1. ## some one please explain

I do not get it
What 4/5=1/x+4 how do I solve please explain step by step my instructions suck

2. ## algebra problem

Originally Posted by sorch
4/5=1/x+4
first arrange so that 1/x is on one side of the quation like so:

1/x = 4/5 - 4

Then simplify the right hand side like so:

1/x = 4/5 - 20/5

1/x = -16/5

Then invert both sides

x/1 = -5/16

or

x = -5/16,

plug it in to the equation to be sure of your answer:

4/5=1/(-5/16)+4

4/5=-16/5+4

4/5=-16/5+20/5

4/5=4/5

3. Here is one way.
It is my favorite way in dealing with fractions.

4/5=1/x+4

I assume that is
4/5 = 1/x +4 ----(i)

[Not 4/5 = 1/(x+4) ----(ii)]

There are two fractions in (i): 4/5 and 1/x.
I always clear/remove/eliminate fractions as my first move.
To "clear/remove/eliminate" here means to make all terms in whole numbers, or no more fractions. For me, it is easier to play with whole numbers than with fractions.

To clear the fractions, multiply both sides of the equation by a common multiplier. This common multiplier is the product (multiplication) of all the denominators, as long as no denominator is a repeater in the common multiplier.

The two denominators in (i) are 5 and x, so, multiply both sides of (i) by 5*x,
4*x = 1*5 +4*5*x
4x = 5 +20x
4x -20x = 5
-16x = 5
x = 5/(-16)

4. ## hmm?

5. No, they both got the same answer just a different way of doing it.

6. Originally Posted by mattballer082
This point comes up a lot. Mathguru and ticbol both used perfectly valid ways of solving the problem by transforming the equation into one of the form x = answer (and then checking that the answer satisfied the original equation). Your teacher may recommend a particular way of transforming the original equation and may even go so far as to call that way 'correct' with the implication that other ways aren't correct. But if a transformation is valid then it's valid whether or not it's useful. Here's a simple example

2x+4 = 10

A. You could
1) Divide both sides by 2
x+2 = 5
2) Subtract 2 from both sides
x = 3

B. You could
1) Subtract 4 from both sides
2x = 6
2) Divide both sides by 2
x = 3

Solutions A and B are equally valid and thus give the same answer. You or your teacher might recommend one over the other.

The transformations 'Subtract the same number from both sides' and 'Divide both sides by the same number' are equally valid. Which order you do them in here is a matter of taste not correctness. Incidentally I much prefer this way of looking at it: it's clear that if two sides are equal, and you apply the same operation to each, then the resulting sides are also equal. Of course in terms of manipulation the step 'Subtract 4 from both sides' is often stated as 'Take the 4 over to the other side' but that's shorthand for what happens when you carry out the operation.

x = -5/16

x = -5/16

If I'd say mine is correct, then MathGuru's answer is correct also.

Our solutions, or ways in finding the answer, are not the same. So if I'd say my solution is correct, then MathGuru's is not correct?
No. MathGuru's solution is correct also.
In Math, there is only one answer, but there are many different ways to solve for the answer. Any of these different ways is correct, as long as the same correct answer is found in using any of these ways.

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That is why I say oftentimes, "Here is one way" before I show my solution.

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In MathGuru's solution, he sticked with the fractions. He solved by using fractions, which, maybe, is closer to your lessons/exercises/homeworks now---fractions. If you are studying fractions now, then you should use fractions to get acquainted with them.

In my solution, I eliminated the fractions. Because, again, for me, its easier to play with whole numbers than with fractions. I am not compelled to use fractions. I am not studying anymore. I can go to easier solutions if I like.

8. ## More on different ways to the same one answer.

I thought I should post this separately.

Oftentimes too, if not always, I post solutions that are "long" or too-detailed. Is this the manner that I solve questions by myserlf? Is this the way that I do solutions to Math questions in my own?

No.
In my own, I find answers very fast. A few scribblings here, some calculator's calculation here---if needed, a sketchy figure there---if needed, and the answer.
Then a fast review, checking, revisions---if needed, and then I'm ready to type a solution to post.
This typing a solution takes longer, much longer, than my getting the answer on my own. [I type with one finger only---my right accusing finger. :-)] It is because of my one-finger typing style, and because of my need to show a different solution that is supposed to be understood by all, or at least understood by the seeker of the answer.
Meaning, my own solution and the solution that I post are, most of the time, different from each other.

That there is one main reason why I love Math. There are many ways to solve a Math problem. And I am free to choose which one way, or two or three ways, to arrive at the answer. If one way supports the other way, I always smile.

Back then, in school, in college, I was told to get out of the classroom, more than once, because I did not follow my professors' solutions/explanations re some problems/topics/subjects. I was a "free bird" even then.

So, if you think you have a way, or two, to solve a problem/lesson that is different from your teacher's, why, go ahead and do it your way!

Math is beautiful.

9. Just a warning that it is possible for invalid methods to (sometimes) give the correct answer! There's a well known example: simplify 16/64. 'Cancelling' the 6 top and bottom happens in this case to give the correct answer 1/4. Of course it isn't a valid method in general (try it on 36/64 for example!).