Question

Simplify this expression:
x^2+7x+10/x+2 divided by x^2+2x-15/2x+4

Answer 1

Do you know what a fraction over a fraction is the same as? It relates to multiplication.

Answer 2

\[\frac{\frac{x^2+7x+10}{x+2}}{\frac{x^2+2x-15}{2x+4}}\]

Answer 3

x-3/2x-4

Answer 4

\[\frac{ x^2+7x+10/x+2 }{ x^2+2x-15/2x+4 }\]

Answer 5

@e.mccormick I got x-3/2x+4

Answer 6

I am not sure you have that right. The lower fraction gets inverted for the cancels, so how would the 2x+4 stay on the bottom?

Answer 7

Oh 1/ (2x+4)(x-3)?

Answer 8

\[\frac{\frac{x^2+7x+10}{x+2}}{\frac{x^2+2x-15}{2x+4}}\implies
\frac{x^2+7x+10}{x+2}\cdot \frac{2x+4}{x^2+2x-15}\implies \\
\frac{(x+5)(x+2)}{x+2}\cdot \frac{2x+4}{(x+5)(x-3)}\implies
\frac{1}{1}\cdot \frac{2x+4}{(x-3)}\]

Answer 9

You answer was just inverted.

Answer 10

Oh thank you!

Answer 11

I have more problems if you dont mind helping me?

Answer 12

Just remember that a fraction divided by a fraction is the same as the top fraction times the reciprocal of the bottom fraction. That is what lead me to the answer.

Answer 13

I could really use the help with these!

Answer 14

Do you know what the lowest common demonimatior is like for rational equations? Like the first one.... 18.

Answer 15

I tried it out and I got 12x-2/(3x+1)(3x-1)

Answer 16

How did you get the -2 on the top?

Answer 17

Im sorry it was 12x-6*

Answer 18

Do you have your work? So I can see where your making the mistake?

Answer 19

Sad excuse... my left arm is fractured I cant type that much sorry

Answer 20

was that right?

Answer 21

Neither of those is right.

Answer 22

Factor the bottom of the second fraction. Find what factor is missing in the first fraction. Multiply through by that missing factor over itself. Do NOT multiply out the bottom, but be sure to distribute the top properly. The 5 distributes to both parts of the new left fraction's top. Distribute the negative to both parts of the right fraction's top. Combined like terms. Factor. Cancel.

Answer 23

I redid it agin, 4/3x-1

Answer 24

Yes.

Answer 25

b is correct. Let me guess, you had distributed something wrong before. =) I know because I have made that mistake myself!

Answer 26

yes exactly!

Answer 27

19 I got no solution

Answer 28

So, you got 0=8 or something fun like that?

Answer 29

Yah, it looks like a classic division by 0 error to me. So no real solutions.

Answer 30

and for 18 B was 4/3x+1 not minus 1 that was A

Answer 31

So for 18 was it A or B?

Answer 32

Sorry, got called away for a min.

Answer 33

THose notes are scribbled by now... let me rework.

Answer 34

ok

Answer 35

a. The (3x+1) cancels so (3x-1) is left.

Answer 36

Dunno why I said b. LOL. Looking at the old notes and the new, and I canceled the same thing...

Answer 37

ok so A!

Answer 38

Yes. Absolutely.

Answer 39

ok 20. I got 3 1/3 hours

Answer 40

Not sure... I was never good at setting those up right. I know Lori is faster than Elissa. If your math supports 3 and 1/3, I would say it is good.

Answer 41

what did you get for 21?

Answer 42

I got D) 3

Answer 43

Did you start with this:
\[36^{(3x-1)}=6^{(4x+2)}\implies (6^2)^{(3x-1)}=6^{(4x+2)}\implies \\
2(3x-1)=(4x+2)\]

Answer 44

No, ill try again now

Answer 45

OK, do you se what I used to get there? Exponet rules:\[\large (x^m)^n=x^{m\cdot n}\]

Answer 46

Yes

Answer 47

Then they are exponents of the same number, and they must be equal. Log rules let me drop them, set them equal, and solve.

Answer 48

I got 1/2

Answer 49

\[2(3x-1)=(4x+2)\implies 6x-2=4x+2\implies \\ 6x-2+2=4x+2+2\]Do you see your mistake that would get you 1/2 when I do it that way?

Answer 50

You subtracted off a 2 rather than add in a 2.

Answer 51

Or something like that.

Answer 52

Yes

Answer 53

I got 2 now

Answer 54

=)

Answer 55

Now, on 22, do you know the big thing you need to be careful of?

Answer 56

No what's that?

Answer 57

\[\Large \log_b(y)=x\Leftrightarrow y=b^x\]If x is positive, can y be negative?

Answer 58

No

Answer 59

Exactly, so you need to test your answers in the originl functions.

Answer 60

Ok

Answer 61

I tried it and got 5 then I checked and got no solution

Answer 62

hmmm... I got something else. What did you start with?

Answer 63

I meant -5

Answer 64

I got -5 and 2. When I checked, one of those went by-by.

Answer 65

2?

Answer 66

When you put 2 into the original equation, are you ending up with a + or a - number in the log?

Answer 67

I understand

Answer 68

\[5-3x=x^2-5\implies 0=x^2-5-5+3x\implies x^2+3x-10=0\implies \\
(x-2)(x+5)=0\implies x=\{-5,2\} \\ \text{OK, so two to test.}\\
5-3(2)\implies 5-6=-1\\
2^2-5\implies 4-5=-1\\
5-3(-5)\implies 5+15=20\\
(-5)^2-5\implies 25-5=20
\]Do you see what I did there? That is the entire process. You have to be careful of the small mistakes.

Answer 69

23 I got A. 11

Answer 70

Now, can I take a log, any log, of -1? 20? If yes to -1, then the answer is 2. If yes to 20, then the answer is -5. If the answer is yest to both it is both, no to both is nnone.

Answer 71

I got -5,2

Answer 72

Rememeber what I said earlier about the log of a negative number? It does not exist. Well, it is a complex number... so if you try \(\log_6(-1)\) it does not exist.

Answer 73

That is the whole point of testing things. If you get a negative value, there is no real log.

Answer 74

23 I got A. 11

Answer 75

Yes, and I am wondering how.

Answer 76

wait... let me make sure I did not make a mistake on one thing there.

Answer 77

I did.. but I still did not get 11.

Answer 78

Ok I'll check too

Answer 79

WHat are you doing with the fractions in front of the logs?

Answer 80

Wait I see my mistake

Answer 81

OK.

Answer 82

14

Answer 83

yah. THat is what I got.

Answer 84

On the last one, have they told you what base they standardly use for antilog? Because people tend to use either 10 or e, but sometimes they just say log or antilog and mean base e rather than 10!

Answer 85

Find what base it is in this book, then 24 is a calculator problem.

Answer 86

Ok

Answer 87

how do I find the antilogrithm?

Answer 88

I got 817

Answer 89

@e.mccormick is that right?