Expand and simplify the following equation: 3(2a+2) + 4(b+4)

This problem is best split into two parts either side of the '+' sign seen as they are independent of each other, so the first part: 3(2a+2), as the 3 is outside of the bracket we have to multiply everything inside the brackets by 3. So this comes out as: 6a + 6Now the same for the second bracket, 4(b+4) becomes 4b + 16 So written out fully we have 6a + 6 + 4b + 16, as the VARIABLES (a & b) are different they cannot be combined but 6 + 16 are CONSTANTS (as in proper numbers) so can be. So we get 6a + 4b + 22. It might be tempting to stop here however there is one more step. As all CONSTANTS, including those infront of the a & b, are divisible by 2, we can put in brackets and take out a factor of 2 like so: 2(3a+2b+11)