10 years ago, the father was 6 times as old as his daughter . After 10 years he will be twice as old as his daughter. what is their present age? (linear equations)
Answers
Application of linear equation in two variable.
Answer: Present age of daughter is 15years and present age of father is 40 years.
Explanation:
Given that
10 years ago, the father was 6 times as old as his daughter.
After 10 years he will be twice as old as his daughter.
Need to determine present age of father and daughter
lets assume present age of daughter = x years
And assume present age of father = y years
10 YEARS AGO
Age of daughter 10 years ago = present age of daughter - 10 = x - 10
Age of father 10 years ago = present age of father - 10 = y - 10
As given that 10 years ago , father was 6 times as old as daughter
⇒ Age of father 10 years ago = 6 × Age of daughter 10 years ago
⇒ ( y - 10 ) = 6 ( x - 10 )
⇒ y - 10 = 6x - 60
⇒ y = 6x - 60 + 10
⇒ y = 6x - 50 ------------eq (1)
AFTER 10 YEARS
Age of daughter after 10 years = present age of daughter + 10 = x + 10
Age of father after 10 years = present age of father + 10 = y + 10
As given that after 10 years father will be twice as old as his daughter
⇒Age of father after 10 years = 2 × ( Age of daughter after 10 years )
⇒ y + 10 = 2 ( x + 10 )
⇒ y + 10 = 2x + 20
⇒ y = 2x + 20 - 10
⇒ y = 2x + 10 ----------eq(2)
substituting value of y = 6x - 50 from equation (1) in equation ( 2) we get
6x - 50 = 2x + 10
⇒ 6x - 2x = 10 + 60
⇒ 4x = 60
⇒ x = 60/4 = 15
⇒ x = 15
On substituting x = 15 in eq (1) that is y = 6x - 50 we get
y = (6 × 15 ) = 50 = 90 - 50 = 40
⇒ y = 40
Hence present age of daughter = x = 15years and present age of father = y = 40 years.
#answerwithqualty
#BAL
Answer:
byeee
Step-by-step explanation:
hope you understand
