Algebra Question

- anonymous

Algebra Question

- katieb

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- anonymous

If \[\frac{ a }{ 2 } + \frac{ b }{ 2 } = 3 \] what is the value of 2a +2b?

- anonymous

I haven't really been able to get anywhere on it, any form of help will help :D

- anonymous

u still there

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## More answers

- anonymous

what do u think

- anonymous

yeah im back sorry

- anonymous

@slade well I don't really have any idea on how to start

- freckles

I dare you to multiply both sides by 4

- anonymous

thats how u do it right there

- freckles

double dog dare you

- anonymous

what would that do? um okaay 3 x 4 = 12 on the right but i don't know how to multiply the fractions :/

- freckles

do you know what 4/2=?

- anonymous

yes 2

- freckles

\[4(\frac{a}{2}+\frac{b}{2}) \\ =\frac{4}{2}a+\frac{4}{2}b=? \]

- anonymous

a/2 and b/2 x 4?

- anonymous

yes

- anonymous

ah gotcha

- anonymous

2a + 2b is what i get! and that means it eqauls 3 !!

- freckles

no

- freckles

3(4)

- anonymous

but what would make me multiply each side by 4?

- anonymous

oh yeah 12

- freckles

to solve for 2a+2b

- anonymous

I like your approach @freckles

- anonymous

2 * 4 =?

- anonymous

I dare you all to learn math :-D double dog dare at that :-D

- anonymous

to solve yes but where did you think of 4 why not 3? like is there a specific reason or thats just how algebra works? common factor what was it?

- freckles

well 4(a/2+b/2)
would give me what I want which is 2a+2b

- anonymous

i'm trying nixy lol

- freckles

but if I multiply one side by 4 I have to multiply the other side by 4

- anonymous

ah yes that is true that is how to get the end result

- freckles

do you want to give you a problem that is similar ?

- anonymous

sure

- anonymous

@kevinhunterwood69 it is hard work but if you keep at it, it will get easier :-D

- anonymous

yeah i am

- freckles

solve for 6a+9b
given
\[2a+3b=-1\]

- anonymous

im out

- anonymous

no one is listening to me

- anonymous

lol there

- freckles

goodbye @slade :(

- freckles

@kevinhunterwood69 do you notice how to find the value of 6a+9b yet?

- anonymous

okay so I did:
solve for 6a + 9b
given: 2a + 3b = -1
work:
2a + 3b = -1
3 (2a + 3b) = -1(3)
6a + 9b = -3
answer: -3

- anonymous

was that correct?

- freckles

omg brilliant

- anonymous

yay :D

- anonymous

math is fun when i can finally understand algebra

- freckles

good job
I bet if that was the problem you had you would have had no problems
do you think the fractions throw you off ?

- anonymous

yes def fractions have always been a struggle to understand

- anonymous

square roots too but mostly fractions

- freckles

try this one:
\[\text{ If } a+\frac{b}{3}=4 \text{ then what is the value of } 21a+7b\]
(this will has a fraction in it; let's see if it throws you off any this time )

- anonymous

hmm

- anonymous

if i move 3 to the other side will it be on top?

- anonymous

actually wait i can't do that right ugh um if i move b let me see

- freckles

so you want to multiply both sides 3 to remove the fraction
you can do that
3a+b=12
but we still are not done

- anonymous

all i know is im trying to get constants on one side and variables on the other right?

- anonymous

sorry what

- anonymous

how did you get 3a and b lost its 3? oh you moved the 3 from b next to a?

- freckles

\[a+\frac{b}{3}=4 \\ \text{ if you don't like the fraction there you can always multiply both sides by } \\ \text{ what is \in the denominator of that fraction } \\ 3(a+\frac{b}{3})=3(4) \\ 3(a)+3(\frac{b}{3})=3(4) \\ 3a+b=12\]

- anonymous

yes that explains it! thanks so much

- anonymous

well 3a i understand actually haha 3(b/3) = b?

- freckles

\[3(\frac{b}{3})=3 \cdot \frac{b}{3}=\frac{3}{1} \cdot \frac{b}{3}= \frac{3}{3} \cdot \frac{b}{1}=1 \cdot b =b \]

- freckles

basically in summary 3/3=1 so you are left with 1*b which is just b

- freckles

like in your problem we had above we had:
\[\frac{a}{2}+\frac{b}{2}=3 \\ \text{ we could have cleared the fractions before finding what we wanted } \\ \text{ by multiplying both sides by 2} \\ 2(\frac{a}{2}+\frac{b}{2})=2(3) \\ 2 \frac{a}{2}+2 \frac{b}{2}=2(3) \\ a+b=6\]

- anonymous

yes i follow up to the point 3/3 = b/1 why does b not become 3b? like 3/3 = 3b/1 ?

- freckles

anything over itself is 1
(well except 0)

- freckles

well 3/3 *b =b

- freckles

first do you know that 3/3=1 ?

- anonymous

yes i do

- anonymous

no time for jokes lol

- freckles

\[\frac{3}{3}=1 \\ \text{ so if I multiply both sides by } b \text{ we have } \\ \frac{3}{3}b=1 b\]

- anonymous

yes

- freckles

is this not answering your question?

- anonymous

no it did

- anonymous

so we were at 3a + b = 12

- anonymous

and m trying to figure out what is 21a + 7b so

- freckles

right and we want 21a+7b 's value

- anonymous

yes

- freckles

any if you multiply both sides by s_v__ you will get?

- freckles

I didn't put the full word there

- anonymous

ah seven

- anonymous

wait a second

- anonymous

then we multiply 12 by 7 that means :O

- freckles

yep! :)

- freckles

just so you know we could have figured out 21a+7b in one step

- freckles

instead of two

- anonymous

ya so 21a + 7b = 84

- freckles

\[a+\frac{b}{3}=4 \\ \text{ we wanted } 21a+7b \\ \text{ so we could have just multiplied 21 on both sides \to \begin with } \\ 21a+21(\frac{b}{3})=21(4) \\ 21a+7b=84\]

- anonymous

wow

- freckles

anyways keep practicing
and numbers will become easier to play with

- anonymous

thanks for all the extra practice!

- freckles

np :)
these can get a lot funnier I promise
\[\text{ assume } a \text{ and } b \text{ are positive } \\\text{ if } a^2+2ab+b^2=4 \text{ then what is } (a+b)^4 \text{ or what is } a+b \]
I think they can even get funnier than that

- anonymous

uh lol

- freckles

\[a^2+2ab+b^2=(a+b)^2 \text{ is the trick there }\]

- anonymous

ah

- anonymous

welp i'll go back to this sat book

- freckles

k k
have fun :)

- anonymous

and ponder on about the funny stuff in math lol

- anonymous

thx frecks

- freckles

np :)

- mathstudent55

@freckles Great explanations!

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