Two Algebraic Expressions

[mathwannabe]

Hello to everyone

I am completely new to this forum and this is my first post.

So I am gonna get straight to the point.

I have two algebraic expressions that need evaluation and/or simplification. I did them and now I'm going to post them along with my results and workflow so you guys could tell me if I am on the right track (I haven't done any math in at least 10 years, and now I'm trying to prepare myself for an exam to enter math faculty).

1) If \(\displaystyle ay=bx\) then the value of the expression \(\displaystyle \frac{x^2}{x^2+y^2}+\frac{b^2}{a^2+b^2}\) equals: ?

This is how I did it:

If \(\displaystyle ay=bx\) then \(\displaystyle y=\frac{bx}{a}\):

Then I switched \(\displaystyle y\) from the expression with \(\displaystyle \frac{bx}{a}\)

\(\displaystyle \frac{x^2}{x^2+(\frac{bx}{a})^2}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{x^2}{x^2+\frac{b^2x^2}{a^2}}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{x^2}{\frac{a^2x^2+b^2x^2}{a^2}}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{x^2}{\frac{x^2(a^2+b^2)}{a^2}}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =x^2*\frac{a^2}{x^2(a^2+b^2)}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{a^2+b^2}{a^2+b^2}= 1\)

Is this correct? Is the workflow process good? Is there any other way to do this?

2) This Latex code is too tiresome to type and the next one took me 2 notebook pages so I'm just gonna put in some final steps.

For \(\displaystyle |a|\ \ne \ |b|\) the value of the expression \(\displaystyle (\frac{a+b}{a-b}+\frac{a-b}{a+b})^2-(\frac{a+b}{a-b}-\frac{a-b}{a+b})^2=\)?

So, I eventually got to:

\(\displaystyle \frac{4a^4-8a^2b^2+4b^4}{a^4-2a^2b^2+b^4}=\)

\(\displaystyle =\frac{2a^2-2b^2}{a^2-b^2}=\)

\(\displaystyle =\frac{2(a^2-b^2)}{a^2-b^2}=\)

\(\displaystyle =2\)

Is this correct?

Thank you all in advance.

 

[Deleted member 4993]

Hello to everyone

I am completely new to this forum and this is my first post.

So I am gonna get straight to the point.

I have two algebraic expressions that need evaluation and/or simplification. I did them and now I'm going to post them along with my results and workflow so you guys could tell me if I am on the right track (I haven't done any math in at least 10 years, and now I'm trying to prepare myself for an exam to enter math faculty).

1) If \(\displaystyle ay=bx\) then the value of the expression \(\displaystyle \frac{x^2}{x^2+y^2}+\frac{b^2}{a^2+b^2}\) equals: ?

Another way:

x/y = a/b

Then the given equation becomes:


1/[1 + (y/x)²] + 1/[1 + (a/b)²]

= 1/[1 + (y/x)²] + 1/[1 + (x/y)²]

= x²/[x² + y²] + y²/[y² + x²]

= [x² + y²]/[x² + y²]

= 1



This is how I did it:

If \(\displaystyle ay=bx\) then \(\displaystyle y=\frac{bx}{a}\):

Then I switched \(\displaystyle y\) from the expression with \(\displaystyle \frac{bx}{a}\)

\(\displaystyle \frac{x^2}{x^2+(\frac{bx}{a})^2}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{x^2}{x^2+\frac{b^2x^2}{a^2}}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{x^2}{\frac{a^2x^2+b^2x^2}{a^2}}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{x^2}{\frac{x^2(a^2+b^2)}{a^2}}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =x^2*\frac{a^2}{x^2(a^2+b^2)}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=\)

\(\displaystyle =\frac{a^2+b^2}{a^2+b^2}= 1\)

Is this correct? Is the workflow process good? Is there any other way to do this?

2) This Latex code is too tiresome to type and the next one took me 2 notebook pages so I'm just gonna put in some final steps.

For \(\displaystyle |a|!=|b|\) the value of the expression \(\displaystyle (\frac{a+b}{a-b}+\frac{a-b}{a+b})^2-(\frac{a+b}{a-b}-\frac{a-b}{a+b})^2=\)?

So, I eventually got to:

\(\displaystyle \frac{4a^4-8a^2b^2+4b^4}{a^4-2a^2b^2+b^4}=\)

\(\displaystyle =\frac{2a^2-2b^2}{a^2-b^2}=\)

\(\displaystyle =\frac{2(a^2-b^2)}{a^2-b^2}=\)

\(\displaystyle =2\)

Is this correct?

Thank you all in advance.

.

