Given the system:

2t + 3u = 14.............(1)

t- 5u = -3.................(2)

We will use the substitution method to solve.

Let us rewrite (2).

==> t= 5u -3

Now we will substitute with t into (1)

==> 2t + 3u = 14

==> 2 (5u-3) + 3u = 14

==> 10u - 6 + 3u = 14

==> 13u = 14+6

==> 13u = 20

==> u = 20/13

==> t= 5u-3 = 5(20/13) - 3 = (100/13 - 39/13 = 61/13

==> t= 61/3

Then, the solution to the system is:

**u = 20/13.**

**t= 61/13**

We have to find t and u using the equations

2t + 3u = 14 ...(1)

t - 5u = -3 ...(2)

(1) - 2*(2)

=> 2t + 3u - 2t + 10u = 14 + 6

=> 13u = 20

=> u = 20/13

substituting u= 20/13 in (2)

=> t = -3 + 100/13

=> t = 61/13

**Therefore t = 20/13 and u = 61/13**

To solve the system 2t+3u = 14 ...(1) and 1-5u = -3...(2):

(1)-2*(2) : (2t+3u)-2(t-5u) = 14-2*-3 = 14+6 = 20.

=> 3u+2*5u = 20

13u = 20, u = 20/13.

We Substitute u = 20/ 13 in (2): t - 5(20/13) = -3. So t = -3+5*(20/13) = 61/13.

Therefore t = 61/13 and u = 20/13.