# Linear equations

#### abhishekdas

##### New member
1/5(1/3x-5)=1/3(3-1/x)

#### Ackbach

##### Indicium Physicus
Staff member
I'm assuming you're trying to solve this equation. What have you tried? Where are you stuck?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
1/5(1/3x-5)=1/3(3-1/x)
Besides, the notation 1/3x is somewhat ambiguous: it may mean either (1/3)x or 1/(3x). The same pertains to coefficients 1/5 and 1/3: it is recommended to write (1/5) if this number is followed by multiplication. Note that if 1/3x means (1/3)x, then the equation is not linear. So please clarify what you mean by inserting parentheses.

#### HallsofIvy

##### Well-known member
MHB Math Helper
1/5(1/3x-5)=1/3(3-1/x)
Since you titled this "linear equations" I would have assumed that your "1/3x" mean "(1/3)x"- that is, "one third times x" rather than 1/(3x), 1 divided by 3x. But then your "1/x" confuses me. From that I have to conclude that this is NOT, as it stands, a "linear equation" and you intend $$\frac{1}{5}\frac{1}{3x- 5}= \frac{1}{3}\left(3- \frac{1}{x}\right)$$.

The simplest way to handle problems like this is to get rid of all those fractions by multiplying both sides by the "least common denominator". Since there are no "duplicates" here, the "least common denominator" is just the product of all of the denominators (5)(3x- 4)(3)(x). Multiplying on both side by that will cancel the denominators "5" and "3x - 5" on the left and the denominators "3" and "x" on the right leaving 3x= 5(3x- 5)(3x- 1) which still is not linear- it is the quadratic equation $$3x= 45x^2- 60x+ 25$$ or $$45x^2- 63x+ 25= 0$$. But that equation has no real number roots.

#### Tennisgoalie

##### New member
1/(15x) -1=1- 1/(3x)

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6/15x =2

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
1/(15x) -1=1- 1/(3x)

- - - Updated - - -

6/15x =2
If you have no further questions, then it is a good idea to mark the thread as solved. Otherwise, please post your questions. The thing is that contrary to what some people may think, bare formulas almost never constitute a piece of mathematical work. They must be accompanied by plain text explanations saying what we want to do with such formulas (e.g., solve an equation or find a counterexample), whether a given formula is an assumption or something to prove, what the difficulty of the problem is, why should one consider such problem interesting and so on.

#### MarkFL

Staff member
1/(15x) -1=1- 1/(3x)

- - - Updated - - -

6/15x =2
If you have no further questions, then it is a good idea to mark the thread as solved. Otherwise, please post your questions. The thing is that contrary to what some people may think, bare formulas almost never constitute a piece of mathematical work. They must be accompanied by plain text explanations saying what we want to do with such formulas (e.g., solve an equation or find a counterexample), whether a given formula is an assumption or something to prove, what the difficulty of the problem is, why should one consider such problem interesting and so on.
What happened here is that help was given by Tennisgoalie to the OP. Admittedly, this would perhaps have been more clear had the post looked more like:

1/5(1/3x-5)=1/3(3-1/x)
Assuming the equation is to be interpreted as follows:

(1/5)(1/(3x) - 5) = (1/3)(3 - 1/x)

then distribution on both sides yields:

1/(15x) -1 = 1 - 1/(3x)

Multiplying 1/(3x) by 5/5 to get 5/(15x), we may then add 1 + 5/(15x) to both sides to obtain:

6/(15x) = 2

Can you continue?
However, Tennisgoalie just joined MHB within the last 24 hours and may be new to math help sites in general and inexperienced with how to most effectively provide help to others, such as to use plain text to elucidate the steps taken.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Ah, sorry, I should have checked the usernames. Thanks to Tennisgoalie for the help.