SOLUTION: Thanks for the help. I've tried every equation I can think of and I haven't come up with a suitable answer. Word Problem: Brian's present age is one year more than twice She

Question 40340: Thanks for the help. I've tried every equation I can think of and I haven't come up with a suitable answer.
Word Problem:
Brian's present age is one year more than twice Shelly's age. Five years ago, Brian was four times as old as Shelly. How old is Brian now?
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let B = Brian's present age and S = Shelly's present age. From the problem description, we get:
1) B = 2S+1 Brian's age is one year more (+1) than twice Shelly's age (2S).
2) B-5 = 4(S-5) Five years ago (-5) Brian was 4 times Shelly's age 4(S-5)
Substitute equation 1) into equation 2) and solve for S.
(2S+1)-5 = 4(S-5) Simplify.
2S-4 = 4S-20 Subtract 2S from both sides of the equation.
-4 = 2S-20 Add 20 to both sides.
16 = 2S Divide both sides by 2.
S = 8 This is Shelly's age.
B = 2S+1
B = 2(8)+1
B = 16+1
B = 17 This is Brian's age.
Check:
B = 2S+1
17 = 2(8)+1
17 = 17
B-5 = 4(S-5)
17-5 = 4(8-5)
12 = 4(3)
12 = 12

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