Simultaneous Linear Equation in Two Variables
Solving Simultaneous Linear Equations
Let us consider a problem related to our daily life,
The total cost of 2 kg of tea and 3 kg of coffee is Rs. 725. If 3 kg of tea and 2 kg of coffee are bought, the cost becomes Rs. 75 less. What is the per kg price of each?
Here,
\begin{align*} i.\: \text{Combined cost of 2kg tea and 3 kg coffee} &= Rs. 725 \\ ii. \: \text{Total cost of 3 kg tea and 2 kg coffee} &= Rs. 725  Rs. 75 \\ &= Rs. 650 \\ \end{align*}
Here are two prices which should be found.
Let the cost of 1 kg of tea = Rs. x
& the cost of 1 kg of coffee = Rs. y
Then, 2x + 3y = 725 ........... (i)
3x + 2y = 650 ...........(ii)
There are two variables x and y in the above pair of equations. The given pair of linear equations represents a pair of straight lines while plotting on the graph. The method of obtaining the value of the two variables (order pairs) which satisfy both the equation is known as 'solution of equations'. If the pair of values of two variables involved in a pair of equations satisfies both the equations, those equations are called linear simultaneous equation.
 Simultaneous equations are broadly used to find two unknown quantities.
 By solving the equations we can find the unknown quantities.
 In the application of simultaneous equations, the coefficient of x is the same in both equations.
 We subtract one equation from other to eliminate x.
 We can make a pair of simultaneous equations under the two given conditions.
Let, the cost of the book = Rs x, the cost of the pen = Rs y
From 1^{st} condition , x + y = 475 (i)
From 2^{nd} condition , x  y = 75 (ii)
Adding (i) and (ii),
(x + y) + (x  y) = 475 + 75
or, 2x = 550
or x = \( \frac {550}{2}\)
∴ x = 275
Then, Putting the value of x in equation (i)
x + y = 475
or, 275 + y = 475
or, y = 475  275
∴ y = 200
∴ Cost of book = Rs 275 and cost of pen = Rs 200
\(
Let, the age of father = x years, the age of son = y years
From 1^{st} condition , x + y = 32 (i)
From 2^{nd} condition , x  y = 24 (ii)
Adding (i) and (ii),
(x + y) + (x  y) = 32 + 24
or, 2x = 56
or x = \( \frac {56}{2}\)
∴ x = 28
Then, Putting the value of x in equation (i)
x + y = 32
or, 28 + y = 32
or, y = 32  28
∴ y = 4
∴ Age of father = 28 years and age of son = 4 years
\)
Let, thetwo numbers are x and y.
From 1^{st} condition , x + y = 80(i)
From 2^{nd} condition , x  y = 40(ii)
Adding (i) and (ii),
(x + y) + (x  y) =80 + 40
or, 2x =120
or x = \( \frac {120}{2}\)
∴ x =60
Then, Putting the value of x in equation (i)
x + y =80
or, 60+ y = 80
or, y =80  60
∴ y =20
∴ The required numbers are 60 and 40.
\(
Let, thetwo numbers are xand y.
From 1^{st} condition , x+ y = 45^{o}(i)
From 2^{nd} condition ,x y = 15^{o}(ii)
Adding (i) and (ii),
(x + y) + (x  y) =45^{o}+15^{o}
or, 2x =60^{o}
or x = \( \frac {60^o}{2}\)
∴ x =30^{o}
Then, Putting the value of x in equation (i)
x + y = 45^{o}
or, 30^{o}+ y =45^{o}
or, y = 15^{o}
∴ y =15^{o}
∴ The required anglesare 45^{o} and 15^{o.}
^{\)}
\(
Let, thetwo numbers are x and y.
From 1^{st} condition , x =3y(i)
From 2^{nd} condition , x +y = 32(ii)
Putting value of x in(ii),
3y + y=32
or, 4y=32
or y= \( \frac {32}{4}\)
∴ x =8
Then, Putting the value of yin equation (i)
x = 3y
or, x = 3× 8
or, y = 24
∴ y =20
∴ The required numbers are 8 and 24.
\)
Let the two numbers be x and y and x be the greater one .
From 1^{st} condition, x + y = 14 (i)
From 2^{nd} condition, x= y + 4 (ii)
Putting value of x in equation (i)
x + y = 14
or, y + 4 + y = 14
or, 2y = 14  4
or, 2y = 10
or y = \(\frac {10}{2}\)
∴y = 5
Now, putting value of y in equation (ii)
x = y + 4 = 5 + 4 = 9
Hence, the required numbers are 9 and 5.
Let, the number be x.
According to question,
x^{2}  10 = 54
or, x^{2} = 54 + 10
or,x^{2} = 64
or,x^{2} = (±8)^{2}
∴ x = ±8
Let, breadth of the pond (b) = x m
Then, length of the pond = (x + 10) m
Perimeter of the pond (P) = 120m
