Problems on Simple Simultaneous Equations

Problems related with numbers.
word problems appear confusing, and it is difficult to know where to begin.Read the problem.Form the equation, then solve it.

Example: If one number is thrice the other and their sum is 40, find the numbers.

Let the numbers be x and y.
x is 3 times y
x = 3y , Equation(1)
Sum of x and y is 40
x + y = 40 , Equation (2)

Putting the value of x from (1) in (2), we get,
3y + y = 40
4y/4 = 40/4 (dividing both sides by 4)
y = 10

Substituting y = 10 in (1), we get,
x = 3 x 10
x= 30

The required numbers are 10 and 30.

Practice :
The sum of 2 numbers is 50 and their difference is 16.find the numbers
The sum of two numbers is 43.if the larger is doubled and the smaller is tripled, the difference is 36. find the two numbers.

Example:
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 Equation(1)

3 years ago, their ages were (x - 3) years and (y - 3) years.
x - 3 = 9(y - 3)
x - 3 = 9y – 27(adding 3 to both sides)
x = 9y – 24
x – 9y = -24 equation (2)

Subtracting (1) from (2) -6y = -36(dividing both sides by -6)
-6y/-6 =-36/-6
y = 6
Substituting y = 6 in (1), we get
x - 3*6 = 12
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.

Practice:
The present age of the father is four times that of his son. Six years hence the age of the father will be thrice that of his son. Find their present ages.
The age of a man is three times the age of his daughter and five years hence his age will be double of his Son's age.find their present age