Algrebra 2 Homework Help

[PurpleScar]

I have 3 problems I am having trouble with in Algebra 2. If anyone could explain to me how to work them out and the answer so I know how to do them in the future, that would be great. My teacher doesn't explain to us the work....just gives it to us and expects us to know it already. Anyway!

1) a(y+c)=b(y-c)-----Solve for y

2) y+3/t=t^2------Solve for y

3) u/5+u/10-u/6=1 (I was thinking this was u=9, not sure.)----Solve the equation

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My work:

For number one I distributed a to y and c, and b to y and c.
1) ay+ac=by-bc
Added bc from the right side
2) ay+ac+bc=by
Stuck from there....

Number two, no clue on how to do.

Number three, I think the answer is u=9, I just need confirmation. Added all fractions
together and came up with u/9. Do the reciprocal and times by 9. Gives you the
answer. Is that correct?

Thanks.

 

[PurpleScar]

What is the question(s)?

Edited my original post. Please help.

 

[Yogi]

1) a(y+c)=b(y-c)-----Solve for y
2) y+3/t=t^2------Solve for y
3) u/5+u/10-u/6=1 (I was thinking this was u=9, not sure.)----Solve the equation

My work:

For number one I distributed a to y and c, and b to y and c.
1) ay+ac=by-bc good
Added bc from the right side
2) ay+ac+bc=by The goal is solving for y so you need all the y terms on the same side and terms without y on the opposite side of the =. Since you started by moving the non-y's to the left, get the y's to the right. Once that is done you can factor out a y and isolate y.
Stuck from there....

Number two, no clue on how to do. Solve for y effectively means, get y on one side of the equation by itself. y is on the left but it is not by itself, get rid of the +3/t using opposites.

Number three, I think the answer is u=9, I just need confirmation. Added all fractions
together and came up with u/9. Do the reciprocal and times by 9. Gives you the
answer. Is that correct? Why not test u=9? Once you get an answer you can always plug it back in.
9/5 + 9/10 -9/6 = 1
(9*60)/5 + (9*60)/10 -(9*60)/6 = 60
108 + 54 - 90 = 60
72 = 60 nope

Thanks.

.

 

[PurpleScar]

.

Number one I solved correctly: y=-c(a-b)/a-b

Number two: Not sure how to do the reciprocal of a variable when there's a t on the other side.
Like if I did t/1 times y+3/t, it would cancel out the t. So I just have y+3. But I still have to
times the t^2 on the other side. Would it stay the same or...?

Number three: I'm confused with this one.

 

[Yogi]

Number one I solved correctly: y=-c(a-b)/a-b If that were correct the a-b on top would cancel with the one on bottom and you would be left with -c, maybe you meant y = -c(a+b)/(a-b)

Number two: Not sure how to do the reciprocal of a variable when there's a t on the other side.
Like if I did t/1 times y+3/t, it would cancel out the t. So I just have y+3 (you would actually have yt+3 which doesnt help). But I still have to times the t^2 on the other side. Would it stay the same or...?
t^2 is t*t so if you multiply it by t it becomes t*t*t = t^3

You are talking about:
y+3/t=t^2
yt + 3 = t^3 (multiply by t on both sides)
which doesn't really help us unless you now subtract 3 to both sides and then divide by t, but then why are we multiplying t and then dividing by t later...too much unnecessary work.

You are over-thinking it, y is on the left, they are ADDING 3/t to y so we have to _________ 3/t from both sides and then y will be by itself.

Number three: I'm confused with this one.
Need to multiply something on both sides to get rid of the fractions. This could be 300 since 5*10*6=300 or if you happen to notice that the LCD is 30 that works too. Show your work on this and you finish it up.

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