How do I do these two math problems

3(y-2)=6(y-1)-3y

and

3a+21 = 7-4A

3y-6 = 6y - 6 - 3y
3y-6 = 3y-6
y = whatever you want. There are infinite solutions.

3a + 21 = 7 - 4a
3a + 4a = 7 - 21
7a = -14
a = -2

for the first one, you distribute the 3 to (y-2) and distribute the 6 to (y-1) then you just combine like terms

for the second one, you subtract the 7 from the 21 then you subtract the 3a from the -4a then just do simple math

1.
3(y-2)=6(y-1)-3y
divide by 3
y-2=2(y-1)-y
distribute the 2
y-2=2y-2-y
subtract the y on the right
y-2=y-2
add 2
y=y
your answer is all real numbers :)

2.
3a+21=7-4a
add 4a
7a+21=7
subtract 21
7a=-14
divide by 7
a=-2

distribute to get rid of the parentheses and solve for y and a algebraically

1) 3y - 6 = 6y - 6 - 3y
y cancels out and leaves -6 = -6 so all real numbers would work

2) 7a = -14 divide both sides by 7
a = -2

3(y-2)=6(y-1)-3y
3y-6=6y-6-3y..bring the numbers to one side and the ys to the other
3y-6y+3y=6-6
0y=0
y=0


3a+21=7-4a
3a+4a=7-21
7a=-14
a=-14/7
a=-2