It can be either of those. I would hesitate to write the first one, though, because a standard convention is to write constants first. Also, you might get someone thinking that the 3 is an exponent, which is not the case.
First question:
From the book... whomever wrote this doesn't know HOW to explain things.
Solve:
From the second two terms, pull out a ‘-1’. You can double check that this is correct by distributing the “-1” back in.
Inside the parenthesis on the far right, rearrange terms.
Answer:Now, in the second group of terms, we have a “-1” times a “-1” which gives a “+1”.
Can someone explain how the answer is obtained better? I'm fumbling on factoring with -1 on these types of equations.
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Second question:
Solve:
Answer:
If you apply the order of distribution it comes out to
Why can't the answer be
or
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Anyone have a better explanation? I'm starting to cram for a COMPASS test this coming Friday and still need to refresh my math skills from algebra, adv. algebra, geo, and trig.
Usually when you multiply a number by a variable or expression you write the number first, so that there is no ambiguity. I could look at
and thought that you forgot to make the 3 an exponent, like this:
... which is certainly wrong. So keep the 3 in front.
This:
is just fine. Commutative property of multiplication says that the order you multiply doesn't matter. So it doesn't matter if you wrote or as your answer.
EDIT: Ack! Too slow!
No.
First, you between the 3d and x you changed from a multiplication to an addition. Can't do that.
You can only distribute the 3 over EITHER the (c - 3d) factor OR the (x + 2y) factor, NOT both. If you try to distribute over both, you would be multiplying by 3 TWICE. Since in the original problem I see a 3x and a 6y, that tells me that I should distribute the 3 over the (x + 2y) factor:
Now distribute the (c - 3d) over the (3x + 6y):
...which what we started with.
Hi pychon,
think in terms of numbers
3(4)5=60 as it's either 3(20) or (12)5, however it's also 4(3)5=4(15)
You can multiply them in any order you like.
3(4)5=3(4)+3(4)+3(4)+3(4)+3(4) or 4(5)+4(5)+4(5) or 3(5)+3(5)+3(5)+3(5).
Alternatively 3(6-2)(8-3) must also equal 60.
We can "distribute" that..
3[(6-2)(8-3)]=3[6(8)+6(-3)-2(8)-2(-3)]=3[48-18-16+6]=3[30-16+6]=3[36-16]=3[20]=60
That's the idea, then the algebra is easy from there.