Step-by-step explanation:

from difference of two squares:

therefore:

factorise out ¾ :

Answer:
y=2x-4
Step-by-step explanation:
well first you have to move the -8x to the other side where the -16 is. by doing so, the -8x will turn into a POSTIVE 8x. than your problem should look like (4y=8x-16). now that your problem is like that, you want "y" to be alone. so, you divide the 4 by every number. (so you would do 4y/4, 8x/4, and -16/4). than your final product would be: y=2x-4
The answer is x-1 hope that helps not sure if it does