Nine years ago Sally was three times older Bob. In eight years time, the sum of their ages will be 78. What are their current ages?

To solve this problem, let's assign variables to Sally's and Bob's current ages. Let's say Sally's current age is S, and Bob's current age is B.

According to the given information, nine years ago, Sally was three times older than Bob. So, Sally's age nine years ago would have been S - 9, and Bob's age nine years ago would have been B - 9.

According to this information, we can write the first equation:

S - 9 = 3(B - 9)

Now, let's solve the second part of the problem. In eight years time, the sum of their ages will be 78.

In eight years, Sally's age will be S + 8, and Bob's age will be B + 8.

So, we can write the second equation:

(S + 8) + (B + 8) = 78

Now, we have two equations:

S - 9 = 3(B - 9)
(S + 8) + (B + 8) = 78

Let's solve these equations to find their current ages.

From the first equation, we can simplify it:

S - 9 = 3B - 27
S - 3B = -18 ----(Equation 3)

Now, let's substitute the value of S - 3B from Equation 3 into Equation 2:

(-18) + (B + 8) + (B + 8) = 78

Combine like terms:

2B + (-10) = 78
2B = 88
B = 44

Now, substitute the value of B into Equation 1 to find Sally's age:

S - 9 = 3(44 - 9)
S - 9 = 3(35)
S - 9 = 105
S = 114

So, Sally's current age is 114, and Bob's current age is 44.

if their ages now are s and b, then we are told that

s-9 = 3(b-9)
s+8 + b+8 = 78
Now just solve for s and b.