How many solutions does each equation have...
2 answers:
A. the two 4y cancel each other out due to taking away a 4y from a 4y leaving -12=14-2 which would be -12
12 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.
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.
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