Getting back to the original question, if that is permitted:
Rather than solve one of
5x+4y=6
-2x-3y=-1
for x or y and then substitute we can also
1) multiply the first equation by 2 to get 10x+ 8y= 12
2) multiply the seond equation by 5 to get -10x- 15y= -5
3) add those two equations to eliminate x and get -7y= 7 and divide both sides of that by 7 so that y= -1.
4) put that value of y back into either of the original equations to get either 5x+ 4(-1)= 5x- 4= 6 so, adding 4 to both sides 5x= 10 and then, dividing both sides by 5, x= 2, or -2x- 3(-1)= -2x+ 3= -1 so, subtracting 3 from both sides, -2x= -4, x= 2.
Or:
1) multiply the first equation by 3 to get 15x+ 12y= 18
2) multiply the second equation by 4 to get -8x- 12y= -4
3) add those two equations to eliminate y and get 7x= 14 and divide both sides of that by 7 to get x= 2.
4) put that value of x back into either of the original equations to get either 5(2)+ 4y= 10+ 4y= 6 so that, subtracting 10 from both sides 4y= -4 so that y= -1, or -2(2)- 3y= -4- 3y= -1 so that, adding 4 to both sides, -3y= 3 and, dividing both sides by -3, y= -1.
There are, in fact, many other ways to solve such equations, but however you do it, you will get x= 2, y= -1 and you can check those in the original
equations: with x= 2, y= -1,
5x+4y= 5(2)+ 4(-1)= 10- 4= 6
-2x-3y= -2(2)- 3(-1)= -4+ 3= -1