Proving equality of expressions with cyclic terms in numerator and denominator

[Miike012]

Homework Statement

If: (x/l)/(mb+nc-ax) = (y/M)/(nc+la-mb) = (z/n)/(la+mb-nc)
Prove: l/x(by+cz-ax)=m/y(cz+ax-by)=n/z(ax+by-cz)

The next line in the book says...
1. (ny+mz)/a = 2. (lz+nx)/b = 3. (mx+ly)/c

I don't understand how they got to this point...
My guess is that they added them seperatly then simplifyed somehow?
This is what I came up with after adding them...

The Attempt at a Solution

1. (y/m + z/n)/2la = 2. (z/n+x/l)/2mb = 3. (y/m+x/l)/2nc
it looks like in 1. they cross mult in the numerator whish would then give us (ny+mz/nm)/2la = (ny+mz)/(nm)2la... but where did the nm2l go in the denominator?

 

[{~}]

If: $$\frac{\frac{x}{l}}{mb+nc-ax} = \frac{\frac{y}{M}}{nc+la-mb} = \frac{{z}{n}}{la+mb-nc}$$
$$Prove: \frac{l}{x}\left(by+cz-ax\right) = \frac{m}{y}\left(cz+ax-by\right)=\frac{n}{z}\left(ax+by-cz\right)$$

the way you wrote it I can't tell whether the parenthesized terms in the prove statement are part of the numerators or denominators

 

[Miike012]

Good point... sorry... it is
l/(x(by+cz-ax))=m/(y(cz+ax-by))=n/(z(ax+by-cz))

 

[nrqed]

Homework Statement

If: (x/l)/(mb+nc-ax) = (y/M)/(nc+la-mb) = (z/n)/(la+mb-nc)
Prove: l/x(by+cz-ax)=m/y(cz+ax-by)=n/z(ax+by-cz)

You have an "m" and an "M". You mean that both are "m", right?

The next line in the book says...
1. (ny+mz)/a = 2. (lz+nx)/b = 3. (mx+ly)/c

I am not sure what this means. Do you mean "(ny+mz)/a = 2 (lz+nx)/b = 3 (mx+ly)/c ?

 

[Miike012]

No, I was dividing them by number 1. 2. and 3.

I ment "(ny+mz)/a = (lz+nx)/b = (mx+ly)/c

 

[Miike012]

If: (x/l)/(mb+nc-la) = (y/m)/(nc+la-mb) = (z/n)/(la+mb-nc)
Prove: l/(x(by+cz-ax))=m/(y(cz+ax-by))=n/(z(ax+by-cz))

The next line in the book says...
(ny+mz)/a = (lz+nx)/b = (mx+ly)/c


3. The Attempt at a Solution
(y/m + z/n)/2la = (z/n+x/l)/2mb = (y/m+x/l)/2nc
it looks like in 1. they cross mult in the numerator whish would then give us (ny+mz/nm)/2la = (ny+mz)/(nm)2la... but where did the nm2l go in the denominator?

 

[Miike012]

There I re-edited it... that should be correct now.

 

[Miike012]

does it still not look right?

 

[{~}]

$$if: \frac{\frac{x}{l}}{mb +nc - ax} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb - nc}$$

$$Prove: \frac{l}{x\left(by + cz - ax\right)} = \frac{m}{y\left(cz + ax - by\right)}=\frac{n}{z\left(ax + by - cz\right)}$$

The next line of the book says...

$$\frac{ny + mz}{a} = \frac{lz + nx}{b} = \frac{mx + ly}{c}$$

Your attempt at a solution

$$\frac{\frac{y}{m} + \frac{z}{n}}{2la} = \frac{\frac{z}{n} + \frac{x}{l}}{2mb} = \frac{\frac{y}{m} + \frac{x}{l}}{2nc}$$

it looks like in 1. they cross mult in the numerator whish would then give us

$$\frac{ny + \frac{mz}{nm}}{2la} = \frac{ny+mz}{(nm)2la}$$

more readable no?

