Mr.Edwards is now three times as old as his daughter, in 15 years' time the sum of their ages will be 86.

a.)How old are they both now?

b.)How old was Mr.Edwards when his daughter was born?

Write and solve an algebraic equation or inequality for this problem.

To solve this problem, let's assign variables to represent the ages of Mr. Edwards and his daughter.

Let's say the current age of Mr. Edwards is 'x' and the current age of his daughter is 'y.'

According to the problem, Mr. Edwards is now three times as old as his daughter. This can be written as:

x = 3y (Equation 1)

In 15 years' time, the sum of their ages will be 86. We can set up the following equation:

(x + 15) + (y + 15) = 86 (Equation 2)

Now, let's solve for the values of x and y.

a) How old are they both now?

To solve for their current ages, we need to determine the values of x and y that satisfy both Equation 1 and Equation 2.

Substituting the value of x from Equation 1 into Equation 2, we get:

(3y + 15) + (y + 15) = 86
4y + 30 = 86
4y = 86 - 30
4y = 56
y = 56 / 4
y = 14

Substituting the value of y back into Equation 1, we can find the value of x:

x = 3y
x = 3(14)
x = 42

So, Mr. Edwards is currently 42 years old, and his daughter is currently 14 years old.

b) How old was Mr. Edwards when his daughter was born?

To find out how old Mr. Edwards was when his daughter was born, we need to subtract the daughter's age from Mr. Edwards' current age.

Mr. Edwards' current age is 42, and his daughter's current age is 14. Therefore, Mr. Edwards was 42 - 14 = 28 years old when his daughter was born.

Let E=age of Mr. Edwards.

(E+15 + E/3+15) = 86
Solve for E.