How many solutions does each equation
a) 4y – 4y – 12 = 14 – 2
b) –3y + 3y + 4 = 4
c) 3x + 6 = –1 – 3 + 4x

A. the two 4y cancel each other out due to taking away a 4y from a 4y leaving -12=14-2 which would be -1212 or, no solution

B. the two 3y cancel eachother out due to adding a negative one to a positive one leaving 4=4 (i forgot the term used for this, sorry)

C. take 3y away from each side  to have only one x leaving 6=-1-3+x subtract 3 from the negative 1 to lesson the amount of times we add/subtract from each side, ending up with 6=-4+x. add 4 to each side to get rid of the -4 leaving 10=X.  

The correct answer is $90