 

[mathwannabe]

If you want to follow your logic

\(\displaystyle \dfrac{4a^4 - 8a^2b^2 + 4b^4}{a^4 - 2a^2b^2 + b^4} = \dfrac{(2a^2 - 2b^2)^2}{(a^2 - b^2)^2} = \dfrac{4(a^2 - b^2)^2}{(a^2 - b^2)^2} = 4 \)

Yes, I see the mistake in that step, I totally forgot that I was dealing with squares there. Such a foolish mistake. Also, the substitution method does make it a LOT easier Thank you for your help.

 

[mathwannabe]

If you want to follow your logic

\(\displaystyle \dfrac{4a^4 - 8a^2b^2 + 4b^4}{a^4 - 2a^2b^2 + b^4} = \dfrac{(2a^2 - 2b^2)^2}{(a^2 - b^2)^2} = \dfrac{4(a^2 - b^2)^2}{(a^2 - b^2)^2} = 4 \)

OMG how could have I overlooked that \(\displaystyle \frac{a+b}{a-b}=\frac{1}{\frac{a-b}{a+b}}\) ???

Your post was an eye opener for sure To think that I filled two pages in my notebook to finish that one and have done it wrong because of a stupid mistake.

 

[Deleted member 4993]

Hello to everyone



2) This Latex code is too tiresome to type and the next one took me 2 notebook pages so I'm just gonna put in some final steps.

For \(\displaystyle |a|\ \ne \ |b|\) the value of the expression \(\displaystyle (\frac{a+b}{a-b}+\frac{a-b}{a+b})^2-(\frac{a+b}{a-b}-\frac{a-b}{a+b})^2=\)?

So, I eventually got to:

\(\displaystyle \frac{4a^4-8a^2b^2+4b^4}{a^4-2a^2b^2+b^4}=\)

\(\displaystyle =\frac{2a^2-2b^2}{a^2-b^2}=\)

\(\displaystyle =\frac{2(a^2-b^2)}{a^2-b^2}=\)

\(\displaystyle =2\)

Is this correct?

Thank you all in advance.

slightly different way:

\(\displaystyle (\frac{a+b}{a-b}+\frac{a-b}{a+b})^2-(\frac{a+b}{a-b}-\frac{a-b}{a+b})^2=\)?

Let

x = (a+b)/(a-b) and y = (a-b)/(a+b)

then

(x+y)² - (x-y)² = 4x*y = 4 * (a+b)/(a-b) * (a-b)/(a+b) = 4

 

[mathwannabe]

Thank you all, your help is greatly appreciated. I am just curious about one more thing: how hard are, in your opinion, those two problems on a scale of 1 to 10?

 

[soroban]

Hello, mathwannabe!

\(\displaystyle 2)\;\text{For }\,|a|\ \ne \ |b|,\,\text{ the value of the expression }\,\left(\dfrac{a+b}{a-b}+\dfrac{a-b}{a+b}\right)^2 - \) \(\displaystyle \left(\dfrac{a+b}{a-b}-\dfrac{a-b}{a+b}\right)^2\;=\;?\)

Did you notice that we have a Difference of Squares.

\(\displaystyle \left[\left(\dfrac {a+b}{a-b} + \dfrac{a-b}{a+b}\right) - \left(\dfrac{a+b}{a-b} - \dfrac{a-b}{a+b}\right)\right]\) \(\displaystyle \left[\left(\dfrac{a+b}{a-b} + \dfrac{a-b}{a+b}\right) + \left(\dfrac{a+b}{a-b} - \dfrac{a-b}{a+b}\right) \right]\)

. . . \(\displaystyle =\;\left[\underbrace{\left(\dfrac{a+b}{a-b} - \dfrac{a+b}{a-b}\right)}_{\text{zero!}} + \left(\dfrac{a-b}{a+b} + \dfrac{a-b}{a+b}\right)\right]\) \(\displaystyle \left[\left(\dfrac{a+b}{a-b} +\dfrac{a+b}{a-b}\right) + \underbrace{\left(\dfrac{a-b}{a+b} - \dfrac{a-b}{a+b}\right)}_{\text{zero!}}\right] \)


. . . \(\displaystyle =\;\left(\dfrac{2(a-b)}{a+b}\right)\,\left(\dfrac{2(a+b)}{a-b}\right) \;=\;\dfrac{4(a-b)(a+b)}{(a-b)(a+b)} \;=\;4\)

 

[mathwannabe]

Difficulty like beauty is in the eye of the beholder.

Yes, I understand that difficulty is relative, so I think I have asked the question in a wrong way. I have to prepare high school math in order to pass the test for getting into math faculty, so, what level of difficulty for high school math are those two problems considered?

