anonymous
  • anonymous
Algebra Question
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
If \[\frac{ a }{ 2 } + \frac{ b }{ 2 } = 3 \] what is the value of 2a +2b?
anonymous
  • anonymous
I haven't really been able to get anywhere on it, any form of help will help :D
anonymous
  • anonymous
u still there

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anonymous
  • anonymous
what do u think
anonymous
  • anonymous
yeah im back sorry
anonymous
  • anonymous
@slade well I don't really have any idea on how to start
freckles
  • freckles
I dare you to multiply both sides by 4
anonymous
  • anonymous
thats how u do it right there
freckles
  • freckles
double dog dare you
anonymous
  • 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
  • freckles
do you know what 4/2=?
anonymous
  • anonymous
yes 2
freckles
  • freckles
\[4(\frac{a}{2}+\frac{b}{2}) \\ =\frac{4}{2}a+\frac{4}{2}b=? \]
anonymous
  • anonymous
a/2 and b/2 x 4?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ah gotcha
anonymous
  • anonymous
2a + 2b is what i get! and that means it eqauls 3 !!
freckles
  • freckles
no
freckles
  • freckles
3(4)
anonymous
  • anonymous
but what would make me multiply each side by 4?
anonymous
  • anonymous
oh yeah 12
freckles
  • freckles
to solve for 2a+2b
anonymous
  • anonymous
I like your approach @freckles
anonymous
  • anonymous
2 * 4 =?
anonymous
  • anonymous
I dare you all to learn math :-D double dog dare at that :-D
anonymous
  • 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
  • freckles
well 4(a/2+b/2) would give me what I want which is 2a+2b
anonymous
  • anonymous
i'm trying nixy lol
freckles
  • freckles
but if I multiply one side by 4 I have to multiply the other side by 4
anonymous
  • anonymous
ah yes that is true that is how to get the end result
freckles
  • freckles
do you want to give you a problem that is similar ?
anonymous
  • anonymous
sure
anonymous
  • anonymous
@kevinhunterwood69 it is hard work but if you keep at it, it will get easier :-D
anonymous
  • anonymous
yeah i am
freckles
  • freckles
solve for 6a+9b given \[2a+3b=-1\]
anonymous
  • anonymous
im out
anonymous
  • anonymous
no one is listening to me
anonymous
  • anonymous
lol there
freckles
  • freckles
goodbye @slade :(
freckles
  • freckles
@kevinhunterwood69 do you notice how to find the value of 6a+9b yet?
anonymous
  • 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
  • anonymous
was that correct?
freckles
  • freckles
omg brilliant
anonymous
  • anonymous
yay :D
anonymous
  • anonymous
math is fun when i can finally understand algebra
freckles
  • 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
  • anonymous
yes def fractions have always been a struggle to understand
anonymous
  • anonymous
square roots too but mostly fractions
freckles
  • 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
  • anonymous
hmm
anonymous
  • anonymous
if i move 3 to the other side will it be on top?
anonymous
  • anonymous
actually wait i can't do that right ugh um if i move b let me see
freckles
  • 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
  • anonymous
all i know is im trying to get constants on one side and variables on the other right?
anonymous
  • anonymous
sorry what
anonymous
  • anonymous
how did you get 3a and b lost its 3? oh you moved the 3 from b next to a?
freckles
  • 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
  • anonymous
yes that explains it! thanks so much
anonymous
  • anonymous
well 3a i understand actually haha 3(b/3) = b?
freckles
  • 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
  • freckles
basically in summary 3/3=1 so you are left with 1*b which is just b
freckles
  • 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
  • anonymous
yes i follow up to the point 3/3 = b/1 why does b not become 3b? like 3/3 = 3b/1 ?
freckles
  • freckles
anything over itself is 1 (well except 0)
freckles
  • freckles
well 3/3 *b =b
freckles
  • freckles
first do you know that 3/3=1 ?
anonymous
  • anonymous
yes i do
anonymous
  • anonymous
no time for jokes lol
freckles
  • freckles
\[\frac{3}{3}=1 \\ \text{ so if I multiply both sides by } b \text{ we have } \\ \frac{3}{3}b=1 b\]
anonymous
  • anonymous
yes
freckles
  • freckles
is this not answering your question?
anonymous
  • anonymous
no it did
anonymous
  • anonymous
so we were at 3a + b = 12
anonymous
  • anonymous
and m trying to figure out what is 21a + 7b so
freckles
  • freckles
right and we want 21a+7b 's value
anonymous
  • anonymous
yes
freckles
  • freckles
any if you multiply both sides by s_v__ you will get?
freckles
  • freckles
I didn't put the full word there
anonymous
  • anonymous
ah seven
anonymous
  • anonymous
wait a second
anonymous
  • anonymous
then we multiply 12 by 7 that means :O
freckles
  • freckles
yep! :)
freckles
  • freckles
just so you know we could have figured out 21a+7b in one step
freckles
  • freckles
instead of two
anonymous
  • anonymous
ya so 21a + 7b = 84
freckles
  • 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
  • anonymous
wow
freckles
  • freckles
anyways keep practicing and numbers will become easier to play with
anonymous
  • anonymous
thanks for all the extra practice!
freckles
  • 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
  • anonymous
uh lol
freckles
  • freckles
\[a^2+2ab+b^2=(a+b)^2 \text{ is the trick there }\]
anonymous
  • anonymous
ah
anonymous
  • anonymous
welp i'll go back to this sat book
freckles
  • freckles
k k have fun :)
anonymous
  • anonymous
and ponder on about the funny stuff in math lol
anonymous
  • anonymous
thx frecks
freckles
  • freckles
np :)
mathstudent55
  • mathstudent55
@freckles Great explanations!

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