Worked Examples on Simultaneous Equations

If one number is thrice the other and their sum is 60, find the numbers.

Let the numbers be x and y.

x is 3 times y x = 3y …(1)

Sum of x and y is 60

x + y = 60 …(2)

Putting the value of x from (1) in (2), we get,

3y + y = 60

4y = 60 y = 15

Substituting y = 15 in (1), we get,

x = 3 x 15

= 45

The required numbers are 15 and 45.

Find the fraction which becomes when the denominator is increased by 5 and is equal to when the numerator is diminished by 4.

Let the fraction by

When y is increased by 5, fraction becomes

or 2x = y + 5

2x - y = 5 …(1)

When x is diminished by 4, fraction becomes

or 3x - 12 = y

3x - y = 12 …(2)

2x - y = 5 …(1)

Subtracting (1) from (2), we get x = 7

Substituting x = 7 in (1), we get,

2 7 - y = 5

- y = 5 - 14

- y = - 9

y = 9

The fraction is .

Six years hence a man's age will be three times his son's age, and three years ago he was nine times as old as his son. Find their present ages.

Let the present age of the man be x years, and the present age of his son be y years.

6 years hence their ages will be (x + 6) years and (y + 6) years.

x + 6 = 3(y + 6)

x + 6 = 3y + 18

x - 3y = 18 - 6

x - 3y = 12 …(1)

3 years ago, their ages were (x - 3) years and (y - 3) years.

x - 3 = 9(y - 3)

x - 3 = 9y - 27

Subtracting (1) from (2) -6y = -36

y = 6

Substituting y = 6 in (1), we get

x - 3 6 = 12

or x - 18 = 12

x = 12 + 18 = 30

The present age of the man is 30 years and the present age of his son is 6 years.

The sum of the digits of a two digit number is 7. If the digits are reversed, the new number increased by 3 equals four times the original number. Find the original number.

Let ten's digit be x and unit's digit be y. Then,

Sum of the digits = x + y

The number is = 10x + y

These steps will be repeated in digit sums

Sum of the digits is 7,

x + y = 7 …(1)

When the digits are reversed, the new number becomes

= 10y + x

New number increased by 3 = 4 times the original number

(10y + x) + 3 = 4 (10x + y)

or 10y + x + 3 = 40x + 4y

10y + x - 40x - 4y = -3

-39x + 6y = -3

- 13x + 2y = -1 (Dividing by 3) …(2)

Multiplying (1) by 2, we get,

2x + 2y = 14 …(3)

Subtracting (3) from (2),

-15x = - 15 x = 1

Substituting x = 1 in (1)

1 + y = 7 y = 6

The number is 10x + y = 10 x 1 + 6

= 16

A boat goes 40 km down stream in 2 hours and returns in 4 hours. Find the speed of the boat in still water and the speed of the current.

Let the speed of the boat in still water = x km/h, and

The speed of the current be = y km/h.

Time taken to go down stream = 2 hours

Time x Speed = Distance

2(x + y) = 40

x + y = 20 (Divided by 2) …(1)

Again, the speed of the boat upstream

= (x - y) km/h

Time take to go up stream = 4 hours

Distance covered = 40 km

4(x - y) = 40

x - y = 10 (Dividing by 4) …(2)

Adding (1) and (2), we get

x + y = 20 …(1)

x - y = 10 …(2)

2x = 30 x = 15

Substituting x = 15 in (1),

15 + y = 20 y = 5

The speed of the boat in still water is 15 km/h and the speed of the current is 5 km/h.

A lady has 50 cents and $ 2 coins in her purse. She has 90 coins in all and their total value is $105. How many $ 2  coins does she have?

Let the number of 50 cents coins be x and

the number of $ 2 coins be y.

x + y = 90 …(1)

The total amount is $ 105

…(2)

From (2)

x + 4y = 210 …(3)

x + y = 90 …(1)

3y = 120 Subtracting (1) from (2)

y = 40

The lady has 40 coins of $ 2.