 

[srmichael]

Yes, I understand that difficulty is relative, so I think I have asked the question in a wrong way. I have to prepare high school math in order to pass the test for getting into math faculty, so, what level of difficulty for high school math are those two problems considered?

Hi Mathwannabe. Most of us that help out on this forum are pretty good at math and most may think that these problems are "not hard". However, we are obviously biased in our opinion. I, however, have been tutoring kids in high school math for the last 11 years and I can tell you that these kind of problems are always a challenge for the high school student. Most of the kids understand what they have to do conceptually on problems like these but it is usually the careless error(s) that they do that makes them not get the answer as there is ample opportunity for a careless error on these kind of problems.

Hope this helps.

 

[mathwannabe]

Hello, mathwannabe!


Did you notice that we have a Difference of Squares.

\(\displaystyle \left[\left(\dfrac {a+b}{a-b} + \dfrac{a-b}{a+b}\right) - \left(\dfrac{a+b}{a-b} - \dfrac{a-b}{a+b}\right)\right]\) \(\displaystyle \left[\left(\dfrac{a+b}{a-b} + \dfrac{a-b}{a+b}\right) + \left(\dfrac{a+b}{a-b} - \dfrac{a-b}{a+b}\right) \right]\)

. . . \(\displaystyle =\;\left[\underbrace{\left(\dfrac{a+b}{a-b} - \dfrac{a+b}{a-b}\right)}_{\text{zero!}} + \left(\dfrac{a-b}{a+b} + \dfrac{a-b}{a+b}\right)\right]\) \(\displaystyle \left[\left(\dfrac{a+b}{a-b} +\dfrac{a+b}{a-b}\right) + \underbrace{\left(\dfrac{a-b}{a+b} - \dfrac{a-b}{a+b}\right)}_{\text{zero!}}\right] \)


. . . \(\displaystyle =\;\left(\dfrac{2(a-b)}{a+b}\right)\,\left(\dfrac{2(a+b)}{a-b}\right) \;=\;\dfrac{4(a-b)(a+b)}{(a-b)(a+b)} \;=\;4\)

Wow When I started doing this one, I barely stopped and gave it a thought, instead I just started brute forcing everything, adding terms, squaring... This solution is very elegant, as are the others I received on this thread, but this one, at least for me makes most sense and hits on the point of this problem. Thank you.

 

[mathwannabe]

Hi Mathwannabe. Most of us that help out on this forum are pretty good at math and most may think that these problems are "not hard". However, we are obviously biased in our opinion. I, however, have been tutoring kids in high school math for the last 11 years and I can tell you that these kind of problems are always a challenge for the high school student. Most of the kids understand what they have to do conceptually on problems like these but it is usually the careless error(s) that they do that makes them not get the answer as there is ample opportunity for a careless error on these kind of problems.

Hope this helps.

Yes, it does. Thanks.

 

[Deleted member 4993]

Well SK, that is very elegant. less dependent on a flash of insight, but still eliminating a bunch of unneeded complexity, and far more general. There was good reason that I was encouraged to take up qualitative studies like history.

These are tricky problems. Lots of meaningless hints. The factorial and absolute value elements of this problem were entirely irrelevant.

Thank you so much for that compliment - I wish I can forward your letter to my three sons. Incidentally, my pattern recognition capability came from those three guys - my sons - purely from self-defense. They would compete in various math competitions - and throw down challenge to me - who can do those pratice problems (correct and fastest). Well if I was late - I had to hear about it - till next month's challenge. And if I were the first one to solve those - look out world (my wife can attest to that!).

Now those guys are all grown and have their own kids to "torture" ......

By the way, that factorial sign (in a flash of brilliant insight) - I realised that is actually for \(\displaystyle \ne\) sign. I had corrected the orifginal post later.

 

[mathwannabe]

By the way, that factorial sign (in a flash of brilliant insight) - I realised that is actually for \(\displaystyle \ne\) sign. I had corrected the orifginal post later.

Yes, that wasn't supposed to be factorial sign, but I didn't know how to write "not equal to" it in latex so I just used the one I use while programming in C ( != ) XD

 

[Deleted member 4993]

Yes, I understand that difficulty is relative, so I think I have asked the question in a wrong way. I have to prepare high school math in order to pass the test for getting into math faculty, so, what level of difficulty for high school math are those two problems considered?

Depends on which grade and which curricula stream (i.e. STEM,Liberal arts, etc.). For a 12 th. grader in STEM, these should not be difficult (not easy as pie either - I'll say difficulty level of 6).

For most 9 th graders - taking Algebra II (except for those taking part in math competition) - these would be 9-10 difficulty,

 

[mathwannabe]

Yes, I understand everything now, thanks to the help and guidance that you guys provided. Thank you very much