if 2a - 3c = 8 and a/2 = (2c+1)/3, find a and c.

Given that :

2a - 3c = 8
.................(1)

a/2 =
(2c+1)/3

Let us multiply by
6:

==> 6a/2 =
6(2c+1)/3

==> 3a =
2(2c+1)

==> 3a = 4c +
2

==> 3a - 4c = 2
...................(2)

Now we will solve the system using
the elimination method.

-3*(1)
+ 1*(2):

==> -6a + 9c =
-24

==> 6a - 8c =
4

==> c =
-20.

Now we will substitute in
(1).

==> 2a- 3c =
8

==> 2a - 3(-20) =
8

==> 2a + 60 =
8

==> 2a =
-52

==> a = -
26

Then, the answer is: a =
-26 and c = -20