Question

Need help figuring out how to solve this problem

Answer 1

\[\frac{ 3 }{ m-4 } + \frac{ 1 }{ 3(m-4) } = \frac{ 6 }{ m }\]

Answer 2

You should notice that both the fractions on the left have a similar denominator. This is a way to simplify your problem.

Remember to add fractions they need to have the same denominator

Hint:
You will need this rule to solve the problem:

-You can multiply the denominator and the numerator by the same number and not change the value of the fraction


For example:

\[\frac{1}{2} = \frac{1}{2}*\frac{2}{2} = \frac{2}{4}\]

This is because,

\[\frac{2}{2} = 1\]

and as you know multiplying by 1 doesn't change a number.

Answer 3

Well actually you dont need to use that rule I showed you but it is good to know how to manipulate numbers

Answer 4

can you simplify your problem using the rule I just gave you?

Answer 5

I know. But I don't understand how to solve the problem. Because when I solve it always get different answers. I try to solve it the way the book tells me, but I never get an answer

Answer 6

Alright,

show me the first step you take

Answer 7

in solving this problem

Answer 8

The first step I do, is to find the LCD. The denominators are -4, m, and 3 if I keep the problem the same. If I use distributive property on the second fraction, the denominators become -4, 3, m, and 12

Answer 9

Using LCD is over complicating a simple algebra problem, to be honest I never use that stuff.

say we have the problem


\[\frac{3}{3+x} + \frac{1}{3(3+x)} = 32\]

First step I do is multiply both sides of the equation by one of the denominators, so lets do that:

\[(\frac{3}{3+x} + \frac{1}{3(3+x)})\frac{(3+x)}{1} = \frac{32}{x}\frac{(3+x)}{1}\]


Now we multiply it out and we get,

\[\frac{3(3+x)}{1(3+x)} + \frac{(3+x)1}{3(1)(3+x)} = \frac{32(3+x)}{x(1)}\]
=
\[\frac{3(3+x)}{(3+x)} + \frac{(3+x)}{3(3+x)} = \frac{32(3+x)}{x}\]

Now notice that,

\[\frac{3+x}{3+x} = 1\]

so we can split these numbers up more

\[\frac{3}{1}(\frac{3+x}{3+x}) + (\frac{1}{3})\frac{3+x}{3+x} = \frac{32(3+x)}{x}\]

so now we have,

\[\frac{3}{1}(1)+ (\frac{1}{3})(1) = \frac{32(3+x)}{x}\]

=

\[\frac{3}{1}+ \frac{1}{3} = \frac{32(3+x)}{x}\]

you can now easily solve just doing the same thing again only with x

Answer 10

Can you solve the original equation? I am really confused??!

Answer 11

Just to note you can divide any number by 1 or show any number as a fraction by dividing by 1

I just use it to show the fraction multiplication

For example:

\[\frac{3}{1} = 3\]

\[\frac{3333333 + x + y + z}{1} = 3333333 + x + y + z\]

etc..

Answer 12

What is confusing you exactly?

Answer 13

please take the time to try to understand what I did, read it step by step

Answer 14

f you learn the simple concepts I'm putting forth you wont have a problem with questions such as these anymore

Answer 15

M would then = 9