SOLUTION: 5 years ago Christian was twice as old as Alec. The sum of their ages in 6 years will be 40 years. What is the present age of each? Note:Please use RESAC (Representation Equation

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Question 936399: 5 years ago Christian was twice as old as Alec. The sum of their ages in 6 years will be 40 years. What is the present age of each?
Note:Please use RESAC (Representation Equation Solution Answer Checking)
Answer by srinivas.g(540) About Me  (Show Source):
You can put this solution on YOUR website!
let x, y be the current ages of Christian, Alec
5 years ago Christian was twice as old as Alec.
x-5 = 2*(y-5)
x-5 =2*y+2*(-5)
x-5 =2y-10
move 2y to the left
x-2y-5 = -10
move -5 to the right
x-2y =-10+5
x-2y =-5 ................eq(1)
The sum of their ages in 6 years will be 40 years
(x+6)+(y+6)=40
x+y+12 =40
x+y= 40-12
x+y= 28 -----------eq(2)
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++1%5Cx+%2B+-2%5Cy+=+-5%2C%0D%0A++++1%5Cx+%2B+1%5Cy+=+28+%29%0D%0A++We'll use substitution. After moving -2*y to the right, we get:
1%2Ax+=+-5+-+-2%2Ay, or x+=+-5%2F1+-+-2%2Ay%2F1. Substitute that
into another equation:
1%2A%28-5%2F1+-+-2%2Ay%2F1%29+%2B+1%5Cy+=+28 and simplify: So, we know that y=11. Since x+=+-5%2F1+-+-2%2Ay%2F1, x=17.

Answer: system%28+x=17%2C+y=11+%29.