Solve Linear Equations: 1/5(1/3x-5)=1/3(3-1/x)

In summary: And yes, it is always helpful to include plain text explanations for those who may not be familiar with the notation being used. Thank you for pointing that out.
  • #1
abhishekdas1
1
0
1/5(1/3x-5)=1/3(3-1/x)
 
Mathematics news on Phys.org
  • #2
I'm assuming you're trying to solve this equation. What have you tried? Where are you stuck?
 
  • #3
abhishekdas said:
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.
 
  • #4
abhishekdas said:
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 [tex]\frac{1}{5}\frac{1}{3x- 5}= \frac{1}{3}\left(3- \frac{1}{x}\right)[/tex].

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 [tex]3x= 45x^2- 60x+ 25[/tex] or [tex]45x^2- 63x+ 25= 0[/tex]. But that equation has no real number roots.
 
  • #5
1/(15x) -1=1- 1/(3x)

- - - Updated - - -

6/15x =2
 
  • #6
Tennisgoalie said:
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.
 
  • #7
Evgeny.Makarov said:
Tennisgoalie said:
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:

Tennisgoalie said:
abhishekdas said:
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.
 
  • #8
Ah, sorry, I should have checked the usernames. Thanks to Tennisgoalie for the help.
 

FAQ: Solve Linear Equations: 1/5(1/3x-5)=1/3(3-1/x)

How do I solve linear equations?

To solve a linear equation, you need to isolate the variable on one side of the equation and simplify the other side. You can do this by using inverse operations, such as addition, subtraction, multiplication, and division, to get the variable by itself. In this equation, you can start by simplifying the left side of the equation by multiplying 1/5 and 1/3x, then subtracting 5 from both sides to isolate the variable.

What is the first step in solving this equation?

The first step in solving this equation is to simplify the left side of the equation by multiplying 1/5 and 1/3x. This will give you 1/15x - 5 = 1/3(3-1/x).

How do I handle fractions in linear equations?

To handle fractions in linear equations, you can use the distributive property to simplify the equation. This involves multiplying both sides of the equation by the least common multiple (LCM) of the fractions. In this equation, the LCM of 1/5 and 1/3 is 15, so you would multiply both sides by 15 to eliminate the fractions.

Can I check my solution for this equation?

Yes, you can check your solution by substituting the value you found for the variable back into the original equation. If the equation is true, then your solution is correct. In this equation, you can plug in your solution for x and see if the left side of the equation equals the right side.

Is this equation in standard form?

No, this equation is not in standard form. Standard form for a linear equation is ax + by = c, where a, b, and c are constants. To put this equation in standard form, you would have to simplify it and rearrange the terms to have the variable on one side and the constants on the other.

Similar threads

Replies
2
Views
1K
Replies
7
Views
2K
Replies
10
Views
2K
Replies
7
Views
3K
Replies
1
Views
920
Replies
4
Views
1K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
Back
Top