A father who is 42 years old has a son and a daughter. The daughter is three times as old as the son. In 10 years the sum of all their ages will be 100 years. How old are the two siblings at present?
Please help quick!
D = 3S
(42+10) + S+10 + D+10 = 100
Substitute 3S for D in the second equation and solve for S. Insert that value into the first equation to solve for D. Check by putting both values into the second equation.
(42+10)+(s+10)+(d+10)=100
72+s+d=100
s+d=28
d=3s so you can plug it in
s+3s=28
4s=28
s=7
d=22
d=21* sry
To find the ages of the son and daughter, we can set up a system of equations based on the given information.
Let's assume the son's age is x years.
According to the problem, the daughter is three times as old as the son, so her age would be 3x years.
In 10 years, the son's age will be x + 10, and the daughter's age will be 3x + 10.
The sum of their ages in 10 years will be 100, so we have the equation:
(x + 10) + (3x + 10) + (42 + 10) = 100
Simplifying the equation, we have:
4x + 72 = 100
Subtracting 72 from both sides:
4x = 28
Dividing both sides by 4:
x= 7
Now, we know that the son is 7 years old.
The daughter is three times as old as the son, so she would be 3 * 7 = 21 years old.
Therefore, the son is currently 7 years old and the daughter is currently 21 years old.