SOLUTION: Celia is 2 times Jacque’s age now. Celia was 6 times Jacque’s age 9 years ago. What system of equations can be solved to find their ages?

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Question 1087117: Celia is 2 times Jacque’s age now. Celia was 6 times Jacque’s age 9 years ago. What system of equations can be solved to find their ages?

Found 3 solutions by Boreal, josmiceli, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
J age 9 years ago was x, and it is now x+9
C's age 9 years ago was 6x, and it is now 6x+9
But 6x+9=2(x+9), since C's age is twice J's age
6x+9=2x+18
4x=9
x=2.25
6x=13.5
today,
x=11.25, 9+2.25
2x=22.5, 9+13.5

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +c+ = Celia's age now
Let +j+ = Jacque's age now
----------------------------
(1) +c+=+2j+
(2) +c+-+9+=+6%2A%28+j+-+9+%29+
----------------------------
(2) +c+-+9+=+6j+-+54+
(2) +6j+-+c+=+45+
Plug (1) into (2)
(2) +6j+-+2j+=+45+
(2) +4j+=+45+
(2) +j+=+11.25+
and
(1) +c+=+2j+
(1) +c+=+2%2A11.25+
(1) +c+=+22.5+
Jacque is 11 and 3 months
Celia is 22 and 6 months
---------------------
check:
(2) +c+-+9+=+6%2A%28+j+-+9+%29+
(2) +22.5+-+9+=+6%2A%28+11.25+-+9+%29+
(2) +13.5+=+6%2A2.25+
(2) +13.5+=+13.5+

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
This formulation is DEFECTIVE, since has no integer solution for ages in years.