or, 2 (l + b) = 120
or, l + b = \( \frac{120}{2}\)
or, l + b = 60
or, ( x + 10) + x = 60
or, 2x + 10 = 60
or, 2x = 60  10
or, 2x = 50
or, x = \( \frac{50}{2} \)
or, x = 25
∴Breadth = 25m
and Length = (x +10) = 35 m
Hence, the length and the breadth of the pond are 35m and 25m respectively.
Let the present age of father = x years and present age of son = y years
From 1^{st} condition, x  2 = 9(y  2)
or, x  2 = 9y  18
or, x  9y 18 + 2
or, x = 9y  16 (i)
From 2^{nd} condition, x + 3 = 5(y +3)
or, x + 3 = 5y + 15
or, x = 5y + 15  3
or, x = 5y + 12 (ii)
Putting the value of x in equation (i)
5y + 12 = 9y  16
or, 16 + 12 = 9y  5y
or, 28 = 4y
or, y = \(\frac{28}{4}\)
or, y = 7
Hence, the present age of son is 7 years.
Putting the value of y in equation (ii)
x = 5× 7 + 12
or, x = 47
Hence, the present age of father is 47 years.
Let the present age of father = x years and present age of son = y years
From 1^{st} condition, x  5= 4(y  5)
or, x  5= 4y 20
or, x =4y 20 + 5
or, x = 4y  15 (i)
From 2^{nd} condition,x + y = 45(ii)
Putting the value of x in equation (ii)
4y  15 + y = 45
or, 5y = 45 +15
or, 5y = 60
or, y = \(\frac{60}{5}\)
or, y =12
Hence, the present age of son is 12years.
Putting the value of y in equation (ii)
x + 12= 45
or, x = 45 12
or, x = 33
or, x = 33
Hence, the present age of father is 33years.
Let the present age of father = x years and present age of son = y years
From 1^{st} condition, x = 3y (i)
From 2^{nd} condition, x 4= 4(y  4)
or, x 4=4y  16
or, x =4y  16 + 4
or, x =4y  12(ii)
Putting the value of x in equation (i)
3y = 4y  12
or, 4y  3y = 12
or y = 12
Hence, the present age of son is 12 years.
Putting the value of y in equation (i)
x = 3 × 12 = 36 years
Hence, the present age of father is 36 years.
Let the present age of father = x years and present age of son = y years
From 1^{st} condition, x 6= 6(y  6)
or, x  6= 6y  36
or, x  6y 36 +6
or, x = 6y  30(i)
From 2^{nd} condition, 3(x + 4)= 8(y +4)
or, 3x + 12 = 8y + 32
or, 3x  8y = 32  12
or, 3x  8y = 20 (ii)
Putting the value of x in equation (ii)
3(6y  30)  8y = 20
or, 18y  90  8y = 20
or, 10y = 20 + 90
or, 10y = 110
or, y = \( \frac{110}{10}\)
or, y = 11
Hence, the present age of son is 11 years.
Putting the value of y in equation (i)
x =6× 11 30
or x = 66  30
or, x = 36
Hence, the present age of father is 36 years.
Let, present age of father is x years and son is y years
From 1^{st} condition, x  4 + y  4 = 48
or, x + y = 48 + 8
or, x + y = 56 (i)
From 2^{nd} condition, x + 4 = 3(y + 4)
or, x + 4 = 3y + 12
or, x  3y = 12  4
or, x  3y = 8
Putting value of x from (i) in (ii)
56  y  3y = 8
or, 56  8 = 4y
or, 48 = 4y
or, y = \(\frac{48}{4}\)
or, y = 12
Putting the value of y in equation (i)
x + y = 56
or, x + 12 = 56
or, x = 56  12
or, x = 44
Hence, the present ages of father and son are 44 years and 12 years respectively.
Let, the present age of mother = x years and present age of daughter = y years
From 1^{st} condition, x + 3 = 4(y + 3)
or, x + 3 = 4y + 12
or, x  4y = 12  3
or, x = 4y + 9 (i)
From 2^{nd}condition, x 2 = 3(y + 4)
or, x = 3y + 12 + 2
or, x = 3y + 14 (ii)
Putting value of x from equaiton (i) in equation (ii)
4y + 9 = 3y + 14
or, 4y  3y = 14  9
or, y = 5
Putting value of y in equation (i)
x = 4× 5 + 9 = 29
Hence, the present ages of mother and daughter are 29 years and 5 years respectively.
Let, the present age of mother = x years and present age of daughter = y years
From 1^{st} condition, x + 2= 4(y + 2)
or, x + 2= 4y + 8
or, x  4y = 8  2
or, x = 4y + 6 (i)
From 2^{nd}condition, x 2 = 3(y + 4)
or, x = 3y + 12 + 2
or, x = 3y + 14 (ii)
Putting value of x from equaiton (i) in equation (ii)
4y + 6 = 3y + 14
or, 4y  3y = 14 6
or, y = 8
Putting value of y in equation (i)
x = 4× 8+ 6= 38
Hence, the present ages of mother and daughter are 38years and 8years respectively.
Let, the ages of two boys be x years and y years.
From first condition,
