Perform indicated operations and simplify

[ScienceJen]

Hi everyone - I think this is considered beginning algebra. I'm brushing up on this because I'm taking calculus, and I need help on number 28 - please and thank-you. If you also feel like doing #27, I would appreciate that.

- Jen

 

[pka]

Multiply each term by $(c-1)$

 

[topsquark]

For 28 where are you having problems? We can help you better if you tell us where your problem is. For example, do you know how to simplify
$\dfrac{1}{ 1 + \dfrac{1}{1 + x} }$

-Dan

 

[ScienceJen]

For 28 where are you having problems? We can help you better if you tell us where your problem is. For example, do you know how to simplify
$\dfrac{1}{ 1 + \dfrac{1}{1 + x} }$

-Dan

Hi Dan,

I'm not sure where to begin. The fraction in the denominator is throwing me off. Do I multiply the reciprocal? I think what I did was multiply the denominator by the denominator and the numerator - same as if I were trying to rationalize the denominator, but I forgot to change the + to a - sign.

Thanks again,

Jen

 

[lookagain]

For # 27, see what you get by multiplying each term by (c - 1).

 

[ScienceJen]

For # 27, see what you get by multiplying each term by (c - 1)

Hi Lookagain,

Thank you. I get 1. Is that right?

 

[lookagain]

Hi Lookagain,

Thank you. I get 1. Is that right?

It is not correct.

Did you get it to $\displaystyle \ \dfrac{c - 1 \ + \ 1}{c - 1 \ - \ 1} \$?

I cannot respond more this evening.

 

[Steven G]

Simplify the numerator, then simplify the denominator then and only then simplify the fraction.

 

[topsquark]

Hi Dan,

I'm not sure where to begin. The fraction in the denominator is throwing me off. Do I multiply the reciprocal? I think what I did was multiply the denominator by the denominator and the numerator - same as if I were trying to rationalize the denominator, but I forgot to change the + to a - sign.

Thanks again,

Jen

Big hint: $\dfrac{1}{1 + \dfrac{1}{x + 1} } \cdot \dfrac{x + 1}{x + 1} =$

-Dan

 

[ScienceJen]

It is not correct.

Did you get it to $\displaystyle \ \dfrac{c - 1 \ + \ 1}{c - 1 \ - \ 1} \$?

I cannot respond more this evening.

Oh, yes. I get that now, thank you. My "final answer" is -1/2.

 

[lookagain]

Oh, yes. I get that now, thank you. My "final answer" is -1/2.

That is not correct. What does the numerator simplify to? What does the denominator simplify to?

 

[Deleted member 4993]

Oh, yes. I get that now, thank you. My "final answer" is -1/2.

Please share "steps" you took to reach your answer. We would like to guide you to the correct answer - once we know where to "fix" our attention.

 

[Steven G]

Oh, yes. I get that now, thank you. My "final answer" is -1/2.

Do you really think that no matter what the value of c is, the fraction will always be -1/2? You have to believe that if you think that the answer is -1/2.

 

[blamocur]

Do you really think that no matter what the value of c is, the fraction will always be -1/2? You have to believe that if you think that the answer is -1/2.

This reminds me of a simple "sanity check": whichever answer you get, pick an arbitrary $c$ and check whether both the original and the simplified expressions get the same value.

 

[ScienceJen]

Simplify the numerator, then simplify the denominator then and only then simplify the fraction.

I never thought of multiplying as simplifying, but I guess that's what it is sometimes. Thank you.

That is not correct. What does the numerator simplify to? What does the denominator simplify to?

I get c-1+1 for the numerator and c-1-1 for the denominator. Next I got c-1+1 = c for the numerator and c-1-1 = c-2 for the denominator. I thought then that the c in the numerator and the c in the denominator would cancel out, leaving me with -1/2. I'm not sure what I'm doing wrong. Thank you for all of your help! My "tutor" (online instructor for introductory calculus) has informed me that he has 160 students and told me I need to "'dig my heels in more and try to figure it out on my own."' I just need to know the basic concepts and then I'm sure it will "click," as math needs to do.
Thank you so much for your help, everybody!

Please share "steps" you took to reach your answer. We would like to guide you to the correct answer - once we know where to "fix" our attention.

I get c-1+1 for the numerator and c-1-1 for the denominator. Next I got c-1+1 = c for the numerator and c-1-1 = c-2 for the denominator. I thought then that the c in the numerator and the c in the denominator would cancel out, leaving me with -1/2. I'm not sure what I'm doing wrong. Thank you for all of your help! My "tutor" (online instructor for introductory calculus) has informed me that he has 160 students and told me I need to "'dig my heels in more and try to figure it out on my own."' I just need to know the basic concepts and then I'm sure it will "click," as math needs to do.
Thank you so much for your help, everybody!

Do you really think that no matter what the value of c is, the fraction will always be -1/2? You have to believe that if you think that the answer is -1/2.

Hmmm....never thought of that before...well, no. It looks like "should be" -1, but I don't know if that's right or how to get there. I know this is a simple equation and that I'm missing some sort of step or concept.
Thank you.

Wait a seo

Hmmm....never thought of that before...well, no. It looks like "should be" -1, but I don't know if that's right or how to get there. I know this is a simple equation and that I'm missing some sort of step or concept.
Thank you. ....

Wait a second....if it's c-1+1/c-1-1...then, the c on numerator and c on denominator cancel out and -1s will cancel out...leaving us with -1???

Big hint: $\dfrac{1}{1 + \dfrac{1}{x + 1} } \cdot \dfrac{x + 1}{x + 1} =$

-Dan

Thank you.

 

[topsquark]

Oh, yes. I get that now, thank you. My "final answer" is -1/2

Wait a seo

Wait a second....if it's c-1+1/c-1-1...then, the c on numerator and c on denominator cancel out and -1s will cancel out...leaving us with -1???

Please use parenthesis. $c - 1 + 1/c - 1 -1 = c - 1 + \dfrac{1}{c} -2$.

$\dfrac{c - 1 + 1}{c - 1 - 1} = \dfrac{c}{c - 2}$

You can only cancel the c's if both the numerator and denominator have factors of c in them. $\dfrac{c}{c - 2} \neq \dfrac{1}{-2}$. Try it for a couple of values of c if you don't see this.

-Dan

 

[ScienceJen]

Please use parenthesis. $c - 1 + 1/c - 1 -1 = c - 1 + \dfrac{1}{c} -2$.

$\dfrac{c - 1 + 1}{c - 1 - 1} = \dfrac{c}{c - 2}$

You can only cancel the c's if both the numerator and denominator have factors of c in them. $\dfrac{c}{c - 2} \neq \dfrac{1}{-2}$. Try it for a couple of values of c if you don't see this.

-Dan

Yeah, I see if I substitute c = 3 into the equation, I can see that it will not equal -1/2. Thank you.

 

[ScienceJen]

Yeah, I see if I substitute c = 3 into the equation, I can see that it will not equal -1/2. Thank you.

So, I guess the answer is (c/c-2) then?

 

[Deleted member 4993]

So, I guess the answer is (c/c-2) then?

No!

The correct expression is c/(c-2)

Those parentheses are SUPER IMPORTANT and must be included and must be placed in correct position.

Now tackle #28 in a similar way.

 

[ScienceJen]

No!

The correct expression is c/(c-2)

Those parentheses are SUPER IMPORTANT and must be included and must be placed in correct position.

Now tackle #28 in a similar way.

Ok! Thank you. Got it.