Karen L.

# equation with variables

hi,
my answer isnt jiving...is it possible the issue is negative sign?
the equation is: 3x-4y=8
then i take 0 for x and 2 for y
what i did was : 3(0)-4(2) =8     0-8=8 is false...unless when i multiply -4(2) because its part of an equation when the 2 comes out of the parentheses its a positive?

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Karen L.

Sorry for not being more specfic...Topic is solving systems
The two equations were: 3x-4y=8  and 2y=2x-4 The order pair that the book suggest is common to both is (0,2) On the first equation I get -8=8 on the second equation I get 4=-4 How are they correct /common?
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03/18/14

Kay G.

That answer is incorrect.  I'm not sure what method you're learning, but I'll do elimination:

First rearrange 2y = 2x - 4 to standard form:
Subtract 2x from both sides:
2y - 2x = 2x - 2x - 4
2y - 2x = -4
Then just turn the x and y terms around (don't forget the - sign is on the x one, and an understood + on the y one):
-2x + 2y = -4

So now we have:
3x - 4y = 8
-2x + 2y = -4

Multiply the 2nd one through by 2:
(2)(-2x + 2y) = -4(2)
-4x + 4y = -8

So we now have:
3x - 4y = 8
-4x + 4y = -8
--------------------
-x           =  0

(You can actually ignore the - sign on that, but just to prove it, you can multiply both sides by -1 and of course you'll still have 0.)

Plug back in to either original equation, but I'll do both:

3(0) - 4y = 8
- 4y = 8   -->  divide both sides by -4
y = -2  (not positive)

2y = 2(0) - 4
2y = -4  --> divide both sides by 2
y = -2

I even ran the whole thing through a second time from scratch, just to make sure I didn't make some silly mistake.  And the positive 2 wasn't working for you either.
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03/18/14

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