#### TamaPATCHI

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a squared = 3x-25a, and 2x=2(a-3)

She doesn't even know if it is possible.

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- Thread starter TamaPATCHI
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a squared = 3x-25a, and 2x=2(a-3)

She doesn't even know if it is possible.

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Well, according to her, you need to know the **formula** and know how to **combine equations.**

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3x+2(5x+6)=25

Have to find what x equals.

For the second equation you sent...

3x+2(5x+6)=25

3x+10x+12=25

13x+12=25

..... -12 -12

13x=13

13 .. 13

x=1

3x+2(5x+6)=25

3x+10x+12=25

13x+12=25

..... -12 -12

13x=13

13 .. 13

x=1

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lol, i messed up one stepFor the second equation you sent...

3x+2(5x+6)=25

3x+10x+12=25

13x+12=25

-12 -12

13x=13

13 ... 13

x=1

And why are you asking us to solve these problems? I don't mind because deep inside I'm just a big math nerd, but do you just want to see how to solve them?

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Umm, it's just for preparation for the big test I have on March 9th, so I asked A to give me some problems that'd help.

I got x = 1, too. :]For the second equation you sent...

3x+2(5x+6)=25

3x+10x+12=25

13x+12=25

..... -12 -12

13x=13

13 .. 13

x=1

^ How did you do that? Because I don't understand it a bit.

I did the same thing that tamacrazy_101 did.^ How did you do that? Because I don't understand it a bit.

You have to subtract the number that's not multiplying with the variable.

And then you do the same thing to the number on the other side of the equal sign.

Then, you have to get the variable by itself, so you have to do the opposite of what the number is doing to the variable. (Like adding, multiplying, subtracting, and dividing.)

And since the number is multiplying with the variable, and the opposite of multiplying is dividing, you divide it by the number the variable is multiplying with.

I highly doubt that made any sense. xP

I'm sorry. I can't explain it without sounding like my teacher. (My teacher uses like...stories? or something like that to explain the problems to us)

And I have no clue how to solve that o.o

What are you solving for? a or x?

Okay, to start off you have to get rid of the parantheses. () So think of it like this (I changed x to n)

3n [stays the same] + 2 x 5n + 2 x 6 = 25

So basically you take the number attached to the outside of the paranthese (in this case 2) and then multiply it by each number inside (2 x 5n and 2 x 6)

So the equation should look like this now:

3n+10n+12=25

Next add the N's together. (3n and 10n) Now you have:

13n+12=25

Since you want to get the N by itself, you have to get rid of anything that doesn't have an N first. (This case it is the 12) Subtract 12 from both sides so on the left side it cancels out. The answer should turn out like this:

13n=13

Final step is to get rid of what is attached to the N and you do that by dividing both sides by 13. The final answer is now N=1.

I got that, too. ;]

3x+2(5x+6)=25

3x+10x+12=25

13x+12=25

..... -12 -12

13x=13

13 .. 13

x=1

I did something wrong XD

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x_x

Holy.

Oh My.

Ugh.

What The.

I suck at all math.

Period.

Holy.

Oh My.

Ugh.

What The.

I suck at all math.

Period.

I took Algebra last year. I remember these.a squared = 3x-25a, and 2x=2(a-3)

They are called Systems of Equations, I believe. You have to either use Substituition, something-else-I-can't-remember, or graph it. That's because there are two variables.

I THINK I know how to Substitute. Like I said, I took Algebra 1 last year. I'm in Geometry right now, so I haven't really had to apply much more than basic Algebra this year.

Uhm... I'm trying to job my memory here. D: Think Katie. We did these. I remember loving Subsituting. It was so much fun.

I LOVE Algebra. I love numbers, and words. I fail at Geometry - I hate diagrams and shapes.

Okay, ANYWAY -

Oh! The-something-I-don't-remember had to do with cancelling things out.

Ahaha! I did a quick Google search to jog my memory. The three ways to solve Systems of Equations are Graphing, Subsitiution, and Elimination.

Graphing = fail. Don't graph. You can do that for very elementary equation systems. If you don't have a nice integer, it doesn't work. And it's much harder to do.

In Subsitution, the idea is to find out one of the variables first, and then plug it in to find the other one.

Elimination is the one my Algebra teacher pushed. It's where you look for things in common between the two equations, and cancel things out.

Here's a link to a GREAT little site that can help you learn how to do these: https://www.scribd.com/doc/3215442/Algebra-...ms-of-Equations

These are advanced Algebra 1 problems - we didn't get to them until April or May of last year. They may even fall into Algebra 2 in some cases.

I'm not going to solve them for you - what good what that do you? And besides, I'm lazy and it would take me a while to remember how to exactly solve them. -fail-

For the second equation:

I got the same thing as Celestia and tama_crazy did. You have to use the Distributive Property and distribute the '2'.

I should be in Algebra 2 this year. D8< I was recommended to take Algebra 1 in 7th grade, but my parents wouldn't let me take it. They said I wasn't capable. I'll never let them live it down. [/rant]

EDIT: Whoops. Typo.

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