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Algebra 1 
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Added a second half

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Mon Feb 08, 2010 1:05 am
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It is nearly impossible, it took me 50 minutes just to get somewhere and i figured out that i made a mistake. I am sure this is not Algebra 1, make it easier. (at least erase that square symbol at the end of the whole thing.)

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Mon Feb 08, 2010 1:50 am
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Fine then

Give me a minute to come up with a new one

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Mon Feb 08, 2010 2:18 am
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Make it grade 9 stuff
I was never good at algebra
Geometry is better

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Mon Feb 08, 2010 4:09 am
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No way!

I'm in year 11 so I'm going to make the most of it

Two roots of x^3 +ax^2 +bx - 5 = 0 are equal and rational. Find m

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Mon Feb 08, 2010 4:28 am
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There's no squared in Algebra I, and at most two variables
I'm taking it right now


Also, some people in my grade are taking algebra II.
8th graders taking Algebra II. Geez.

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Mon Feb 08, 2010 5:46 pm
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Goemetry is way harder, but that's only because of that classic saying: Those who excel at Algebra will most likely fail at Geometry and vice versa.

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Mon Feb 08, 2010 5:47 pm
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ok

But screw Algebra 1, just answer the question.

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Mon Feb 08, 2010 5:56 pm
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geez
I don't even know how

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Mon Feb 08, 2010 7:04 pm
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|~DVDV~| wrote:
No way!

I'm in year 11 so I'm going to make the most of it

Two roots of x^3 +ax^2 +bx - 5 = 0 are equal and rational. Find m

old joke

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Mon Feb 08, 2010 7:19 pm
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|~DVDV~| wrote:
ok

But screw Algebra 1, just answer the question.


READ THE TITLE, IT SAYS ALGEBRA 1 NOT YEAR 11 MATH! GEEZ... some people are just morons.
I'm taking Algebra 1 right now, and I know that that problem is no where near Algebra 1, again read the title before posting BS. You would like your teacher giving you a test about Calculus when you're in Algebra 1? I think not.

@Eraser
And yes there is squared but not cubed or square rooted.

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Mon Feb 08, 2010 8:54 pm
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Alright, someone make another if nobody can solve it. It'd also be nice to have a step-by-step solution from the maker.

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Mon Feb 08, 2010 9:03 pm
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There isn't squared, as far as my gifted Algebra I book says.
Then again, I haven't covered everything.


Sure. Here's one:
2x+3y=20
y+3x=2

Solve with substitution. Or if you want to play with the numbers in the second one a while, you could do multiplication with elimination.

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Mon Feb 08, 2010 9:15 pm
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X=-6
y=8

ok my turn,

y=3x/4+21

-9x+3y=66

How many solutions does this linear equation have?

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Mon Feb 08, 2010 10:34 pm
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x² is part of Algebra 1. You probably aren't that far, though.

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Jin wrote:
I support the bombing of Israel.


Tue Feb 09, 2010 1:16 am
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