SOLUTION: My mom even tried to work this problem out. My teacher got it from edHelper.com. 1. In ten years, Eric's age will be seventy-three less than four times the age of Alexander. 2.

Question 115629: My mom even tried to work this problem out. My teacher got it from edHelper.com.
1. In ten years, Eric's age will be seventy-three less than four times the age of Alexander.
2. Kevin is sixty-three years younger than Jonathan.
3. Elizabeth is forty-one years older than Kevin.
4. The sum of of the ages of Elizabeth and Kevin is sixty-seven.
5. Eric's age is sixteen less than three times the age of Alexander.
This is how far my mom and I got on this question:
1. 4a - 73 a=Alexander
2. e - 63 b=Eric
3. d + 41 c=Elizabeth
4. c + d = 67 d=Kevin
5. 3a - 16 e=Jonathan
3. 41 + 34 = 75
4. 34 + 33 = 67
My mom and I would appreciate any help with this problem. Thank you for taking the time out to help us.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1. In ten years, Eric's age will be seventy-three less than four times the age of Alexander.
(E+10) = 4(A+10) - 73
--------------------
2. Kevin is sixty-three years younger than Jonathan.
K = J - 63
--------------------
3. Elizabeth is forty-one years older than Kevin.
E = K + 41
--------------------
4. The sum of of the ages of Elizabeth and Kevin is sixty-seven.
E + K = 67
-------------------
5. Eric's age is sixteen less than three times the age of Alexander.
E = 3A - 16
-------------------
#3: E-K=41
#4: E+K=67
Add to get:2E = 108
E = 54 years old
K = 13 years old
------------------
#5: E =3A-16
But E = 54
54 = 3A-16
3A = 70
A = 70/3 years old
----------------------------
#1: (E+10) = 4(A+10) - 73
(54+10) = 4A+40 - 73
64 = 4A+40 - 73
4A = 97
A = 97/4
Comment: Result contradicts A = 70/3
==================
Cheers,
Stan H.

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