Elimination method of finding solution of a pair of Linear Equations:
Consider x + 2 y = 8 and 2 x –3 y = 2
Step 1: Make the coefficients of any variable the same by multiplying the equations with constants. Multiplying the first equation by 2, we get,
2 x + 4 y = 16
Step 2: Add or subtract the equations to eliminate one variable, giving a single variable equation.
Subtract second equation from the previous equation
2 x + 4 y = 16
2 x – 3 y = 2
(–) ( +) (–0)
———————
0( x ) + 7 y = 14
Step 3: Solve for one variable and substitute this in any equation to get the other variable.
y = 2,
x = 8 – 2 y ⇒ x = 8 – 4 ⇒ x = 4
( 4, 2 ) is the solution.
EXERCISE 3.4
1.Solve the following pair of linear equations by the elimination method
and the substitution method :
x + y = 5 and 2 x – 3 y = 4
Solution : x + y = 5 and 2 x – 3 y = 4
By elimination method
x + y = 5 ……. ( i )
2 x – 3 y = 4 …….. ( ii )
Multiplying equation ( i ) by ( ii ), we get
2 x + 2 y = 10 ……….. ( iii )
2 x – 3 y = 4 … ( ii )
Subtracting equation ( ii ) from equation ( iii ), we get
5 y = 6 ⇒ y = 6 / 5
Putting the value in equation ( i ), we get
x = 5 – ( 6 / 5 ) = 19 / 5
Hence, x = 19 / 5 and y = 6 / 5
By substitution method x + y = 5 … ( i )
Subtracting y from both sides, we get
x = 5 – y … ( iv )
Putting the value of x in equation ( ii ) we get
2 ( 5 – y ) – 3 y = 4 ⇒ – 5 y = – 6
y = – 6 / – 5 = 6 / 5
Putting the value of y in equation ( iv ) we get
x = 5 – 6 / 5 ⇒ x = 19 / 5
Hence, x = 19 / 5 and y = 6 / 5 .
(ii). 3 x + 4 y = 10 and 2 x – 2 y = 2
Solution : 3 x + 4 y = 10 and 2 x – 2 y = 2
By elimination method
3 x + 4 y = 10 …. ( i )
2 x – 2 y = 2 … ( ii )
Multiplying equation ( ii ) by 2, we get
4 x – 4 y = 4 … ( iii )
3 x + 4 y = 10 … ( i )
Adding equation ( i ) and ( iii ), we get
7 x + 0 = 14
Dividing both sides by 7, we get
x = 14 / 7 = 2
Putting in equation ( i ), we get
3 x + 4 y = 10 ⇒ 3( 2 ) + 4 y = 10
6 + 4 y = 10 ⇒ 4 y = 10 – 6
4 y = 4 ⇒ y = 4 / 4 = 1
Hence, x = 2, y = 1
(iii). 3 x – 5 y – 4 = 0 and 9 x = 2 y + 7
Solution : By elimination method
3 x – 5 y – 4 = 0
3 x – 5 y = 4 ……… ( i )
9 x = 2 y + 7
9 x – 2 y = 7 ……. … ( ii )
Multiplying equation ( i ) by 3, we get
9 x – 15 y = 11 …… ( iii )
9 x – 2 y = 7 ….. ( ii )
Subtracting equation ( ii ) from equation ( iii ), we get
– 13 y = 5
y = – 5 / 13
Putting value in equation ( i ), we get
3 x – 5 y = 4 … ( i ) ⇒ 3 x – 5 ( – 5 / 13 ) = 4
Multiplying by 13 we get
39 x + 25 = 52 ⇒ 39 x = 27
x = 27 / 39 = 9 / 13
Hence x = 9 / 13 and y = – 5 / 13
By substitution method
3 x – 5 y = 4 … ( i )
Adding 5 y to both sides we get
3 x = 4 + 5 y
Dividing by 3 we get
x = ( 4 + 5 y ) / 3 … ( iv )
Putting this value in equation ( ii ) we get
9 x – 2 y = 7 … ( ii )
9 ( ( 4 + 5 y ) / 3 ) – 2 y = 7
Solve it we get
3( 4 + 5 y ) – 2 y = 7
12 + 15 y – 2 y = 7
13y = – 5
y = – 5 / 13
Put this value in eq (iv)
x = [4 + 5 (– 5 / 13 ) ]/ 3
= [4 – 25 / 13 ] / 3
= [4 × 13 – 25 / 13 ] / 3
= (27 / 13) × 3
= 27 / 39
= 9 / 13
Hence we get x = 9 / 13 and y = – 5 / 13 .
(iv). x / 2 + 2 y / 3 = –1 and (x – y )/ 3 = 3
Solution : By elimination method
x / 2 + 2 y / 3 = –1 …..(i)
(x – y) / 3 = 3 … ( ii )
Multiplying equation ( i ) by 2, we get
(x + 4 y) / 3 = – 2 … ( iii )
(x – y) / 3 = 3 … ( ii )
Subtracting equation ( ii ) from equation ( iii ), we get
5 y / 3 = – 5
Dividing by 5 and multiplying by 3, we get
y = – 15 / 5
y = – 3
Putting this value in equation ( ii ), we get
x – y / 3 = 3 … ( ii )
x – ( – 3 ) / 3 = 3
x + 1 = 3
x = 2
Hence our answer is x = 2 and y = − 3.
