ALGEBRAIC SOLUTIONS:
Finding solution for consistent pair of Linear Equations.
The solution of a pair of linear equations is of the form ( x,y ) which satisfies both the equations simultaneously. Solution for a consistent pair of linear equations can be found out using
i ) Elimination method ii ) Substitution Method
iii ) Cross – multiplication method iv ) Graphical method
Substitution Method of finding solution of a pair of Linear Equations:
y – 2 x = 1 and x + 2 y = 12
( i ) Express one variable in terms of the other using one of the equations.
In this case, y = 2 x + 1.
( ii ) Substitute for this variable ( y ) in the second equation to get a linear equation in one variable, x.
x + 2 × ( 2 x + 1 ) = 12 ⇒ 5 x + 2 = 12
( iii ) Solve the linear equation in one variable to find the value of that variable.
5 x + 2 = 12 ⇒ x = 2
( iv ) Substitute this value in one of the equations to get the value of the other variable.
y = 2 × 2 + 1 ⇒ y = 5
So, ( 2, 5 ) is the required solution of the pair of linear equations y – 2 x = 1 and x + 2 y = 12.
EXERCISE 3.3
- Solve the following pair of linear equations by the substitution method.
( i ) Given: x + y = 14 and x – y = 4
Solution :
⇒ x + y = 14 … ( I )
⇒ x – y = 4 . .. ( II )
From equation ( I ), we get
⇒ x = 14 – y . .. ( III )
Putting this value in equation ( ii ), we get
⇒ ( 14 – y ) – y = 4
⇒ 14 – 2 y = 4
⇒ 10 = 2 y
⇒ y = 10 / 2
⇒ y = 5 . .. ( IV )
Putting this in equation ( III ), we get
⇒ x = 9
Hence, x = 9 and y = 5.
( iv ) Given: 0.2 x + 0.3 y = 1.3 and 0.4 x + 0.5 y = 2.3 .
Solution : From 1st equation, we get,
x = ( 1.3 – 0.3 y ) / 0.2 _______ ( 1 )
Now, substitute the value of x in the given second equation to get,
0.4 ( 1.3 – 0.3 y ) / 0.2 + 0.5 y = 2.3
⇒ 2 ( 1.3 – 0.3 y ) + 0.5 y = 2.3
⇒ 2.6 – 0.6 y + 0.5 y = 2.3
⇒ 2.6 – 0.1 y = 2.3
⇒ 0.1 y = 0.3
⇒ y = 3
Now, substitute the value of y in equation ( 1 ), we get,
x = ( 1.3 – 0.3 (3) ) / 0.2 = ( 1.3 – 0.9 ) / 0.2 = 0.4 / 0.2 = 2
∴ x = 2 and y = 3.
( v ) Given: √2 x + √3 y = 0 and √3 x – √8 y = 0
Solution : From 1st equation, we get,
x = – ( √3 / √2 ) y ______ ( 1 )
Putting the value of x in the given second equation to get,
√3 ( – √3 / √2 ) y – √8 y = 0 ⇒ ( –3 / √2 ) y – √8 y = 0
⇒ y = 0
Now, substitute the value of y in equation ( 1 ), we get,
x = 0
Therefore, x = 0 and y = 0.
( vi ) Given: ( 3 x / 2 ) – ( 5 y / 3 ) = – 2 and ( x / 3 ) + ( y / 2 ) = 13 / 6 .
Solution : From 1st equation, we get,
( 3 / 2 ) x = – 2 + ( 5 y / 3 )
⇒ x = 2 (– 6 + 5 y ) / 9 = (– 12 + 10 y ) / 9 ………………………( 1 )
Putting the value of x in the given second equation to get,
[–12 + 10 y ) / 9 ] /3 + y / 2 = 13 / 6
⇒ y / 2 = 13 / 6 – (– 12 + 10 y ) / 27 ) + y / 2 = 13 / 6
Now, substitute the value of y in equation ( 1 ), we get,
( 3 x / 2 ) – 5 ( 3 ) / 3 = – 2
⇒ ( 3 x / 2 ) –5 = – 2
⇒ x = 2
Therefore, x = 2 and y = 3.
2. Solve 2 x + 3 y = 11 and 2 x – 4 y = – 24 and hence find the value of ‘m’
for which y = mx + 3.
Solution:
2 x + 3 y = 11 …….( i )
2 x – 4 y = – 24 ……( ii )
Now, from ( ii ), we get,
2 x – 4 y = – 24
2 x = – 24 + 4 yx = (–24 + 4 y ) / 2
x = – 24 / 2 + 4 y / 2
x = – 12 + 2 y ………..( iii )
Put this value of x in ( i ), we get, ( By substitution method)
2 (– 12 + 2 y ) + 3 y = 11
– 24 + 4 y + 3 y = 11 ⇒ 7 y = 11 + 24
7 y = 35 ⇒ y = 35 / 7
y = 5
now put this value of y in ( iii )
= –12 + 2 y = −12 + 2 ( 5 )
= –12 + 10
= – 2
y = m x + 3 ( y = 5 and x = – 2 )
5 = m (– 2 ) + 3
5 = – 2 m + 3 ⇒ 5 – 3 = – 2 m
2 = – 2 m ⇒ – m = 2 / 2
– m = 1 ⇒ m = –1
3.Form the pair of linear equations for the following problems and find their solution by substitution method.
