Simplify (1/3)(dz/dx - 2) = -(z - 1)/(z + 2)

[Evil]

\(\displaystyle \left(\dfrac{1}{3}\right)\left(\dfrac{dz}{dx}\, -\, 2\right)\, =\, -\dfrac{z\, -\, 1}{z\, +\, 2}\)

\(\displaystyle \dfrac{dz}{dx}\, =\, \dfrac{-z\, +\, 7}{z\, +\, 2}\)

Hi, please simplify this with steps.

Thanks.

 

[Ishuda]

\(\displaystyle \left(\dfrac{1}{3}\right)\left(\dfrac{dz}{dx}\, -\, 2\right)\, =\, -\dfrac{z\, -\, 1}{z\, +\, 2}\)

\(\displaystyle \dfrac{dz}{dx}\, =\, \dfrac{-z\, +\, 7}{z\, +\, 2}\)


Hi, please simplify this with steps.

Thanks.

What are your thoughts? What have you done so far? Please show us your work so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

 

[Steven G]

\(\displaystyle \left(\dfrac{1}{3}\right)\left(\dfrac{dz}{dx}\, -\, 2\right)\, =\, -\dfrac{z\, -\, 1}{z\, +\, 2}\)

\(\displaystyle \dfrac{dz}{dx}\, =\, \dfrac{-z\, +\, 7}{z\, +\, 2}\)

Hi, please simplify this with steps.

Thanks.

You want to write the equation in the form f(z)dz = f(x)dx.
f(z) only has z's and #s in it and f(x) only has x's and #s in them. Note that f(z) can be a fraction or only be 'in the denominator'
Here is an example that I will separate for you.
dy/dx = [(y+3)(x-9)]/(x+2)
dy/(y+3)=[(x-3)/(x+2)]dx
Now you would integrate both sides

 

[Evil]

You want to write the equation in the form f(z)dz = f(x)dx.
f(z) only has z's and #s in it and f(x) only has x's and #s in them. Note that f(z) can be a fraction or only be 'in the denominator'
Here is an example that I will separate for you.
dy/dx = [(y+3)(x-9)]/(x+2)
dy/(y+3)=[(x-3)/(x+2)]dx
Now you would integrate both sides

1/3 + 2 dz/dx = -z-1 / z+2

7 /3+ z+2/-z-1 dz = dx

Did not get how to simplify now for getting z+2/-z+7, can you please explain ??