Problem 2(x + 5)

The Distributive property can be a pain in the but especially when it is like this 2(x + 5). When it is like this it is super hard if you do not know to do distributive property. But if you know how to solve it isn’t that hard. So first you multiply the number outside of the parentheses by what is inside. Example 2(x + 5) you would multiply the 2 by x and 2 times 5. Depending on the equation inside the parentheses you would add the product of the x and 2 and x and 5 or subtract if the x and 5 were being subtracted. In the end you would be able to simplify your equation but if we knew the value of x we could just solve the problem like a normal problem with parentheses. For example x = 14 the problem would then be 2(14 + 5) so you could then solve it without distributive property. Distributive has the word distribute because of  you would solve  a problem with parentheses and a variable.

Problem

12c + 3b

If you have two variables you can use GCF, greatest common factor. For example 12c + 3b. The first thing you do is look for the greatest factor that both can be multiplied by. Such as 3 in this case 12 can be multiplied by 4 and 3, 3 can only be multiplied by one to get 3 so you would put the other factor other than 3 to then get 3(4b + b).