A little help needed

Algernon

Hello, I am working through some problem sets and my answers are coming out a littler different to the given solution. I reckon I must be missing a little rule or trick somewhere, so I will provide my attempt below, and the correct solution and maybe someone can guide me?

Ok, I will just come in at the point where its going wrong.

Given 1) 2P2X2 = P1X1 (which is correct)

Solve the above for X2 and plug into:

2) M = P1X1 + P2X2

So, 1) becomes X2 = P1X1/2P2

and plugs into 2):

M = P1X1 + P2(P1X1/2P2)

The P2's cancel, giving:

M = P1X1 + P1X1/2

2M = P1X1 + P1X1

2M = 2P1X1

2M/2P1 = X1

The answer given in the notes is:

2M/3P1 = X1

So where did I go wrong? I will post my other one in a sec.

blueshoes

Algernon wrote: »

Hi

I think its a small error on the 2nd line below...

> The P2's cancel, giving:
> M = P1X1 + P1X1/2

you multiple both sides of the equation by 2, to get rid of the 2 under the line...

that gives...

2M = 2P1X1 + P1X1

think it works out...

good luck

Algernon

The other is a similar problem:

Given 1) 2X1 = 3X2

Solve for X1, X2 for:

2) M = P1X1 + P2X2

So, starting with 1) X1 = 3X2/2

Plugging into 2):

M = P1(3X2/2) + P2X2

2M = P1(3X2) + P2X2

2M/3P1 + P2 = 2X2

2M/6P1 + P2 = X2

And the given solution is:

X2 = 2M/3P1 + 2P2

Similarly, I get the numerator correct, but the denominator wrong for X1.

SOLVED.

Algernon

blueshoes wrote: »

Algernon wrote: »

Hi

I think its a small error on the 2nd line below...

> The P2's cancel, giving:
> M = P1X1 + P1X1/2

you multiple both sides of the equation by 2, to get rid of the 2 under the line...

that gives...

2M = 2P1X1 + P1X1

think it works out...

good luck

Ah right, that makes sense.

Thanks!

blueshoes

Similar mistake..
> 2M = P1(3X2) + P2X2
should be
2M = P1(3X2) + 2P2X2

with that i got this answer..

X2 = 2M / (3P1 + 2P2)