Need help with variables and fractions

[Vampire_Rex]

i'm starting algebra on my own and i need a little help with adding fractions with variables. one of the questions is 9/16t + 5/12. how exactly does the variable end up in both the numerator and denominator? i know it's probably so basic but some explanation would help a lot. i know the answer i just don't know how to get to the answer on my own. answer is 27+20t/48t
and i'm sorry if i'm doing this all wrong maybe i'm posting something i'm not supposed to but i have yet to read the rules of the forum. just wanted help right away if possible.

 

[Otis]

i'm starting algebra on my own and i need a little help with adding fractions with variables.

one of the questions is 9/16t + 5/12. how exactly does the variable end up in both the numerator and denominator?

You need a common denominator, before adding fractions. In order to get a factor of t into the denominator of 5/12, we need to multiply both the top and bottom by t. That's how t gets on top.

i know it's probably so basic but some explanation would help a lot. i know the answer i just don't know how to get to the answer on my own. answer is (27+20t)/48t

Note the grouping symbols, added around the numerator above. It's very important to type grouping symbols, when a numerator or denominator contains more than one term. Without the grouping symbols, the expression above would indicate that the number 27 is not part of the ratio.

Do you remember how to add fractions, when none of them contain any algebraic symbols? If you don't, let us know, and we will provide link(s) to lessons. The process is much the same, when adding algebraic fractions (ratios).

Adding 9/16t + 5/12 requires a common denominator. That would be 48t.

We multiply 9/16t * 3/3 to get 27/48t

We multiply 5/12 * 4t/4t to get 20t/48t

Now we have 27/48t + 20t/48t, and the addition happens with the numerators; the common denominator stays the same. Just like when adding regular fractions.

(27 + 20t)/48t

 

[lookagain]

one of the questions is 9/16t + 5/12.

how exactly does the variable end up in both the numerator and denominator?
answer is 27+20t/48t

That first expression could mean \(\displaystyle \ \ \dfrac{9}{16}t \ + \ \ \dfrac{5}{12}.\)

So, you must use grouping symbols around the first denominator:

9/(16t) + 5/12

That is the same for the answer, so you must use them in two places:

(27 + 20t)/(48t)

So, it continues like this:

Adding 9/(16t) + 5/12 requires a common denominator. That would be 48t.

We multiply 9/(16t)*3/3 to get 27/(48t).

We multiply 5/12*4t/(4t) to get 20t/(48t).

Now we have 27/(48t) + 20t/(48t), and the addition happens with the numerators;
the common denominator stays the same. Just like when adding regular fractions.

(27 + 20t)/(48t)

 

[Mauro]

That first expression could mean \(\displaystyle \ \ \dfrac{9}{16}t \ + \ \ \dfrac{5}{12}.\)

A correction:
No :\(\displaystyle \ \ \dfrac{9}{16}t \ + \ \ \dfrac{5}{12}.\)

Yes:\(\displaystyle \ \ \dfrac{9}{16} \cdot\dfrac{1}{t} \ + \dfrac{5}{12}\)


We can say?\(\displaystyle \ \ \dfrac{9}{16} \cdot\dfrac{1}{t} \ + \dfrac{5}{12} \ \ \Longleftrightarrow \ \ \dfrac{1}{t} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)\)

 

[HallsofIvy]

A correction:
No :\(\displaystyle \ \ \dfrac{9}{16}t \ + \ \ \dfrac{5}{12}.\)

Yes:\(\displaystyle \ \ \dfrac{9}{16} \cdot\dfrac{1}{t} \ + \dfrac{5}{12}\)


We can say:\(\displaystyle \ \ \dfrac{9}{16} \cdot\dfrac{1}{t} \ + \dfrac{5}{12} \ \ \Longleftrightarrow \ \ \dfrac{1}{t} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)\)

NO! Those are not at all the same thing. For example it t= 2 then
\(\displaystyle \dfrac{9}{16}\cdot\dfrac{1}{t}+ \dfrac{5}{12}= \dfrac{9}{16}\cdot\dfrac{1}{2}+ \dfrac{5}{12}= \dfrac{9}{32}+ \dfrac{5}{12}= \dfrac{27}{96}+ \frac{40}{96}= \frac{67}{96}\)

But \(\displaystyle \dfrac{1}{t} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)= \dfrac{1}{2} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)= \dfrac{1}{2} \cdot \bigg ( \dfrac{27}{49} \ + \dfrac{20}{48} \bigg)= \dfrac{47}{48}\)

 

[Mauro]

NO! Those are not at all the same thing. For example it t= 2 then
\(\displaystyle \dfrac{9}{16}\cdot\dfrac{1}{t}+ \dfrac{5}{12}= \dfrac{9}{16}\cdot\dfrac{1}{2}+ \dfrac{5}{12}= \dfrac{9}{32}+ \dfrac{5}{12}= \dfrac{27}{96}+ \frac{40}{96}= \frac{67}{96}\)

But \(\displaystyle \dfrac{1}{t} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)= \dfrac{1}{2} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)= \dfrac{1}{2} \cdot \bigg ( \dfrac{27}{49} \ + \dfrac{20}{48} \bigg)= \dfrac{47}{48}\)

Probable Common error
\(\displaystyle \dfrac{1}{t} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)= \dfrac{1}{t} \cdot \bigg ( \dfrac{9}{16} \ + \dfrac{5}{12} \bigg)= \bigg ( \dfrac{9}{16t} \ + \dfrac{5}{12t} \bigg)= \dfrac{27}{48t} \ + \dfrac{20}{48t} =\dfrac{27 + 20}{48t} = \dfrac{47}{48t}\).



Solution
\(\displaystyle \ \dfrac{9}{16t} + \dfrac{5}{12}= \dfrac{27}{48t}+ \dfrac{20t}{48t}= \dfrac{27 + 20t}{48t}\)

Look:
Step 1.: Least Common Multiple (16t, 12)= 48t.
Step 2.: 48t / 12 = 4t, 4t times 5 equal 20t.
Step 3.: 48t / 16t= 3 , 3 times 9 equal 27.

This is a great charade case 2: OK

forum friends,
thank!

Edited by reason: clarity

 

[Mauro]

:shock::shock::shock::shock: I was not clear.

After editing
I believe to be sufficient.