How to solve using substitution y=3/4x+5 4x-3y=-1

Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as y=(3/4)X+5

Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:

4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)

Let 's now solve the system   

4y=3x + 20
4x-3y=-1

We are to solve this system using subtraction.  To accomplish this, multiply the first equation by 3 and the second equation by 4.  Here's what happens:

12y = 9x + 60  (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)

Then we have 

 12y = 9x + 60
-12y =-16x - 4

If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x:  7x = 56; x=8

We were given equations   
y=3/4X+5
4x-3y=-1

We can subst. x=8 into either of these eqn's to find y.  Let's try the first one:

y = (3/4)(8)+5 = 6+5=11

Then x=8 and y=11.

You should check this result.  Subst. x=8 and y=11 into the second given equation.  Is this equation now true?