\(\frac{x}{y}\) = \(\frac{3}{4}\)
or, x = \(\frac{3y}{4}\) (i)
From second condition,
\(\frac{x 8}{y8}\) = \(\frac{1}{2}\)
or, 2x  16 = y  8
or, y = 2x  16 + 8
or, y = 2x  8(ii)
Putting value of y in equation (i)
4x = 3y
or, 4x = 3(2x  8)
or, 4x = 6x  24
or, 24 = 6x  4x
or, 24 = 2x
or, \(\frac{24}{2}\) = x
or, x = 12
Putting value of x in equation (ii)
y = 2× 12  8
or, y = 24  8
or, y = 16
The present ages of boys are 12 and 16 years.
Let the present ages of father and son be x years and y years respectively.
From first condition, x + 2y = 50 (i)
From second condition, 2x + y = 76
or, y = 76  2x (ii)
Putting the value of y in equation (i)
x + 2(76  2x) = 50
or, x + 152  4x = 50
or, 3x = 102
or, x = \(\frac{102}{3}\)
or, x = 34
Putting value of x in (ii)
y = 76  2× 34 = 76  68 = 8
Hence, the present ages of father and son are 34 years and 8 years respectively.
Let, the two digit number be 10x + y, where x is ten's place and y is one's place.
From first condition, x + y = 10 (i)
From second condition, 10x + y  18 = 10y + x
or, 10x x + y  10y = 18
or, 9x  9y = 18
or, 9 (x y) = 18
or, x  y = \(\frac{18}{9}\)
or, x  y = 2(ii)
Adding euqation (i) and (ii)
x + y + (x  y) = 10 + 2
or, 2x = 12
or, x = \( \frac{12}{2}\)
or, x = 6
Putting value of x in equation (i)
6 + y = 10
or, y = 10  6
or, y = 4
Hence, the required two digit number is 10x + y = 10× 6 + 4 = 60 + 4 = 64
Let the two digit number be 10x + y where x is ten's place and y is one's place
From first cndition, x + y = 8 (i)
From second condition, 10y + x = 10x + y + 18
or, 18 = 10x  x + y  10y
or, 18 = 9x  9y
or, \(\frac{18}{9}\) = x  y
or, x  y = 2 (ii)
Adding equation (i) and (ii)
x+y +(x  y) = 8  2
or, 2x = 6
or, x = \(\frac{6}{2}\)
or, x = 3
Putting value of x in equation (i)
3 + y = 8
or, y = 8  3
or, y = 5
Hence, the required two digit number is 10x + y = 10× 3 + 5 = 35
Let, the present ages of father and son is x years and y years respectively.
From first condition, x + y = 50 (i)
From second condition, {x + (x  y) }+ {y + (x  y) } = 110
or, x + x + x + y  y  y = 110
or, 3x  y = 110
or, 3x  110 = y (iI)
Putting value of y in equation (i)
3x  110 + x = 50
or, 4x = 110+ 50
or, 4x = 160
or, x = \(\frac{160}{4}\)
or, x = 40
Putting the value of x in equation (i)
40 + y = 50
or, y = 50  40
or, y = 10
Hence, the present ages of father and son are 40 years and 10 years respectively.
Let, the two digit number = 10x + y
From first condition, 10x + y = 4(x + y) + 3
or, 10x + y = 4x + 4y + 3
or, 10x  4x + y  4y = 3
or, 6x  3y = 3
or, 3 (2x  y) = 3
or, 2x  y = 1 (i)
From second condition, 10x + y + 36 = 10y + x
or, 10x  x + y  10y = 36
or, 9x  9y =  36
or, 9 ( x  y) = 36
or, x  y = \(\frac{36}{9}\)
or, x  y =  4
or, x = y  4(ii)
Putting value of x in equation (i)
2(y  4)  y = 1
or, 2y  y = 1 + 8
or, y = 9
Putting the value of y in equation (ii)
x = 9  4 = 5
Therefore, the required number is 10x + y = 10× 5 + 9 = 59
Let, the two digit number = 10x + y
From first condition, 10x + y = 6(x + y) + 3
or, 10x + y = 6x + 6y + 3
or, 10x  6x + y  6y = 3
or, 4x  5y = 3(i)
From second condition, 10y + x = 10x + y  18
or, 18 = 10x  x + y  10y
or, 18 = 9(x  y)
or, x  y = \(\frac{18}{9}\)
or, x  y = 2
or, x = y + 2(ii)
Putting the value of x in equation (i)
4(y +2)  5y = 3
or, 4y +8  5y = 3
or, y = 8  3
or, y = 5
Putting the value of y in equation (ii)
x = 5 + 2 = 7
Hencel the required two digit number is 10× 7 + 5 = 75
Let, the two digit number = 10x + y
From first condition, 10x + y = 4(x + y)
or, 10x + y = 4x + 4y
or, 10x  4x + y  4y = 0
or, 6x  3y = 0
or, 3 (2x  y) = 0
or, 2x  y = 0
or, y = 2x(i)
From second condition, 10x + y + 18 = 10y + x
or, 10x + y  10y  x = 18
or, 9x  9y = 18
or, x  y = \(\frac{18}{9}\)
or, x  y = 2 (ii)
Putting value of y from (i) in (ii)
x 2x = 2
or, 2 = x
Putting value of x in equation (i)
y = 2 x 2 = 4
Hence, the required number is 10 x 2 + 4 = 24.

