Perform indicated operations and simplify

[ScienceJen]

Ok! Thank you. Got it.

Here's what I got for #28: First, I multiplied the numerator and the denominator by (1+x) and I got: (1+x+1+x)/(1+x+1+x+1). Then canceling out the 2x/2x portion, I end up with 2/3. ?

 

[Deleted member 4993]

Here's what I got for #28: First, I multiplied the numerator and the denominator by (1+x) and I got: (1+x+1+x)/(1+x+1+x+1). Then canceling out the 2x/2x portion, I end up with 2/3. ?

That is correct idea but incorrect execution.

\(\displaystyle 1 + \ \frac{1}{1 + \frac{1}{1 + x}}\)

= \(\displaystyle 1 + \ \left[ \frac{1}{1 + \frac{1}{1 + x}}\right] \ \ * \ \frac{x+1}{x+1}\)

= \(\displaystyle 1 + \ \frac{1 * (x+1)}{\left(1 + \frac{1}{1 + x}\right) * (x+1)}\ \ \)

= \(\displaystyle 1 + \ \frac{\left( x+1\right) }{ \left( x + 1 + 1 \right) }\ \ \)

Continue and finish it.......

 

[topsquark]

Here's what I got for #28: First, I multiplied the numerator and the denominator by (1+x) and I got: (1+x+1+x)/(1+x+1+x+1). Then canceling out the 2x/2x portion, I end up with 2/3. ?

[math]\dfrac{1}{1 + \dfrac{1}{x + 1} } \cdot \dfrac{x + 1}{x + 1} = \dfrac{x + 1}{(x + 1) + 1} = \dfrac{x + 1}{x + 2}[/math]
Remember you can't cancel anything out unless it is a factor of both numerator and denominator!!

Can you finish from here?

-Dan

 

[ScienceJen]

That is correct idea but incorrect execution.

\(\displaystyle 1 + \ \frac{1}{1 + \frac{1}{1 + x}}\)

= \(\displaystyle 1 + \ \left[ \frac{1}{1 + \frac{1}{1 + x}}\right] \ \ * \ \frac{x+1}{x+1}\)

= \(\displaystyle 1 + \ \frac{1 * (x+1)}{\left(1 + \frac{1}{1 + x}\right) * (x+1)}\ \ \)

= \(\displaystyle 1 + \ \frac{\left( x+1\right) }{ \left( x + 1 + 1 \right) }\ \ \)

Continue and finish it.......

[math]\dfrac{1}{1 + \dfrac{1}{x + 1} } \cdot \dfrac{x + 1}{x + 1} = \dfrac{x + 1}{(x + 1) + 1} = \dfrac{x + 1}{x + 2}[/math]
Remember you can't cancel anything out unless it is a factor of both numerator and denominator!!

Can you finish from here?

-Dan

Thank you for the reminder that I can't cancel anything out unless it is a factor of both the numerator and the denominator - I'm having trouble with knowing when values can cancel out. By "factor" do you mean values that have been factored out? For example, (x^2 - 9) = (x+3)(x-3)...and if there is (x+3) on the denominator, then the (x+3)s can cancel out?

So, from here: 1 + (x+1)/(x+2): I'm wondering if the x can cancel out, leaving us with 1+1/2 = 1 1/2 ?
It doesn't seem right to me, though....I just don't know what else to do.

 

[Deleted member 4993]

Thank you for the reminder that I can't cancel anything out unless it is a factor of both the numerator and the denominator - I'm having trouble with knowing when values can cancel out. By "factor" do you mean values that have been factored out? For example, (x^2 - 9) = (x+3)(x-3)...and if there is (x+3) on the denominator, then the (x+3)s can cancel out?

So, from here: 1 + (x+1)/(x+2): I'm wondering if the x can cancel out, leaving us with 1+1/2 = 1 1/2 ?
It doesn't seem right to me, though....I just don't know what else to do.

NO! you cannot cancel 'x' - 'x' has NOT been factored out.

Now work with 1 + (x+1) / (x+2)

How would you simplify:

1 + (1+3)/(2+ 3) .......................... show all the steps and do not use calculator!

and

1/7 + 4/5 .......................... show all the steps and do not use calculator!

Also re-read response #16 and think about it (with pencil and paper}

 

[topsquark]

Thank you for the reminder that I can't cancel anything out unless it is a factor of both the numerator and the denominator - I'm having trouble with knowing when values can cancel out. By "factor" do you mean values that have been factored out? For example, (x^2 - 9) = (x+3)(x-3)...and if there is (x+3) on the denominator, then the (x+3)s can cancel out?

So, from here: 1 + (x+1)/(x+2): I'm wondering if the x can cancel out, leaving us with 1+1/2 = 1 1/2 ?
It doesn't seem right to me, though....I just don't know what else to do.

Is x a factor of x + 1? If it were then x + 1 = x( some factor). Can you find one?

[imath]1 + \dfrac{x + 1}{x + 2} = \dfrac{1}{1} + \dfrac{x + 1}{x + 2}[/imath]. How do you add fractions?

-Dan

 

[ScienceJen]

NO! you cannot cancel 'x' - 'x' has NOT been factored out.

