how can I be wrong here?

[allegansveritatem]

I tried 4 times to solve this problem so that my solution matched the book solution. No luck. Here is the problem:


Here is the book solution:


Here is my solution and the tortured path I followed to reach it:


I can't for the life of me figure out what I'm doing wrong here.

 

[tkhunny]

I tried 4 times to solve this problem so that my solution matched the book solution. No luck. Here is the problem:
View attachment 10568

Here is the book solution:
View attachment 10569

Here is my solution and the tortured path I followed to reach it:
View attachment 10570

I can't for the life of me figure out what I'm doing wrong here.

Your grouping is a little off, that's all.

You were good to here:

\(\displaystyle \dfrac{4}{(y-1)(y-2)}-\dfrac{y^{2}-y}{(y-1)(y-2)}+\dfrac{2y^{2}-4y}{(y-1)(y-2)}\)

Now, VERY CAREFULLY...

\(\displaystyle \dfrac{4 - (y^{2}-y) + (2y^{2}-4y)}{(y-1)(y-2)}\)

See if that makes a difference. The middle numerator wasn't working quite right.

 

[Dr.Peterson]

I tried 4 times to solve this problem so that my solution matched the book solution. No luck. Here is the problem:
View attachment 10568

Here is the book solution:
View attachment 10569

Here is my solution and the tortured path I followed to reach it:
View attachment 10570

I can't for the life of me figure out what I'm doing wrong here.

First, note that what you showed is not a "problem"; it's an "expression". The problem is to simplify it (carry out the addition and subtraction). But that's just words.

Your error is in the fourth line. Your numerators on the line above are

4 - (y^2 - y) + (2y^2 - 4y)
​

which is

4 - y^2 + y + 2y^2 - 4y
​

What you did is apparently

4 - (y^2 - y + 2y^2 - 4y) = 4 - (3y^2 - 5y) = -3y^2 + 5y + 4
​

That is, you did the addition before the subtraction. Possibly you are under the impression that the order of operations requires that; it doesn't. You are supposed to do additions and subtractions from left to right; or, equivalently, just turn each subtraction into the addition of the negative (as I did in removing the parentheses above).

 

[allegansveritatem]

Your grouping is a little off, that's all.

You were good to here:

\(\displaystyle \dfrac{4}{(y-1)(y-2)}-\dfrac{y^{2}-y}{(y-1)(y-2)}+\dfrac{2y^{2}-4y}{(y-1)(y-2)}\)

Now, VERY CAREFULLY...

\(\displaystyle \dfrac{4 - (y^{2}-y) + (2y^{2}-4y)}{(y-1)(y-2)}\)

See if that makes a difference. The middle numerator wasn't working quite right.

What I did was first add the the 2nd and 3rd numerators and then subtracted result from the first numerator. But, as I see from this post and the one below, that I should have simply performed the operations in the order they appear. Thanks

 

[allegansveritatem]

First, note that what you showed is not a "problem"; it's an "expression". The problem is to simplify it (carry out the addition and subtraction). But that's just words.

Your error is in the fourth line. Your numerators on the line above are

4 - (y^2 - y) + (2y^2 - 4y)
​

which is

4 - y^2 + y + 2y^2 - 4y
​

What you did is apparently

4 - (y^2 - y + 2y^2 - 4y) = 4 - (3y^2 - 5y) = -3y^2 + 5y + 4
​

That is, you did the addition before the subtraction. Possibly you are under the impression that the order of operations requires that; it doesn't. You are supposed to do additions and subtractions from left to right; or, equivalently, just turn each subtraction into the addition of the negative (as I did in removing the parentheses above).

Yes, I added the 2nd and 3rd numerator and then subtracted the result from the first. I will have to reinforce my grasp of the rules for performing operations. Thanks very much for pointing these things out.

 

[JeffM]

I tried 4 times to solve this problem so that my solution matched the book solution. No luck. Here is the problem:
View attachment 10568

Here is the book solution:
View attachment 10569

Here is my solution and the tortured path I followed to reach it:
View attachment 10570

I can't for the life of me figure out what I'm doing wrong here.

Several thoughts.

First, in theory, it makes no difference whether you work with

\(\displaystyle \dfrac{a}{b} - \dfrac{b}{c} + \dfrac{d}{e} \text { or } \dfrac{a}{b} + \dfrac{(-\ 1)b}{c} + \dfrac{d}{e}.\)

They are mathematically the same. However, I find that I make fewer sign errors if I avoid subtracting fractions: I always add fractions. To do so, I turn minus signs for fractions into plus signs by multiplying their numerators by - 1 before combining fractions.

So the very first thing I would do is

\(\displaystyle \dfrac{4}{y^2 - 3y + 2} - \dfrac{y}{y - 2} + \dfrac{2y}{y - 1} = \dfrac{4}{y^2 - 3y + 2} + \dfrac{-\ y}{y - 2} + \dfrac{2y}{y - 1}.\)

I doubt you would have made the error you did had you been adding fractions with the proper signs on the numerators. (I reirerate that this has nothing to do with math, but it seems to accord with human psychology.)

Second, check your work. A good way to check simplifications is to choose a small integer > 2, and see whether it works out. This is not 100% effective, but it will identify a very high percentage of errors. (If it is really important to be correct and you have the time, trying 2 different integers is almost certain to catch any errors.)

\(\displaystyle \dfrac{4}{3^2 - (3 * 3) + 2} - \dfrac{3}{3 - 2} + \dfrac{2 * 3}{3 - 1} =\)

\(\displaystyle \dfrac{4}{9 - 9 + 2} + \dfrac{-\ 3}{1} + \dfrac{2 * 3}{2} = \dfrac{4}{2} - 3 + 3 = 2.\)

\(\displaystyle \dfrac{-\ 3(3^2) + 5(3) + 4}{3^2 - (3 * 3) + 2} = \dfrac{-\ 3(9) + 15 + 4}{2} = \dfrac{-\ 8}{2} = -\ 4.\)

\(\displaystyle \text {AND } 2 \ne -\ 4.\)

So you do not need to look in the back of your book to know your answer is incorrect. Moreover, you know that the evaluation of each step should give the answer of 2 when 3 is substituted for y. This will let you find the exact step where you went wrong.