Solve.
6(a + 3) = 21 – 3(1 – 2a)
–3
–1
no solution
all real numbers

We have the equation 6(a+3) = 21 - 3(1-2a), and we need to solve for a.
Let's first start by the first part of the equation: 6(a+3). We need to expand it.
6(a+3) = 6*a + 6*3 = 6a + 18
In the second part, let's expand 3(1-2a)
3(1-2a) = 3 - 6a

So now we got ride of the parenthesis, we get:
6a + 18 = 21 - (3-6a)
6a + 18 = 21 - 3 + 6a
6a + 18 = 18 + 6a

Let's subtract 18 from both sides to have variables on a side and numbers on the other:
6a + 18 - 18 = 18 + 6a - 18
6a = 6a
Subtract 6a from both sides, and you get:
6a - 6a = 6a - 6a
0 = 0

So based on the reflexive property (property when a number is equal to itself when solving), there's an infinite solution to this equation. So All real numbers are solution to this problem.

Hope this Helps! :)