In four years' time, a father will be three times as old as his son. Six years ago, he was seven times as old as his son. How old are they now? Explain your workings. Thanks
son's present age: x
father's present age : y
4 yrs from now:
son : x+4
father : y+4
y+4 = 3(x+4)
y = 3x + 8
6 yrs ago:
son : x-6
father: y-6
y - 6 = 7(x-6)
y = 7x -36
so 7x -36 =3x+8
4x = 44
x = 11
y = 3(11)+8 = 41
The son is now 11 and the father is now 41
check:
4 yrs from now: son-- 15, father 45
3 times as old? yes
6 yrs ago: son -- 5, father 35
7 times as old? yes
all is good
Thank you very much Reiny!
To solve this problem, let's assign variables to represent the ages of the father and the son. Let's call the father's current age F and the son's current age S.
According to the problem statement, in four years' time, the father will be three times as old as his son. So, we can write the first equation:
F + 4 = 3(S + 4)
Next, the problem states that six years ago the father was seven times as old as his son. We can write the second equation:
F - 6 = 7(S - 6)
Now we have a system of two equations with two unknowns (F and S). Let's solve them simultaneously to find the values of F and S.
Expanding the first equation:
F + 4 = 3S + 12
F = 3S + 12 - 4
F = 3S + 8
Expanding the second equation:
F - 6 = 7S - 42
F = 7S - 42 + 6
F = 7S - 36
Since both equations are equal to F, we can equate them:
3S + 8 = 7S - 36
Now, let's solve this equation for S:
4S = 44
S = 11
Substitute the value of S back into either of the original equations to find F:
F = 3S + 8
F = 3 * 11 + 8
F = 33 + 8
F = 41
Therefore, the father is currently 41 years old, and the son is currently 11 years old.