Add and simplify

[samwertz3]

Sorry to keep bugging you all, but our help tutorial is currently closed.

I have 2 problems that are similar and these are homework problems so you dont have to answer them for me, if you would just help me with the steps I would greatly appreciate it. I am getting stuck with what to do with the negative.

- (6/x-1) + (1-2x/x)

-6x+x-2x^2-1+2x/x(x-1)

-2x^2-3x-1/x(x-1) is this right so far? if it is how do you factor this with the negative?

The other one is similar in that I don't know what to do with the negative sign also.

- (3t+8y/5t) + (5t-7y/5t)

I came up with 2t-15y/5t

but was I supposed to distibute the negative through the entire equation or just the first part?

Anyway direction that can be provided is appreciated.

 

[Loren]

The first thing you need to do is to check for typos before posting your question. The second thing is to use parenthesis to ensure you are saying what you mean to say. For instance...

- (6/x-1) + (1-2x/x) means \(\displaystyle -(\frac{6}{x}-1) + (1 - \frac{2x}{x})\) which can't possibly lead to the next line.

-6x+x-2x^2-1+2x/x(x-1)

With regard to the minus sign preceding a parenthesis, some people just change it to "-1" then follow the rules for multiplying a polynomial by a monomial. For instance, -(2a - 3) becomes -1(2a - 3) = -1(2a) + (-1)(-3) = -2a + 3.

 

[mmm4444bot]

In one of your previous submissions, I wrote the following.

'

Firstly, we must type parentheses around numerators and denominators when they consist of more than one term or factor.

Properly typed, the given equation is:

4 + 3/(y - 3) = -5/(y - 4)

I'm not kidding.

If you do not use grouping symbols to show us your given expressions clearly, then I am not wasting my time trying to decipher your ambiguous typing.

You got the message?

If so, then please fix your post above.

If not, then please tell me what it is that you do not understand about properly typing algbraic ratios using grouping symbols.

 

[Deleted member 4993]

Sorry to keep bugging you all, but our help tutorial is currently closed.

I have 2 problems that are similar and these are homework problems so you dont have to answer them for me, if you would just help me with the steps I would greatly appreciate it. I am getting stuck with what to do with the negative.

- (6/x-1) + (1-2x/x)

\(\displaystyle -\frac{6}{x-1} \, + \, \frac{1-2x}{x}\) .... this should written as: -6/(x+1) + (1-2x)/x

\(\displaystyle = \, \frac{-6x + x - 1- 2x^2 + 2x}{x(x-1)}\)

\(\displaystyle = \, -\frac{2x^2 + 3x + 1}{x(x-1)}\)

\(\displaystyle = \, -\frac{(2x+1)(x+1)}{x(x-1)}\)
______________________________________________________
-6x+x-2x^2-1+2x/x(x-1)

-2x^2-3x-1/x(x-1) is this right so far? if it is how do you factor this with the negative?

The other one is similar in that I don't know what to do with the negative sign also.

- (3t+8y/5t) + (5t-7y/5t)

I came up with 2t-15y/5t

but was I supposed to distibute the negative through the entire equation or just the first part?

Anyway direction that can be provided is appreciated.

 

[samwertz3]

Yes mmm4444bot, I got the message. No mmm4444bot, I did not get what you were getting at. (hence, the reason I am in BASIC Algebra at my age) BUT thanks to the example provided by Subhotosh Khan, I am pretty sure I got it now. Thanks for taking the time to "decipher my ambiguous typing" Subhotosh Khan, it is much appreciated.

 

[mmm4444bot]

… No mmm4444bot, I did not get what you were getting at …

In mathematics instruction, whenever somebody tells us anything that we do not understand, then we should ask questions. (Well, we should, if we are interested in learning something, I suppose.)

I am pretty sure I got it now.

Time will tell.

 

[Denis]

Sam, it's easy:
a+b/c and (a+b)/c are NOT the same;
1st one means divide b by c, then add a
2nd one means add a and b, then divide by c

Try a = 1, b =4 and c=2:
4/2 = 2 + 1 = 3
1 + 4 = 5, divided by 2 = 5/2 ; quite different, right?