Question 267552
Let A = Mr. Atkin's present age (33) and S = Mr. Speyer's present age (27).
x years ago, Mr. Atkin's age was 1.5 times Mr. Speyer's age then.
This can be expressed algebraically as:
(A-x) = 1.5(S-x) Substitute A = 33 and S = 27.
(33-x) = 1.5(27-x) Simplify and solve for x.
33-x = 40.5-1.5x Add 1.5x to both sides.
33+0.5x = 40.5 Subtract 33 from both sides.
0.5x = 7.5 Divide both sides by 0.5
x = 15
15 years ago, Mr. Atkin's age was 1.5 times Mr. Speyer's age then.