Question

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Question below (in comments)

Answer 1

What is \[\frac{ x+9 }{ 2x+3 } + \frac{ x+4 }{ x+2 }\]

Answer 2

@mathmale @ganeshie8 @Mehek14

Answer 3

? it looks like a rational expression... what more do you need?

Answer 4

I need to simplify it.

Answer 5

So I need to add the fractions, but I don't know how.

Answer 6

so you need to combine the terms into a single fraction, yes? if so, you need a common denominator to do that

Answer 7

Okay. \[(2x+3)(x+2) = 2x^2 + 7x + 6,\]

Answer 8

So is that the common denominator?

Answer 9

good, but don't multiply out...
now you need both terms to have that as a denominator

Answer 10

Okay, so the new equation is \[\frac{ x^2 + 11x + 12 }{ 2x^2 + 7x + 6 } + \frac{ 2x^2 + 11x + 12 }{ 2x^2 + 7x + 6 }\]

Answer 11

Am I doing it right so far?

Answer 12

again, don't multiply out right away... just when it's needed. you're going to end up factoring and cancelling, if possible...\[\frac{ x+9 }{ 2x+3 }+\frac{ x+4 }{ x+2 }=\frac{ x+9 }{ 2x+3 }\cdot\left(\frac{ x+2 }{ x+2 }\right)+\frac{ x+4 }{ x+2 }\cdot\left(\frac{ 2x+3 }{2x+3 }\right)\]\[=\frac{ \left(x+9\right) \cdot\left( x+2\right)+\left( x+4\right)\cdot\left(2x+3\right)}{ \left(2x+3\right) \cdot\left(x+2 \right)}\]Now multiply on the top only and combine like terms... then factor the resulting poynomial to see if any factors cancel and you're done!

Answer 13

btw 2*9 = 18

Answer 14

Okay, I did that, and I got \[\frac{ 3x^2+22x+30 }{ 2x^2 + 7x + 6 }. \] IS that the correct answer?

Answer 15

*Is

Answer 16

sorry, but if you're not going to follow my advice I'm not going to be able to help you. good luck!

Answer 17

not that I wouldn't (be able to help) just that you aren't listening (so that I can help you)

Answer 18

I simplified the fraction that you gave me like you said to, by multiplying and combining like terms. What did I do wrong?

Answer 19

@pgpilot326

Answer 20

Jerk