solve 1/t = 1/m - 1/n for m

1/t = 1/m - 1/n

if i move the 1/n such as
1/t + 1/n = 1/m

then do i need to multiply to clear fractions

m(1/t) + m(1/n) = m(1/m)

so would i write it as m/t + m/n = m?

would i need to clear the other fractions?

Thank You

Re: solve for m

There are different approaches to a problem such as this. Anytime there are fractions involved I like to determine the least common denominator. Once I have that, I multiply through (all terms on both sides of the equation) by that lcd. That leaves me with a simple equation with no fractions which will be solved easily. In this case you will get the terms containing t on the same side of the equation and the other term on the other side. Then factor out the t and one more step completes the process.
Remember that you need to get t all by itself on one side of the equation. Getting 1/t on one side doesn't do it.

Re: solve for m

ginny1029 said:

m(1/t) + m(1/n) = m(1/m)
so would i write it as m/t + m/n = m?

Click to expand...

Are you saying that m(1/m) = m ? NOooooooo :shock:

What is the question anyway? Solve for m?

Re: solve for m

yes i am suppose to solve for m

so what do i do

Re: solve for m

ginny1029 said:

so what do i do

Click to expand...

Do your best :wink:

Re: solve for m

i tried that but you said it was wrong im just asking how to do it correctly

ginny1029 said:

1/t = 1/m - 1/n

Click to expand...

To learn, in general, how to solve literal equations, try here. :wink:

Then:

. . . . .\(\displaystyle \frac{1}{t}\, =\,\frac{1}{m}\, -\, \frac{1}{n}\)

You want to get at the \(\displaystyle m\), so multiply through to get it on top:

. . . . .\(\displaystyle \frac{m}{t}\, =\, 1\, -\, \frac{m}{n}\)

Now get all the \(\displaystyle m\) terms together alone on one side of the "equals" sign:

. . . . .\(\displaystyle \frac{m}{t}\, +\, \frac{m}{n}\, =\, 1\)

The trick here (and it is a trick, so take notice) is to factor:

. . . . .\(\displaystyle m\left(\frac{1}{t}\, +\, \frac{1}{n}\right)\, =\, 1\)

Now divide through, simplify, etc, etc....

Re:

stapel said:

. . . . .\(\displaystyle \frac{m}{t}\, +\, \frac{m}{n}\, =\, 1\)

Click to expand...

Or multiply by tn:
mn + mt = tn
m(n + t) = tn ; finish it