Show your work -(3z+4)=6z-3(3z+2) a. 5/9 b. -1/9 c. infinitly many solutions d. no solution

Ok -(3z + 4) = 6z - 3 (3z +2)

so first you bring that minus across so that will be
-3z -4 = 6z -3 (3z + 2)
then what ever is infront of the parentehesis you multiply in this case its -3 not 6 because thats technally not infront well u no
-3times3z is -9z and -3 times 2 is -6 so you new equation is
-3z -4 = 6z -9z + -6 then u add or sutract the varibles and the variables or the constants and the constants thats ON THE SAME SIDE so in this case u subtract 9z and 6z which is -3z because there are more negatives than positives. new equation:
-3z -4 =-3z + -6 ( btw at this point when you have the same variable on the same side and different constants this is no solution but u can keep going if you want to show ALL WORK)
then you move them to each others side add 6 to -4 thats positive 2 and add 3 to -3 they will cancel out so your final anwser will be:
2=0 (which means)
no solution ALWAYS

The operation is suck a c@ck? !

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