stumped by what should be an easy algebra problem

N

nevergiveup

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Mar 9, 2010
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Hi Everyone. Thanks to anyone who has the time to help me with this problem. I am trying to teach myself some algebra and I am stumped
(y/2-6)+ y=? simplified
now I know the answer is supposed to be 3y/2-6 but try as I may I cannot solve it
I tried multiplying numerators by the denominator 2 but ...
I tried putting 2 in th edenominator of the 6 and the y
I think if I saw a dummy-proof solution with each step (no step omitted) I could follow it without explanation--It is embarrassing to say that the author assumed the solution was so easy that he just left it out.
Thanks if someone has time--I also tried substituting 10 for 'Y' and I knowthe solution works but I cannot solve the problem when I am working with variables :-(
 
does y/2 - 6 =3y/2 -6 ? multiply both sides of = sign by 2 [ not neccessary but it makes it easier for me]
does y-12=3y-6 ? add 6 to each side of = sign
does y - 6 =3y ? subtract y from each side
does -6=2y ? divide both sides by 2
does - 3=y only at the one point.

the answer you gave is not true for the general solution
y/2 - 6 is a expression that cant be reduced

Arthur
 
Thank you very much the response.
I will try to explain better
If I substitute 10 for 'y' i will get :
(y/2-6)+y
(10/2-6)+10= 9

and the simplified solution given is 3y/2-6 and if I substitute again I get:
3(10)/2-6=9

But I cannot for the life of me do it with variables and so it means I really do not understand it--If I need to work with numbers, it means I only have a superficial knowledge of what I am doing.
 
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Given

\(\displaystyle \frac{y}{2} \ - \ 6 \ + \ y\)

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Commutative Property of Addition

\(\displaystyle \frac{y}{2} \ + \ y \ - \ 6\)

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Common denominator

\(\displaystyle \frac{y}{2} \ + \ \frac{2}{2} \cdot \frac{y}{1} \ - \ 6\)

\(\displaystyle \frac{y}{2} \ + \ \frac{2y}{2} \ - \ 6\)

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Combine ratios

\(\displaystyle \frac{y + 2y}{2} \ - \ 6\)

\(\displaystyle \frac{3y}{2} \ - \ 6\)

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Sorry back again.
let us assume the problem was just
y/2 +y
now to simplify this I thought I could eliminate the denominator by multiplying each 'y' numerator by 2 and thus getting:
y+2y
now add in the minus 6 from above and we get:
y+2y-6

What am I doing wrong? If I continue in the above way and divide everything by 2 I would still be wrong because the answer would be
y/2+y-3

Pardon the stuidity but where did the2/2 come from? and if it just represents 1 why not multiply the y/2 by 2/2 too?
 
nevergiveup said:
Sorry back again.
let us assume the problem was just
y/2 +y
now to simplify this I thought I could eliminate the denominator by multiplying each 'y' numerator by 2 and thus getting:
y+2y
now add in the minus 6 from above and we get:
y+2y-6
Click to expand...

You could do that if it were an equation like \(\displaystyle \frac{y}{2}+y-6=0\). Here, you could multiply everything by 2 and get \(\displaystyle y+2y-12=0\)

nevergiveup said:
What am I doing wrong? If I continue in the above way and divide everything by 2 I would still be wrong because the answer would be
y/2+y-3

Pardon the stuidity but where did the2/2 come from? and if it just represents 1 why not multiply the y/2 by 2/2 too?
Click to expand...

Given \(\displaystyle \frac{y}{2}+y-6\) to simplify, we would multiply each term by the common denominator 2 and put it all over 2.

\(\displaystyle \frac{y}{2}+y-6 \Longrightarrow \frac{y}{2}+\frac{2y}{2}-\frac{12}{2} \Longrightarrow \frac{y+2y-12}{2} \Longrightarrow \frac{3y-12}{2} \Longrightarrow \frac{3y}{2}-\frac{12}{2} \Longrightarrow \frac{3y}{2}-6\)
 
You folks are very kind to help but another basic question.
Doesn't
(y+2y-12)/2 also equal
y/2+2y/2-12/2???
if so I would then divide each separately and get:
y/2+y-6
which maybe would equal
(2y+-6)/2 ====>y-3
Sorry about this--The more I write, the more I expose my ignorance but I really want to understand my errors. I would rather be ignorant about this for a day than be ignorant for a lifetime :)
 
nevergiveup said:
You folks are very kind to help but another basic question.
Doesn't
(y+2y-12)/2 also equal
y/2+2y/2-12/2??? <===== Yes
if so I would then divide each separately and get:
y/2+y-6 <==== Yes
which maybe would equal
(2y+-6)/2 ====>y-3 <====No!
Sorry about this--The more I write, the more I expose my ignorance but I really want to understand my errors. I would rather be ignorant about this for a day then be ignorant for a lifetime :)
Click to expand...

I have edited my last post to include every detail of this operation. Study it from end to end.
 
Thanks again for your time.
I understand the way you did it and I will do it that way next time--namely you added all the numerators over ONE division bar up like so
(y+2y-12)/2 and then you combined like terms ,divided and then separated again
I ran into a problem trying to keep them separately namely
(y/2)+(2y/2)-(12/2)
in trying to do it that way I lost a 'y' because 2y/2 would equal 'y'

I know what you folks have done but still not sure why I can't do it the way I wanted--The answer might well just be "well dummy you just can't do it that way."
I am sure I tested your patience much --Thanks a lot guys.
 
Keep it SIMPLE; remember that (as example):
1/2 + 4/3 + 2/5
= 15/30 + 40/30 + 12/30
= (15 + 40 + 12) / 30
= 67 / 30

Thassa way "they all" work...
 
Thanks again for the help. I think it is a great that all of you blessed with great math skills take the time to help folks like myself. Denis thanks for keeping the the final S off of the Keep it simple stupid acronym :)

It just hit me how y/2+y =3y/2
my first kneejerk response was to add the numerators and think I had 2y(that's why i thought I was missing a 'y' compared to the answer) but then like a ton of bricks the idea hit me "you have half of a 'y' plus a full 'y' therefore you have one and a half Ys and if they are written as a fraction it must be 3y/2" ----it is a pity I focused on the numbers and process too much and didn't just take a second and THINK about the concept.
Thaks again for all your help folks.
 
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Mathematical knowledge proceeds by making mistakes.

(I must have a "lot" of knowledge!) 8-)

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