Reducing algebraic equations (is what I think it's called)

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Soph

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I'm trying to help my daughter with her algebra, but because it's about 35 years since I did this kind of thing, I'm finding I'm completely lost.

To compound the problem, the question is in Greek, but I understand the solution is to reduce the equations in some way. I hope that makes sense to somebody out there, and I would be glad of any help, especially with the workings so that I might see what's supposed to be done. Thanks in advance.

A = (3(sqrt2)) /((2(sqrt3)) - (3(sqrt2)))

B = (1 + (sqrt5)) / ((sqrt3) + (sqrt5) - (sqrt7))

C = 1 / (1 + (1 / (1 + (sqrt2))))
 


Hi Soph:

These are not equations; they are expressions.

Equations always contain an equals sign.

I'm not sure what the Greeks say, but I'm thinking that "reduce" here means to "rationalize the denominator".

In other words, your daughter needs to rewrite these expressions so that no square roots appear in the denominators.

Let's start with the first one.

3 sqrt(2)/[2 sqrt(3) - 3 sqrt(2)]

Multiply the top and bottom of this ratio by the conjugate of the denominator.

The conjugate of 2 sqrt(3) - 3 sqrt(2) is 2 sqrt(3) + 3 sqrt(2).

Does your daughter know how to use FOIL to do the multiplication on the bottom?

[2 sqrt(3) - 3 sqrt(2)] * [2 sqrt(3) + 3 sqrt(2)]

Let us know how it goes.

Cheers, ~ Mark

PS: The third exercise requires first simplifying the compound ratio; then rationalize the resulting denominator.

The second exercise is going to be messy; are you sure that you typed it properly?

 
Hi Mark,

Thanks for taking the time to answer my question.

My daughter says she knows what to do now, although I have to say, it's still all Greek to me. I can't believe I used to be able to do this stuff!

Cheers and thanks again,

Miriam (Soph's my daughter)

(Yes, I typed the second one correctly - they're all inventions of her teacher.)
 
Hi Soph:

The third one is what looks like a 'continued fraction'. Normally, continued fractions extend indefinitely, but this one is finite so it can be simplified.

\(\displaystyle \frac{1}{1+\frac{1}{\sqrt{2}+1}}\)

Make a substitution by letting \(\displaystyle x=\sqrt{2}+1\). Then, we have:

\(\displaystyle \frac{1}{1+\frac{1}{x}}\)

\(\displaystyle =\frac{1}{\frac{x+1}{x}}\)

\(\displaystyle =\frac{x}{x+1}\)

Now, resub the \(\displaystyle x=\sqrt{2}+1\)

\(\displaystyle \frac{\sqrt{2}+1}{\sqrt{2}+2}\)

Now, we can rationalize the denominator so we can get rid if the radical in the denominator.

Multiply top and bottom by what we call the 'conjugate'. The opposite sign of \(\displaystyle \sqrt{2}+1\). Which is \(\displaystyle \sqrt{2}-1\)

\(\displaystyle \frac{(\sqrt{2}+1)}{(\sqrt{2}+2)}\cdot \frac{(\sqrt{2}-2)}{(\sqrt{2}-2)}\)

Multiply across the top and bottom and get:

\(\displaystyle \frac{\sqrt{2}-2+2-2\sqrt{2}}{2-4}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)

Look how that simplified down to that little thing.
 
\(\displaystyle Here's\ a\ quicker\ way\ for\ C....\)

\(\displaystyle (\sqrt{2}-1)(\sqrt{2}+1)=\sqrt{2}\sqrt{2}+\sqrt{2}-\sqrt{2}-1=2-1=1\)

\(\displaystyle Therefore\)

\(\displaystyle \sqrt{2}-1=\frac{1}{1+\sqrt{2}}\)

\(\displaystyle \sqrt{2}=1+\frac{1}{1+\sqrt{2}}\)

\(\displaystyle Hence\)

\(\displaystyle \frac{1}{1+\frac{1}{1+\sqrt{2}}}=\frac{1}{\sqrt{2}}\)
 
Thanks, chrisr, it's all a lot clearer now (not!) LOL. I'll pass your information on to Soph.

I like your Bruce Lee joke. I don't understand it, but I'm sure it has maths boffins chuckling.

Cheers and thanks again,

Miriam (Soph's Ma)
 
Hi bambam2,

3(-3) means subtract 3 three times

and -2(-3) means the opposite of subtract 3 twice.

\(\displaystyle x-2(x-3)+3(x-3)=x-2(x)-2(-3)+3(x)+3(-3)=x-2x+6+3x-9=4x-2x+6-9=2x-3\)
 
Soph said:
I like your Bruce Lee joke. I don't understand it, but I'm sure it has maths boffins chuckling.

Cheers and thanks again,

Miriam (Soph's Ma)
Click to expand...

No Miriam,

it didn't go down too well at all
 
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