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

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ginny1029

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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

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.... :D
 
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
 
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