a father is 7times as old as his son .two years ago,the father was 13times as old as his son.how old tyey are now?
Let x = son's age, and y = father's age
y = 7x
y-2 = 13(x-2) = 13x - 26
Substitute 7x for y in the second equation.
7x - 2 = 13x - 26
Solve for x, then put that value into first equation. Check by putting both values into the second equation.
28 year,4 year
To solve this problem, let's assign variables to represent the ages of the father and the son. Let's say the father's age is 'F' and the son's age is 'S'.
According to the given information, "a father is 7 times as old as his son." We can express this as an equation: F = 7S.
"Two years ago, the father was 13 times as old as his son." This can be expressed as: F - 2 = 13(S - 2).
Now, we have two equations:
1) F = 7S
2) F - 2 = 13(S - 2)
To solve this system of equations, we can substitute the value of F from the first equation into the second equation:
7S - 2 = 13(S - 2)
7S - 2 = 13S - 26
7S - 13S = -26 + 2
-6S = -24
S = (-24) / (-6)
S = 4
Now that we know the son's age is 4, we can substitute this value into the first equation to find the father's age:
F = 7S
F = 7(4)
F = 28
Therefore, the son is 4 years old and the father is 28 years old.