Solving for x in terms of other variables (with factors)

[clueless666]

Hello everyone, hoping this is in the right forum.
I am having trouble understanding how the teacher solved for x in the following expression:



I am currently stuck at this step:


Thank you in advance <(_ _)>

 

[Steven G]

Move, the (x-l) in the denominator to the numerator on the other side of the equal sign. Then distribute the factors with x in them. Post back with your work so we can offer additional help.

 

[Dr.Peterson]

Since the big expression [MATH]\frac{z\sqrt{np(1-p)}}{n}[/MATH] is effectively a constant, you may find the work easier if you temporarily replace it with a single variable: [MATH]A=\frac{-lp-x(1-p)}{x-l}[/MATH]. Then do as Jomo suggested.

 

[JeffM]

One way to reduce errors on this kind of problem and “relieve the imagination” is called “substitution of variables.

Turn complex factors of x into single variables.

[MATH]\text {Let } u = (1 - p) \text { and } v = \sqrt{np(1 - p)}.[/MATH]
[MATH]\therefore z = \dfrac{- \{Lp + x(1 - p)\}n}{(x - L)\sqrt{np(1 - p)}} = \dfrac{- n(Lp + xu)}{v(x - L)}.[/MATH]
Easier to understand and far less likely to make a transcription error. Now clear fractions as jomo suggested.

[MATH]\therefore zv(x - L) = - n(Lp + xu) \implies xvZ- Lvz = - Lnp - xnu \implies xvz + xnu = Lzv - Lnp \implies[/MATH]
[MATH]x(zv + nu) = L(zv - np) \implies x = \dfrac{L(zv -np)}{zv + nu}. [/MATH]
Pretty basic algebra.

Now substitute back.

[MATH]x = L * \dfrac{z\sqrt{np(1 - p)} - np}{z \sqrt{np(1 - p)} + n(1 - p)}.[/MATH]
That is equivalent to what your teacher got but less easy to screw up.