By substitution method
x – y / 3 = 3 … ( ii )
Add y / 3 to both sides, we get
x = 3 + y / 3 … ( iv )
Putting this value in equation ( i ) we get
x / 2 + 2 y / 3 = – 1 … ( i )
( 3 + y / 3 ) / 2 + 2 y / 3 = –1
3 / 2 + y / 6 + 2 y / 3 = – 1
Multiplying by 6, we get
9 + y + 4 y = – 6
5 y = – 15
y = – 3 Hence x = 2 and y =−3.
2.Form the pair of linear equations in the following problems, and find their solutions (if they exist ) by the elimination method :
( i ) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1 / 2 if we only add 1 to the denominator. What is the fraction.
Solution : Let numerator = x and denominator = y
∴ The fraction = x / y
If we add 1 to the numerator and subtract 1 from the denominator,the fraction reduces to 1
Cross multiplying we get
x + 1 = y – 1
x – y = – 2 …….. ( i )
It becomes 1 / 2 if we only add 1 to the denominator.
Cross multiplying we get
2 x = y + 1
2 x – y = 1 …….. ( ii )
x – y = – 2 ……..( i )
Subtracting equation ( i ) from equation ( ii ) we get
x = 3
Substituting this value in equation ( i ) we get
3 – y = – 2 ⇒ – y = –5
∴ y = 5
Hence our fraction is 3/5
( ii ) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu.
Solution : Let the present age of Nuri = x year
And present age of Sonu = y year
Five years ago
Age of Nuri = x – 5 years
Age of Sonu = y – 5 years
Nuri was thrice as old as Sonu . So the equation will be
x – 5 = 3 ( y – 5 )
x – 5 = 35 – 15
x – 3 y = – 15 + 5
x – 3 y = – 10 ………..( i )
Ten years later,
Age of Nuri = x + 10
Age of Sonu = y + 10
Nuri will be twice as old as Sonu.
x + 10 = 2 ( y + 10 )
x + 10 = 2 y + 20
x – 2 y = 10 ………..( ii )
x – 3 y = – 10 ………..(i)
Subtracting equation ( i ) from equation ( ii ) we get
y = 20
Put this value in equation ( i ) we get
x – 3 (20) = – 10 ⇒ x = 60 – 10
x = 50
Hence age of Nuri = 50 years and age of Sonu = 20 years
( iii ) The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution : Let the unit’s digit = x
Ten’s digit = y
Number will 10 times the tens digit + unit times the unit digit
Hence number will 10 y + x
Sum of digits are 9
So that x + y = 9 ………….( i )
nine times this number is twice the number obtained by reversing the order of the digits
9 ( 10 y + x ) = 2 ( 10 x + y )
90 y + 9 x = 20 x + 2 y
88 y – 11 x = 0
Dividing by 11 we get
8 y – x = 0 …………..( ii )
x + y = 9 ………….( i )
Adding both equations we get
9 y = 9
y = 9 / 9 = 1
Put this value in equation (i) we get
x + y = 9 ⇒ x + 1 = 9
x = 8
So our original number is 10 y + x = 10 (1) + 8 = 18
( iv ) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
Solution : The number of Rs 100 notes = y
According to the given information
Meena went to a bank to withdraw Rs 2000
So cost of 50 rupee notes = 50 x
And cost of 100 rupee notes = 100 y
Total cost is Rs 2000
∴ 50 x + 100 y = 2000
Dividing by 50 we get
x + 2 y = 40 ………( i )
Meena got 25 notes in all so that
x + y = 25 ………( ii )
x + 2 y = 40 ………( i )
Subtracting equation ( i ) from ( ii) we get
– y = –15
y = 15
Put this value in equation first we get
x + 2 (15) = 40 ⇒ x + 30 = 40
x = 10
Hence 50 rupee notes are 10 and 100 rupee notes are 15
( v ) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution : Let fixed change = Rs x
And charge for extra day = Rs y
Saritha paid Rs 27 for a book kept for seven days
So first three days she paid Rs x and for remaining 4 days she will pay 4 y
So total payment will be
x + 4 y = 27 …….. ( i )
Similarly Susy paid Rs 21 for the book she kept for five days
x + 2 y = 21 ……..( ii )
x + 4 y = 27 …….. ( i )
Subtracting equation ( i ) from equation ( ii ), we get
– 2 y = – 6
y = 3
Putting this value in equation ( 1 ) we get
x + 4 y = 27 …….. ( 1 )
x + 4 ( 3 ) = 27 ⇒ x = 27 – 12
x = 15
Hence fixed cost is Rs 15 and cost for extra each day is Rs 3.
Additional Questions:
1.Solve the following pair of linear equations by the elimination method and the substitution method :
( a ) 2 x + 3 y = 15 and x – 3y = 14
( b )4 x + 5 y = 20 and x – y = 12
2.Form the pair of linear equations in the following problems, and find their solutions ( if they exist ) by the elimination method :
( a ) The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
( b ) Abigail went to a bank to withdraw Rs 5000. She asked the cashier to give her Rs 100 and Rs 200 notes only. She got 35 notes in all. Find how many notes of Rs 200 and Rs 100 she received