n = 40 ……………………( 5 )
Now, subtract equation. 4 from 5, to get,
1200 n = 15
n = 15 / 1200 = 1 / 80
Substitute the value of n in equation. 3, to get,
60 m + 3 = 4
m = 1 / 60 ⇒ m = 1 / x = 1 / 60 ⇒ x = 60
And y = 1 / n ⇒ y = 80
∴ Speed of the train = 60 km / h
Speed of the bus = 80 km / h
Additional Questions:
Formulate the following problems as a pair of equations, and hence find their solutions:
1. 7 women and 9 men can together finish an embroidery work in 14 days, while 4 women and 12 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
2. Rohit travels 400 km to his home town partly by train and partly by bus. He takes 5 hours if he travels 80 km by train and the remaining by bus. If he travels 150 km by train and the remaining by bus, he takes 20 minutes longer. Find the speed of the train and the bus separately.