Multiplying algebraic fractions: 2/a + (a+2)/4 I understand that the 2 and the 4 can cancel out to become 1/a + (a+2)/2

[lucyfoxwell]

2/a + (a+2)/4

I understand that the 2 and the 4 can cancel out to become a 1 and a 2 as seen in my working out below:

1/a + (a + 2)/2

I would have thought that I could cancel the a from the numerator and denominator but according to the answer in the textbook I can't cancel out the two a's. My question is: why can't I cancel out the a's????

Thank you!!!

 

[Dr.Peterson]

2/a + (a+2)/4

I understand that the 2 and the 4 can cancel out to become a 1 and a 2 as seen in my working out below:

1/a + (a + 2)/2

I would have thought that I could cancel the a from the numerator and denominator but according to the answer in the textbook I can't cancel out the two a's. My question is: why can't I cancel out the a's????

Thank you!!!

You can only cancel factors from the same fraction. Assuming you meant what you wrote, [imath]\frac{2}{a}+\frac{a+2}{4}[/imath], the 2 in the second fraction is not a factor of the numerator, so it can't cancel with the 4, and the 2 in the first fraction is in the numerator of the wrong fraction, so it too can't cancel with the 4.

And the a's, likewise, are not in the same fraction.

Please show your answer, and the answer in the book (and, ideally, an image of the problem as given) so we can be sure we're talking about the same things.

 

[Otis]

2/a + (a+2)/4

I understand that the 2 and the 4 can cancel out to become…

1/a + (a + 2)/2

Hello. Did you intend to multiply those two ratios? (The subject line says so.) You put a plus sign between them.

If the answer you looked at is (a+2)/(2a) then the two ratios were multiplied. The reason you can't cancel the a in the numerator is because it's attached to the number 2 by addition. In other words, there is no factor a in the numerator to cancel with the factor a in the denominator. (The numerator represents the number that is 2 more than a.)

 

[lucyfoxwell]

 

[lucyfoxwell]

My apologies everybody, I meant to multiply the two fractions as seen in the image above. Sorry for the confusion

 

[lucyfoxwell]

Thank you for the replies. Could someone please explain very simply what you mean by only being able to cancel factors? I'm still struggling to understand why I can't cancel out the two a's.

 

[Otis]

please explain very simply what you mean by only being able to cancel factors

Hi. Factors are what we call quantities that are multiplied together.

If you had the fraction 25/150, you wouldn't cancel 5s just because you see one on top and bottom, right? But if you factor the numerator and denominator, then you may cancel the common factors.

(5x5)/(2x3x5x5) = 1/(2x3)

In algebra, numbers are expressed in many forms. Each of the following three lines represents a single number (value).

[imath]a + 2[/imath]
[imath]\sqrt{5}[/imath]
[imath]\frac{b(a–1)}{3b}[/imath]

If we write those three quantities as multiplied together, then they are factors.

\(\displaystyle \sqrt{5}\;(a + 2)\bigg( \frac{b(a–1)}{3b} \bigg)\)

Let's look at the third factor above.

\(\displaystyle \bigg( \frac{b(a–1)}{3b} \bigg)\)

In this algebraic fraction, we see two factors in the numerator and two factors in the denominator. Can you see that b on top and b on bottom are both factors? They cancel, and the ratio is simplified.

\(\displaystyle \bigg( \frac{a–1}{3} \bigg)\)

 

[lucyfoxwell]

Hi. Factors are what we call quantities that are multiplied together.

If you had the fraction 25/150, you wouldn't cancel 5s just because you see one on top and bottom, right? But if you factor the numerator and denominator, then you may cancel the common factors.

(5x5)/(2x3x5x5) = 1/(2x3)

In algebra, numbers are expressed in many forms. Each of the following three lines represents a single number (value).

[imath]a + 2[/imath]
[imath]\sqrt{5}[/imath]
[imath]\frac{b(a–1)}{3b}[/imath]

If we write those three quantities as multiplied together, then they are factors.

\(\displaystyle \sqrt{5}\;(a + 2)\bigg( \frac{b(a–1)}{3b} \bigg)\)

Let's look at the third factor above.

\(\displaystyle \bigg( \frac{b(a–1)}{3b} \bigg)\)

In this algebraic fraction, we see two factors in the numerator and two factors in the denominator. Can you see that b on top and b on bottom are both factors? They cancel, and the ratio is simplified.

\(\displaystyle \bigg( \frac{a–1}{3} \bigg)\)

Thanks so much Otis. This really helped me

 

[Steven G]

Remember that a is a place holder for a number. It is a number! The thing is that you are given the value.

Suppose you had (2+3)/2. I hope that you that it equals 5/2.
Now if we cancel out the 2's, then we got either 4 or 3.
You get 4, if you cancel out the 2's and replace it is 1 (after all 2/2=1). That is 1+3 = 4. But that not 5/2!
You get 3 if you simply cross out the 2's as if they were not there.


Now consider the following. 2/3 + 5/3. Now this equals (2+5)/3 =7/3. Note that in (2+3)/5 that BOTH the 2 and the 3 are divided by 5! That is why (2+3)/5 = 2/5 + 3/5