 

[Miike012]

I made a website... here it has the problem on the front page... if u don't mind looking at it thank you..

http://mmathhelp.webs.com/index.htm

 

[Miike012]

Any one?

 

[{~}]

Example 2. If $$\frac{x}{l\left(mb + nc -la\right)} = \frac{y}{m\left(nc + la - mb\right)} = \frac{z}{n\left(la + mb - nc\right)}$$,

prove that $$\frac{l}{x\left(by + cz - ax\right)} = \frac{m}{y\left(cz + ax - by\right)}=\frac{n}{z\left(ax + by - cz\right)}$$.

We have $$\frac{ny + mz}{a} = \frac{lz + nx}{b} = \frac{mx + ly}{c}$$

$$= \frac{\frac{y}{m} + \frac{z}{n}}{2la}$$

=two similar expressions;

$$\frac{ny + mz}{a} = \frac{lz + nz}{b} = \frac{mx + ly}{c}$$.​

Multiply the first of these fractions above and below by x, the second by y, and the third by z; then

$$\frac{nxy + mxz}{ax} = \frac{lyz + nxy}{by} = \frac{mxz + lyz}{cz}$$

$$= \frac{2lyz}{by + cz -ax}$$

=two similar expressions;

$$\frac{l}{x\left(by + cz - ax\right)} = \frac{m}{y\left(cz + ax -by\right)} = \frac{n}{z\left(ax + by - cz\right)}$$.​

 

[Miike012]

how did they get from $$= \frac{\frac{y}{m} + \frac{z}{n}}{2la}$$
to $$\frac{ny + mz}{a} = \frac{lz + nz}{b} = \frac{mx + ly}{c}$$
thats the only thing I am stuck on

 

[vela]

You got to

$$\frac{\frac{y}{m} + \frac{z}{n}}{2la} = \frac{\frac{z}{n} + \frac{x}{l}}{2mb} = \frac{\frac{y}{m} + \frac{x}{l}}{2nc}$$

Try multiplying all three by 2lmn:

$$\frac{\frac{y}{m} + \frac{z}{n}}{2la}\times 2lmn= \frac{\frac{z}{n} + \frac{x}{l}}{2mb} \times 2lmn= \frac{\frac{y}{m} + \frac{x}{l}}{2nc}\times 2lmn$$

and then simplify each term.

 

[Miike012]

so your telling me that I can pull a 2lmn out of the blue stick it into the equation for no apparent reason? That just doesn't make sense to me.

 

[romsofia]

so your telling me that I can pull a 2lmn out of the blue stick it into the equation for no apparent reason? That just doesn't make sense to me.

It doesn't come out the blue, look at the equation: $$\frac{\frac{y}{m} + \frac{z}{n}}{2la} = \frac{\frac{z}{n} + \frac{x}{l}}{2mb} = \frac{\frac{y}{m} + \frac{x}{l}}{2nc}$$

Notice how 2l, 2m, 2n (respectively) are in front of the a, b, c? Looks like they wanted to get the equations set over a, b, and c and that's the way to do it.

 

[vela]

You can also do it step by step. For instance, you want to get rid of the 2 in the denominators, so the obvious way would seem to be to multiply through by 2 to get

$$\frac{\frac{y}{m} + \frac{z}{n}}{la} = \frac{\frac{z}{n} + \frac{x}{l}}{mb} = \frac{\frac{y}{m} + \frac{x}{l}}{nc}$$

Now, to get a alone in the denominator of the first fraction, you multiply through by l. When you do this, you'll see parts of the numerator in the other fractions start to look like what you want. So you just keep going and hope it works out.

Sorry if it seemed like I pulled the factor 2lmn out of the air, but after years of doing tons of algebra, I can see these patterns relatively quickly. It's a skill you develop just from working out a bunch of problems.

 

[Miike012]

Wow, after you explained it, it sound easy lol... I think my problem is, I over think the problem and make it harder than it needs to be, then I mess up...