Solve the following simultaneous linear equations: (x-2)/5 = (1-y)/4 and 26x+3y+4=0

I agree with justaguide, the answer is x = -1/2 and y =
3

The step for solving the problem in my way
is:

We have two equations:

(1)
1/5(x-2) = 1/4 (1-y)

(2) 26x + 3y + 4 =
0

Step #1: Process the (1)
equation:

1/5 (x-2) = 1/4
(1-y)

=> (x-2)/5 =
(1-y)/4

=> 4(x-2) =
5(1-y)

=> 4x - 8 = 5 -
5y

=> 4x + 5y -8 - 5 =
0

=> 4x + 5y - 13 =
0

Gime Times 3 to (1) => 3 (4x+5y-13)
=0

=> 12x + 15y - 39 = 0
................(1*)

Give Times 5 to (2) => 5 (26x
+3y + 4) = 0

=> 130x + 15y + 20 = 0
 ......(2*)

(1*) and
(2*)

12x + 15y - 39 = 0

130x +
15y + 20 = 0
 -

__________________

-118 x -
59 = 0  => - 118 x = 59

=> x = 59/-118 =
-1/2

Substitude x = -1/2 to either (1) or
(2)

Here I choose to substitute x = -1/2 to
(1)

=> 4x + 5y - 13 =
0

=> 4(-1/2) + 5y - 13 =
0

=> -2 + 5y - 13 =
0

=>  5y - 15 =
0

=>  5y = 15

=>
 y = 15/5 = 3

Thus, the answer is x = -1/2 & y =
3