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Question from Alissa, a student:

i have more then one question:

1. 2(y minus 3) =
2. 3(4 plus x) =
3. negative 6(2a plus 3b) =
4. negative 2(5x plus 2y) =
5. negative 3(x minus y) =
6. negative 3(3x minus 4w) =

Hi Alissa,

If you have, for example 4(3 + 2) you can evaluate it two ways. First evaluate what is inside the parentheses and then multiply

4(3 + 2) = 4 × 5 = 20

or use the distributive law

4(3 + 2) = 4 × 3 + 4 × 2 = 12 + 8 = 20.

When you have a variable inside the parentheses, as for example

4(3 + a)

then the first option is not available so you are forced to use the distributive law,

4(3 + a) = 4 × 3 + 4 × a = 12 + 4a.

This technique works even if there are minus signs and more than one variable,

-3(5x - y) = -3(5x + (-y)) = -3 × 5x + -3 × (-y) = -15x + 3y

Now try your questions and write back if you have difficulties.

Penny

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