.
Let D be the Dave age and C be the Christine's age.
Then you have these two equations
C-5 = 5*(D-5) (1) ("5 years ago Christine's age was 5 times Dave's)
D + 40 = 15 + (2) ("In 40 years, Dave will be 15 years more than half Christine's age")
Multiply equation (2) by 2 (both sides). You will get
2D + 80 = 30 + C + 40. (3)
Next express C = 5*(D-5) + 5 from equation (1) and substitute it into equation (3). You will get
2D + 80 = 30 + (5*(D-5) + 5) + 40.
Simplify and solve for D.
2D + 80 = 30 + 5D - 25 + 5 + 40
2D + 80 = 5D + 50
80 - 50 = 5D - 2D
30 = 3D
D = 30/3 = 10.
ANSWER. Dave is 10 years old. Christine is C = 5*(D-5) + 5 = 5*(10-5) + 5 = 5*5 + 5 = 30 years old.
You may check it ON YOUR OWN that my answer is correct.
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There is a bunch of lessons on age word problems
- Age problems and their solutions
- A fresh formulation of a traditional age problem
- Really intricate age word problem
- Selected age word problems from the archive
- Age problems for mental solution
- One unusual age word problem
- Age problems with a defective sum of ages
in this site.
Read them and become an expert in solving age problems.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Age word problems".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.