### 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)

(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)

(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

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)

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)

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%

3, 1
2, -1
2, 3
1, 2

2,3
1, 3
4, 3
3,4

34, 30
22, 36
25, 30
26, 33
• ### 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 175, Rs 325
Rs 150. Rs 350
Rs 120, Rs 300

2, 1
3, 1
2, 3
3, 2

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.

35 years, 10 years
37 years, 8 yeras
30 years, 5 years
36 years, 12 years

55, 35
42, 30
45, 32
40, 30

5, 2
2, 1
6, 3
4, 2

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

#### Any Questions on Simultaneous Linear Equation in Two Variables ?

Forum Time Replies Report
##### Deewakar

Narmada was 25 years old when her son was born. If product of their ages now is 600, find their present age.

##### Rakesh

Alina said to alisha " i was twice as old as you were when i was as old as you are." If the sum of their ages is 50 years, find their ages.

##### dkd

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 .

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 .