( i )The difference between two numbers is 26 and one number is three times the other. Find them.
Solution: Let the two numbers be x and y.
x – y = 26 ……… ( i )
and x = 3 y
Substituting the value of x in equation ( i )
3 y – y = 26 ⇒ 2 y = 26 ⇒ y = 13
x = 3 × 13 = 39
So, the numbers are 39 and 13.
( ii ) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution :
Let the two supplementary angles be x° and y°.
x° + y° = 180°
x° = y°+ 18° (Given)
Substituting the value of x:
y° + 18° + y° = 180°
⇒ 2 y° + 18° = 180°
⇒ y° = 81°
⇒ x° = 180° – 81° = 99°
the two angles are 99° and 81°.
( iii )The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
Solution : Let the cost of 1 bat be Rs .x and l ball be Rs. y.
7 x + 6 y = 3800
3 x + 5 y = 1750
By solving we get x = Rs.500 , y = Rs.50.
( iv ).The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km. How much does a person have to pay for travelling a distance of 25 km.
Solution : Let fixed charge be Rs x and charge per km be Rs y.
⇒ x + 10 y = 105 ………. ( i )
⇒ x + 15 y = 155 ……….. ( ii ).
Now From equation (ii)
⇒ 15 y = 155 − x
⇒ y = ( 155 − x )/ 15 ……… ( iii )
Substituting y from equation (iii) in equation (i)
⇒ x + 10 ( 155 − x )/ 15 = 105
⇒ 15 x + 1550 − 10 x = 1575 ⇒ 5 x = 1575 − 1550
⇒ 5 x = 25 ⇒ x = 5
Substituting x in equation (ii)
⇒ 5 + 15 y = 155
⇒ 15 y = 155 − 5
⇒ 15 y = 150
⇒ y = 10
Hence x = 5 and y = 10
For 25 km a person has to pay = 5 + 10 × 25 = Rs 255.
( v ). A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Solution : Let numerator be x and denominator be y.
If 2 is added to both numerator and denominator, the fraction becomes 9 / 11.
x + 2 / y + 2 = 9 / 11
11( x + 2 ) = 9 ( y + 2 ) ⇒ 11 x + 22 = 9 y + 18
9 y – 11 x = 22 – 18 ⇒ 9 y – 11 x = 4
9 y = 4 + 11 x ⇒ y = ( 4 + 11 x ) / 9 ……………. (i)
If 3 is added to both numerator and denominator, the fraction becomes 5 / 6.
x + 3 / y + 3 = 5 / 6
6 ( x + 3 ) = 5 ( y + 3 ) ⇒ 6 x + 18 = 5 y + 15
5 y = 6 x + 18 – 15 ⇒ 5 y = 6 x + 3
y = ( 6 x + 3 ) / 5 ……………. (ii)
From (i) and (ii) we get
( 4 + 11 x ) / 9 = ( 6 x + 3 ) / 5
5 ( 4 + 11 x ) = 9 ( 6 x + 3 ) ⇒ 20 + 55 x = 54 x + 27
55 x – 54 x = 27 – 20 ⇒ x = 7
y = ( 4 + 11 ( 7 ) ) / 9 = ( 4 + 77 ) /9 = 81 / 9 = 9
∴ x = 7 and y = 9
The fraction = 7 / 9
( vi ) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages.
Solution : Let ‘x’ be the present age of Jacob and ‘y’ be the age of his son
As per question
x + 5 = 3 ( y + 5 ) ….. ( i )
x − 5 = 7 ( y − 5 ) …… ( ii )
Adding ( i ) and ( ii )
2 x = 3 y + 15 + 7 y − 35
⇒ 2 x = 10 y − 20
x = 5 y − 10 ………… ( iii )
Putting ( iii ) in ( i )
5 y − 5 = 3 y + 15
⇒ 2 y = 20 ⇒ y = 10
Putting in ( iii ) we get
x = 5 × 10 − 10
x = 40
∴ Present age of Jacob is 40 years and his son is 10 years.
Additional Questions:
1.Solve the following pair of linear equations by the substitution method.
( a ) x + y = 15 : x – y = 8
( b ) 6 x – 3 y = 9 : 9 x – 6 y = 18
2. Solve 3 x + 4 y =14 and 2 x – 4 y = –24 and hence find the value of ‘m’ for which y = m x + 3.
3. Form the pair of linear equations for the following problems and find their solution by substitution method.
( a )The difference between two numbers is 36 and one number is three times the other. Find them. ( b )The larger of two supplementary angles exceeds the smaller by 22 degrees. Find them.
( c ) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 15 km, the charge paid is Rs 150 and for a journey of 18 km, the charge paid is Rs 175. What are the fixed charges and the charge per km. How much does a person have to pay for travelling a distance of 35 km.