Solve:
2x3y = 7, 7x + 27 = 12
3, 1
2, 1
1, 2
2, 3

Solve the following equations:
4x 3y =0, 5x 4y = 1
3,4
2,3
1, 3
4, 3

The sum of two numbers is 59. If the smaller number is less than the bigger one by 7. Find the numbers.
26, 33
22, 36
25, 30
34, 30

The total cost of TV and Radio is Rs 500. If the watch is cheaper than Radio by 150, find their cost.
Rs 125, Rs 375
Rs 120, Rs 300
Rs 150. Rs 350
Rs 175, Rs 325

Solve: x+3y = 6, x +y = 4
3, 1
2, 3
3, 2
2, 1

The sum of the ages of a father and his son is 42 years. Six years ago, the father was 5 times as old as his son was. Find their present ages.
31, 11 years
36, 11 years
30, 10 years
35, 12 years

Ramesh is three times as old as his son. If the differences of their ages is 24 years. Find their present ages.
30 years, 5 years
36 years, 12 years
37 years, 8 yeras
35 years, 10 years

The sum of two numbers is 77 and their difference is 13. Find the numbers.
55, 35
40, 30
42, 30
45, 32

Solve:
x + y = 6, x + 2y = 8
6, 3
5, 2
2, 1
4, 2

The difference of two numbers is 12. If the greater number is three times the smaller one, find the numbers.
6, 18
7, 15
4, 11
5, 12

You scored /10
Any Questions on Simultaneous Linear Equation in Two Variables ?
Please Wait...
Discussions about this note
Forum  Time  Replies  Report 

dkdAsk any queries on this note.Find the number which is as much more than 12 as twice the same number is less than 60 . 
Jan 03, 2017 
0 Replies View Replies 

Ask any queries on this note.find the number which is as much more than 12 as twice the same number is less than 60 . 
Jan 03, 2017 
0 Replies View Replies 

amritthe sum of present age of father and his son is 80 year .when the father 's age was equal to present age of the son,the sum of their age was 40 years.find their present ages. 
Dec 28, 2016 
0 Replies View Replies 