What is the simplified form of the quantity of x plus 7, all over the quantity of x plus 3 + the quantity of x minus 4, all over 3?

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What is the simplified form of the quantity of x plus 7, all over the quantity of x plus 3 + the quantity of x minus 4, all over 3?

Mathematics
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hint: the least common multiple between x+3 and 3 is: \[3\left( {x + 3} \right)\]

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am I right?
would it be x2++3x-28/3(x+3)
Yes you are:)
so we can write this: \[\Large \begin{gathered} \frac{{x + 7}}{{x + 3}} + \frac{{x - 4}}{3} = \hfill \\ \hfill \\ = \frac{{3\left( {x + 7} \right) + \left( {x + 3} \right)\left( {x - 4} \right)}}{{3\left( {x + 3} \right)}} = ... \hfill \\ \end{gathered} \] please continue
(x+7)(x+3)(x-4)/3(X+3)
not exactly, since you have to compute the multiplications first
hOW do I do that?
hint:
\[3\left( {x + 7} \right) = 3x + 3 \times 7 = 3x + 21\]
3(x+7)
I'm sorry :(
\[\Large \left( {x + 3} \right)\left( {x - 4} \right) = x \cdot x - 4x + 3x - 4 \cdot 3 = ...\] here, I have applied the foil procedure
please simplify, what do you get?
x^2-1x-12
ok!
so the numerator is: \[\Large 3x + 21 + {x^2} - x - 12 = ...\]
2x+x^2+9
perfect! so your expression becomes: \[\Large \frac{{x + 7}}{{x + 3}} + \frac{{x - 4}}{3} = \frac{{{x^2} + 2x + 9}}{{3\left( {x + 3} \right)}}\]
ohhh! Thank-you! I need one more please?
ok!
What is the simplified form of the quantity of x plus 9, all over the quantity of 2x plus 3 + the quantity of x plus 4, all over the quantity of x plus 2?
please, make a drawing of your expression
ok hold on
here the least common multiple is: \[\left( {2x + 3} \right)\left( {x + 2} \right)\]
is it right?
Yes
Now what?
ok! So we can write: \[\large \frac{{x + 9}}{{2x + 3}} + \frac{{x + 4}}{{x + 2}} = \frac{{\left( {x + 9} \right)\left( {x + 2} \right) + \left( {x + 4} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right)\left( {x + 2} \right)}}\]
now you have to apply the foil method in order to evaluate this: \[\Large \left( {x + 9} \right)\left( {x + 2} \right) = ...?\]
please continue
x2+11x+18
ok! Now do the same with this expression: \[\Large {\left( {x + 4} \right)\left( {2x + 3} \right)}\]
2x2+11x+12
ok! so the numerator is: \[\Large {x^2} + 11x + 18 + 2{x^2} + 11x + 12 = ...\]
2x3+22x+30
are you sure?
yes. wait no
2x4+22x+30
I got: 3 x^2+...
\[\Large {3{x^2} + 22x + 30}\]
is it right?
yes bcuz 1x2+2x2=3x2. I forgot about the imaginary 1
ok! So your original expression, becomes: \[\Large \frac{{x + 9}}{{2x + 3}} + \frac{{x + 4}}{{x + 2}} = \frac{{3{x^2} + 22x + 30}}{{\left( {2x + 3} \right)\left( {x + 2} \right)}}\]
That's not an answer choice, so I guess we will have to foil the bottom also
ok! then what is: \[\Large \left( {2x + 3} \right)\left( {x + 2} \right) = ...?\]
2x2+7x+6
so we get: \[\Large \frac{{x + 9}}{{2x + 3}} + \frac{{x + 4}}{{x + 2}} = \frac{{3{x^2} + 22x + 30}}{{2{x^2} + 7x + 6}}\]
yes! that's right!
Yay! Ok so I want you to keep helping me, and give u medals, so i AM going to close the question and ask another ok?
I'm sorry, I have to go to dinner :(
:( Ok well thanks anyways!
thanks! :)

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