Multiplying a Rational Expression Problem

[humanbean]

SOLVED: Multiplying a Rational Expression Problem

Hello! I've been brushing up on some algebra and for the most part, I understand multiplying rational expressions rather well. However, I have been stuck on a particular problem all night:

(x-y)²-z² / x-y+z
(x+y)²-z² / x+y-z

...or, if this format poses the question more clearly:

(((x-y)^2-z^2)/((x+y)^2-z^2))/((x-y+z)/(x+y-z))

According to the book, the answer I should be getting is:

x-y-z
x+y+z

or

(x-y-z)/(x+y+z)

I just don't understand how to get there. I understand that I must multiply the dividend by the reciprocal of the divisor, but I don't know how to multiply across such large expressions. I've tried distributing x across the expression, then -y, etc., but I end up with big, scary looking terms that don't get me anywhere close to the correct solution.

This problem is driving me mad and I appreciate any help you can provide! Thanks!

 

[soroban]

Hello, humanbean

\(\displaystyle \text{Simplify: }\;\dfrac{\dfrac{(x-y)^2 - z^2}{(x+y)^2 - z^2}}{\dfrac{x-y+z}{x+y-z}}\)

\(\displaystyle \text{Answer: }\:\dfrac{x-y-z}{z+y+z}\)

\(\displaystyle \text{We have: }\:\dfrac{(x-y)^2-z^2}{(x+y)^2 - z^2} \cdot\dfrac{x+y-z}{x-y+z}\)


The first numerator is a difference of squares:
. . \(\displaystyle (x-y)^2 - z^2 \:=\x-y-z)(x-y+z)\)

The first denominator is a difference of squares:
. . \(\displaystyle (x+y)^2 - z^2 \:=\x+y-z)(x+y+z)\)


The problem becomes: .\(\displaystyle \displaystyle \frac{(x-y-z)(x-y+z)}{(x+y-z)(x+y+z)}\cdot\frac{x+y-z}{x-y+z} \)

And we have: .\(\displaystyle \displaystyle \frac{(x-y-z)(\rlap{////////}x-y+z)}{(\rlap{////////}x+y-z)(x+y+z)}\cdot\frac{\rlap{////////}x+y-z}{\rlap{////////}x-y+z} \;=\;\frac{x-y-z}{x+y+z}
\)

 

[humanbean]

Thank you, Denis! This method really made the problem so much clearer to me! At long last, I've arrived at the correct answer, and I know how I got there!

Thank you too, soroban! Unfortunately, my book hasn't mentioned anything about differences of squares and I can't remember being taught about that at any point. I had to spend a bit of time proving to myself that it would work! I've made a note of it in my notebook and will hopefully remember when I need it!

So, if (x+y)²-z²=(x+y-z)(x+y+z)

is it true that (x+y)²+z²=(x+y+z)(x+y+z) ? Just curious!

 

[srmichael]

is it true that (x+y)²+z²=(x+y+z)(x+y+z) ? Just curious!

Nope.

\(\displaystyle \displaystyle(x+y+z)(x+y+z)= \) \(\displaystyle \displaystyle (x+y)^2+z(x+y)+z(x+y)+z^2= \) \(\displaystyle \displaystyle (x+y)^2+2z(x+y)+z^2\)

 

[Deleted member 4993]

It will help you to memorize some basic patterns AFTER YOU UNDERSTAND THEM.

\(\displaystyle (a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) = a^2 + ab + ab + b^2 =\) \(\displaystyle a^2 + 2ab + b^2 \neq a^2 + b^2\ if ab \neq 0. \)

In general, \(\displaystyle (a + b)^2 = a^2 + 2ab + b^2 \neq a^2 + b^2.\)

Similarly, \(\displaystyle (a - b)^2 = a^2 - 2ab + b^2 \neq a^2 - b^2\ unless\ ab = 0.\) <<< even if ab = 0 (trivial cases b = 0 or a = 0)

\(\displaystyle (a - b)(a + b) = a(a + b) - b(a + b) = a^2 + ab - ab - b^2 = a^2 - b^2.\)

In short \(\displaystyle (a^2 - b^2) = (a + b)(a - b)\)

These patterns or identities come up again and again.

.

 

[Deleted member 4993]

To keep it less wieldy: let a = x + y and b = x - y ; so you have:

[(b^2 - z^2) / (a^2 - z^2)] / [(b + z) / (a - z)]

= [(b^2 - z^2) / (a^2 - z^2)] * [(a - z) / (b + z)]

= [(b + z)(b - z) / ((a + z)(a - z))] * [(a - z) / (b + z)]

Since you've been swearing on this one for hours, I'll let you finish it

I think Denis is getting bizzare...

 

[chockblock12]

need help

-39r⁴+15r³-18r²+2r+2
-3r

 

[chockblock12]

Hey JeffM,
the problem is diving the -3r woth the top one. I divied it and got -13r^3-15r^2+6r-2/3-2/3

 

[srmichael]

Hey JeffM,
the problem is diving the -3r woth the top one. I divied it and got -13r^3-15r^2+6r-2/3-2/3

Holy cow! Slow down when you are typing. Your comment is loaded with misspellings and confusion

AND...as JeffM said, start a new thread.

 

[chockblock12]

I dont know if I did it right.....??

 

[chockblock12]

I am sorry. this is my first time. I am having problems with dividing this problem. it is -39^4+15r^3-18r^2+2r+2 it is Divided by -3r. The answer I got is -13r^3-5r^2+6r-2/3-2/3