Make x the subject of (2xy)/z + 2x - L = 0

[richiesmasher]

Here is an equation, with three varibles x, z, and L

. . .(2xy)/z + 2x - L = 0

I want to make x the subject.

I would think to do this:

. . .(2xy)/z + 2x - L = 0

Bring L across

. . .(2xy)/z + 2x = L

Then factorize the left side:

. . .2x(y/z + 1) = L

Simplify the term (y/z + 1) into ((y + z)/z):

. . .2x ((y + z)/z) = L

Then multiply both sides by z:

. . .2x(y + z) = Lz

Divide by (y + z):

. . .2x = Lz/(y + z)

Divide by 2:

. . .x = Lz/(2y + 2z)

Is this correct?

 

[JeffM]

Yes, it is correct. Good job.

A few minor points of mathematical vocabulary, at least in the US. The verb is "factor" rather than "factorize." You are "solving the equation for x in terms of the other variables" or "making x the dependent variable" are probably more frequent verbalizations of what you are doing.

Finally, here is how to check your work for this kind of problem. Pick some small prime odd numbers for the independent variables.

Say L = 3, Y = 5, and Z = 7.

\(\displaystyle X = \dfrac{3 * 7}{(2* 5) + (2 * 7)} = \dfrac{21}{10 + 14} = \dfrac{21}{24} = \dfrac{7}{8}.\)

Now see if the other formulation of the equation works.

\(\displaystyle \dfrac{2 * \dfrac{7}{8} * 5}{7} + 2 * \dfrac{7}{8} - 3 = \dfrac{\dfrac{70}{8}}{\dfrac{7}{1}} + \dfrac{14}{8} - \dfrac{3 * 8}{8} = \)

\(\displaystyle \dfrac{70}{8} * \dfrac{1}{7} + \dfrac{14}{8} - \dfrac{24}{8} = \dfrac{10 + 14 - 24}{8} = 0.\)

It checks.

 

[richiesmasher]

Yes, it is correct. Good job.

A few minor points of mathematical vocabulary, at least in the US. The verb is "factor" rather than "factorize." You are "solving the equation for x in terms of the other variables" or "making x the dependent variable" are probably more frequent verbalizations of what you are doing.

Finally, here is how to check your work for this kind of problem. Pick some small prime odd numbers for the independent variables.

Say L = 3, Y = 5, and Z = 7.

\(\displaystyle X = \dfrac{3 * 7}{(2* 5) + (2 * 7)} = \dfrac{21}{10 + 14} = \dfrac{21}{24} = \dfrac{7}{8}.\)

Now see if the other formulation of the equation works.

\(\displaystyle \dfrac{2 * \dfrac{7}{8} * 5}{7} + 2 * \dfrac{7}{8} - 3 = \dfrac{\dfrac{70}{8}}{\dfrac{7}{1}} + \dfrac{14}{8} - \dfrac{3 * 8}{8} = \)

\(\displaystyle \dfrac{70}{8} * \dfrac{1}{7} + \dfrac{14}{8} - \dfrac{24}{8} = \dfrac{10 + 14 - 24}{8} = 0.\)

It checks.

Oh ok, thanks thats a good tip for using prime numbers to check answers, was pretty frustrating at first but I have some knowledge now about this kind of thing.