Now work with 1 + (x+1) / (x+2)

How would you simplify:

1 + (1+3)/(2+ 3) .......................... show all the steps and do not use calculator!

and

1/7 + 4/5 .......................... show all the steps and do not use calculator!

Also re-read response #16 and think about it (with pencil and paper}

1 + (1+3)/(2+3) = Common denominator, therefore 5/5 + 4/5 = 9/5

1/7 + 4/5 = common denominator of 35, so multiply 1 and 7 by 5, and 4 and 5 by 7.....giving 5/35 and 28/5 = 33/35.

As for response #16, I will do that and try again...

 

[Deleted member 4993]

1 + (1+3)/(2+3) = Common denominator, therefore 5/5 + 4/5 = 9/5

1/7 + 4/5 = common denominator of 35, so multiply 1 and 7 by 5, and 4 and 5 by 7.....giving 5/35 and 28/5 = 33/35.

As for response #16, I will do that and try again...

Excellent!!

 

[Deleted member 4993]

1+\(\displaystyle \frac{x+2}{x+1} = \ \frac{1}{1} \+ \ \frac{x+2}{x+1}\). How do you add fractions?

1 + {x+2}/{x+1} = {1}/{1} + {x+2}/{x+1} ?

 

[topsquark]

1 + {x+2}/{x+1} = {1}/{1} + {x+2}/{x+1} ?

Ummm... I'm not sure why you seem to be trying to flip the fraction? .................................fixed it above

-Dan

 

[Deleted member 4993]

Ummm... I'm not sure why you seem to be trying to flip the fraction?

-Dan

Sorry mistake ...... transcription error. Eyes went cross-eyed . Fixed it above (just don't tell Steve)

 

[topsquark]

Sorry mistake ...... transcription error. Eyes went cross-eyed . Fixed it above (just don't tell Steve)

Just slip me a cupcake the next time I'm in the corner.

-Dan

 

[ScienceJen]

1 + {x+2}/{x+1} = {1}/{1} + {x+2}/{x+1} ?

I would multiply the (1/1) denominator by (x+1), therefore = (x+1)/(x+1) + (x+2)/(x+1) = (2x+3)/(x+1) ? Now, I'm thinking that should be it, because I didn't factor out any of those values. I don't think I can simplify further, but a part of me thinks the x's could cancel out...apparently this isn't the case. Even though I don't quite get it, I think I'm getting there. Thank you everybody.

 

[ScienceJen]

Excellent!!

Thank you.
Now for #16: (c−1−1)/(c−1+1) = (c-2)/c
I think we went through that before, though. ?

 

[topsquark]

I would multiply the (1/1) denominator by (x+1), therefore = (x+1)/(x+1) + (x+2)/(x+1) = (2x+3)/(x+1) ? Now, I'm thinking that should be it, because I didn't factor out any of those values. I don't think I can simplify further, but a part of me thinks the x's could cancel out...apparently this isn't the case. Even though I don't quite get it, I think I'm getting there. Thank you everybody.

[imath]1 + \dfrac{1}{1 + \dfrac{1}{x + 1}} = 1 + \dfrac{1}{1 + \dfrac{1}{x + 1}} \cdot \dfrac{x + 1}{x + 1} = 1 + \dfrac{x + 1}{x + 2} = \dfrac{2x + 3}{x + 2}[/imath]

Looks good!

As to the factoring problem the only thing I can suggest is that you check one or two number examples before you try to cancel in order to check your work. Note that
[imath]\dfrac{2x + 2}{4x + 2} = \dfrac{2(x + 1)}{2(2x + 1)} = \dfrac{\cancel{2}(x + 1)}{\cancel{2}(2x +1)} = \dfrac{x + 1}{2x + 1}[/imath]

and
[imath]\dfrac{2x + 2}{4x + 3} \neq \dfrac{2 \cancel{x} + 2}{4 \cancel{x} + 3}[/imath]

-Dan

 

[ScienceJen]

[imath]1 + \dfrac{1}{1 + \dfrac{1}{x + 1}} = 1 + \dfrac{1}{1 + \dfrac{1}{x + 1}} \cdot \dfrac{x + 1}{x + 1} = 1 + \dfrac{x + 1}{x + 2} = \dfrac{2x + 3}{x + 2}[/imath]

Looks good!

As to the factoring problem the only thing I can suggest is that you check one or two number examples before you try to cancel in order to check your work. Note that
[imath]\dfrac{2x + 2}{4x + 2} = \dfrac{2(x + 1)}{2(2x + 1)} = \dfrac{\cancel{2}(x + 1)}{\cancel{2}(2x +1)} = \dfrac{x + 1}{2x + 1}[/imath]

and
[imath]\dfrac{2x + 2}{4x + 3} \neq \dfrac{2 \cancel{x} + 2}{4 \cancel{x} + 3}[/imath]

-Dan

Ok, thanks!
I understand factoring out the 2. So, the only time we would be able to factor out the x is if it was, for example, 2x^2 + 2x, then we could factor out 2x. Same as if the denominator was 4x^2 + 6x, the 2x could factor out and become 2x + 3.