# College Algebra refresher

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This is my first math class in a LONG time, I am taking college algebra, whatever that means. I am a bit rusty with the process. How do you get rid of the negative sign on a variable? Solve for y so points can be plotted using a T table
3x-y=6
-y= 6-3x or is it 3x-6?
?// now what?


It is probably real simple but it escapes me at the moment. Any help appreciated!!

1: 3x - y = 6
2: 3x = 6 + y
3: 3x - 6 = y
ans: y = 3x - 6

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Quote:
 Original post by bluefox25-y = 6-3x

-y * -1 = (6-3x) * -1

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Sou would it be
y= 3x-6 or y= 6-3x

It makes a difference right?

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Quote:
 Original post by smurfkiller1: 3x - y = 62: 3x = 6 + y3: 3x - 6 = yans: y = 3x - 6

I had this in mind but wasn't sure if you could switch the sides of the equation. Like i siad its been awhile..........

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Quote:
 Original post by bluefox25Sou would it bey= 3x-6 or y= 6-3xIt makes a difference right?

-y * -1 =   (6-3x) * -1   y    = (6 * -1) - (3x * -1)   y    =   (-6)   -   (-3x)   y    =    -6    +    3x   y    =    3x-6

Introduction to algebra, simplifying equations.

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Would this be right?
x-4y=5
x=5+4y
x-5=4y
x-5
___ = y
4
So if x = -3,-2,-1,0,1,2,3
(-3,-2)
(-2,-7/4)
(-1,2/3)

Come to think of it, this isn't right is it? Canyou have a fraction as coordinate?
Maybe it should be
y= 1/4x+(-5)
?

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Quote:
 Original post by bluefox25So if x = -3,-2,-1,0,1,2,3(-3,-2)(-2,-7/4)(-1,2/3)Come to think of it, this isn't right is it?

You were doing it just right.

(-3,-8/4)
(-2,-7/4)
(-1,-6/4) //not 2/3
( 0,-5/4)
( 1,-4/4)
( 2,-3/4)
...

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Thanks!1
So I am not totally out of my mind yet!1
Wish I hadnt taken this class oer summer
4hrs a day for class * 4 days a week = TOO MUCH MATH!!!

So there is nno need to reduce the fractions? Any particular reason?

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Let see if I can get this one
4x-3y=12
4x=12+3y
4x-12=3y
4x-12
_____ = y
3

y= 4x-4

Am I right?

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Quote:
 Original post by bluefox25Let see if I can get this one4x-3y=124x=12+3y4x-12=3y4x-12_____ = y3y= 4x-4Am I right?

You do the divison incorrectly. When you have:
4x-12=3y
then what you do is divide both sides by 3 and get:
(4x-12)/3=3y/3
Clearly 3y/3=y, so we have:
(4x-12)/3=y
When we have multiple terms we divide every term seperately so we'll get:
4x/3-12/3=y
So:
y = 4x/3-4 is the final solution.

You need to refresh your arithmetical rules a little if you want to solve it successfully.

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If instead of trying to smplfy the equation
and if I left it like this:
4x-12
_____ = y
3
then that would acceptable too right like when I had
x-5
___ = y(in the prev problem)
4

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Quote:
 Original post by bluefox25Let see if I can get this one4x-3y=124x=12+3y4x-12=3y4x-12_____ = y3y= 4x-4Am I right?

The first step you should have followed is divide all by 3;
4x   - 3y = 12  4x/3 -  y = 12/3y = 4x/3 - 4

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(-x-x) squared would be ?
2x squared?

ok thanx

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Quote:
 Original post by bluefox25(-x-x) squared would be ?2x squared?

for (x-3)squared:
Square of a sum

-x-x = 0;
 Yeah! I did write that! Now I'll feel stupid for about a month or so![/edit]

[Edited by - Zanshibumi on June 5, 2007 6:51:37 AM]

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Quote:
 Original post by Zanshibumi-x-x = 0;

No, that is only the case when x=0. -x-x = -2x.

Quote:
 (-x-x) squared would be ?2x squared?

Instead of trying to remember a lot of complicated rules try to consider what you're doing. - x - x must equal -2x, which is a product of two numbers (-2 and x). When squaring a product we can just square its factors separately like so:
(-2x)2 =
(-2)2x2 =
4x2
Remember that when we say 2x we just mean 2 times x. Try to think about what you're doing instead of remembing rules for all kind of stuff. For instance if you've got:
(x+2)2
How would you expand it? Squaring is simply a multiplication by itself so we have:
(x+2)(x+2)
To simplify it a little we introduce another identifier, y which we makes equal to (x+2), then we have:
(x+2)y
We then multiply into the parantheses:
xy+2y
We then expand the y to what it actually stood for:
x(x+2)+2(x+2)
We then multiply into the parantheses again:
xx+2x+2x+2*2
Since 2x+2x=4x, 2*2=4 and xx=x2 we get:
x2+4x+4
Simple if you just give it a little thought and know your basic rules. Also you don't really need to introduce the extra identifier y, but I felt it would help show how you can multiply into parantheses even when you multiply by a sum.

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Thanx that clarifies things alot. I might just hold off on taking this course over the summer. It is insane, the amount of homework there is. We had about 150 problems due today!! I still have about 10 problems to go in the first section and I still got 3 more sections to go!!

And its gona be like that everyday all week long!

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Folks,

I've been a bit more relaxed on forum schoolwork policies lately, for various reasons. But, my basic position is still the same. If someone posts a simple academic type problem (e.g., how do i solve this types of questions), that isn't focused on game development, please do not give them the answer! Give them hints to help them to learn how to better solve the problem on their own. My opinion is that giving answers may harm some (though not all) people who might ask.

As for whether such posts are allowed in the future, I will continue to make that decision as forum moderator. It will in most cases depend on how the question is asked, amonst other things.

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