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 1st condition , x + y = 475 --------------(i)

From 2nd 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 1st condition , x + y = 32 --------------(i)

From 2nd 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 1st condition , x + y = 80--------------(i)

From 2nd 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 1st condition , x+ y = 45o--------------(i)

From 2nd condition ,x- y = 15o--------------(ii)

Adding (i) and (ii),

(x + y) + (x - y) =45o+15o

or, 2x =60o

or x = \( \frac {60^o}{2}\)

∴ x =30o

Then, Putting the value of x in equation (i)

x + y = 45o

or, 30o+ y =45o

or, y = 15o

∴ y =15o

∴ The required anglesare 45o and 15o.

\)

\(

Let, thetwo numbers are x and y.

From 1st condition , x =3y--------------(i)

From 2nd 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 1st condition, x + y = 14 ---------(i)

From 2nd 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,

x2 - 10 = 54

or, x2 = 54 + 10

or,x2 = 64

or,x2 = (±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 1st condition, x - 2 = 9(y - 2)

or, x - 2 = 9y - 18

or, x - 9y -18 + 2

or, x = 9y - 16 -----------(i)

From 2nd 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 1st condition, x - 5= 4(y - 5)

or, x - 5= 4y- 20

or, x =4y -20 + 5

or, x = 4y - 15 -----------(i)

From 2nd 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 1st condition, x = 3y -----------(i)

From 2nd 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 1st condition, x -6= 6(y - 6)

or, x - 6= 6y - 36

or, x - 6y -36 +6

or, x = 6y - 30-----------(i)

From 2nd 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 1st condition, x - 4 + y - 4 = 48

or, x + y = 48 + 8

or, x + y = 56 ---------(i)

From 2nd 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 1st condition, x + 3 = 4(y + 3)

or, x + 3 = 4y + 12

or, x - 4y = 12 - 3

or, x = 4y + 9 --------(i)

From 2ndcondition, 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 1st condition, x + 2= 4(y + 2)

or, x + 2= 4y + 8

or, x - 4y = 8 - 2

or, x = 4y + 6 --------(i)

From 2ndcondition, 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,

\(\frac{x -8}{y-8}\) = \(\frac{1}{2}\)

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.

0%
  • Solve:

    2x-3y = 7, 7x + 27 = 12

    2, 3
    2, -1
    3, 1
    1, 2
  • 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
    34, 30
    22, 36
    25, 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

    2, 1
    3, 2
    2, 3
    3, 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
    30, 10 years
    35, 12 years
    36, 11 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
    35 years, 10 years
    36 years, 12 years
    37 years, 8 yeras
  • The sum of two numbers is 77 and their difference is 13. Find the numbers. 

    45, 32
    55, 35
    42, 30
    40, 30
  • Solve:

     x + y = 6, x + 2y = 8

    4, 2
    5, 2
    2, 1
    6, 3
  • The difference of two numbers is 12. If the greater number is three times the smaller one, find the numbers. 

    4, 11
    6, 18
    7, 15
    5, 12
  • You scored /10


    Take test again

Any Questions on Simultaneous Linear Equation in Two Variables ?

Please Wait...

Discussions about this note

Forum Time Replies Report
Rijen

In two digit number, the product of two digit is 14 and if 45 are sutracted from the numbers, the number will be reversed what is the number find it

Aasesh

If the age of father was 13 times than his son 1 year ago.Now the age of father is equal to cube of his son age.Find their present age.

Resham

the present age of mother is equal to the squre of the age of her daughter after one year . If the age of daughter in 10 year heance will be 1 year less then the age of her mother in 10 years ago then find their present sges

the present age of mother is equal to the square of the age of her daughter after one year. If theage of daughter in 10 year hence will be 1 year less than the age of her mother in 10 year ago then their present agess