Solve for ##x## in the Given Problem

[chwala]

Homework Statement

I normally set and create my own math problems ; This one was a bit interesting ...need your input.

Find $x$ given:



$-2\dfrac{2}{x} - 1\dfrac{4}{x} =20$

Relevant Equations

Equations with unknown variable

In the first approach; which yields the correct solutiuon we have;$-\left[\dfrac{3x+6}{x}\right] =20$

I factored out the negative...solved for $x$ then re-introduced the negative at the end.

$3x+6=20x$

$6=17x$

$x=-\dfrac{6}{17}$

Now to my question: I tried doing this...

$-\left[\dfrac{3x+6}{x}\right] =20$ then i multiplied both sides by $-1$ i.e

$\left[\dfrac{3x+6}{x}\right] =-20$

$3x+6=-20x$

i.e
$-\dfrac{3x+6}{x}=\left[-1 ×\dfrac{3x+6}{x}\right]=20$
$\left[-1× -1 ×\dfrac{3x+6}{x}\right]=-20$

$x=-\dfrac{6}{23}$ which is clearly not correct! following my checking through substitution...something i am missing here?

I know that;

$-\left[\dfrac{3x+6}{x}\right]=-\dfrac{3x+6}{x} = \dfrac{-3x-6}{x} = \dfrac{3x+6}{-x}$ cheers guys!I think i see the mistake the negative only applies to the numerator...was a nice one...time for coffee lol!

 

[BvU]

Hi,

This one was a bit interesting

You can make it a whole lot less interesting if you define $y={1\over x}$ and solve for $y$ !

$\$

 

[chwala]

Hi,

You can make it a whole lot less interesting if you define $y={1\over x}$ and solve for $y$ !

$\$

I'll check this out...later mate. Cheers...

 

[Steve4Physics]

@chwala, neither of your answers are correct!
$-\left[\dfrac{3x+6}{x}\right] =20$ is wrong.

And even if the above equation were correct, your next line
$3x+6=20x$
would also be wrong; it should be
$-3x-6=20x$ or
$3x+6=-20x$

I would solve your original equation by multiplying both sides by $x$. Or use @BvU's method.

Also, always check your answer by substituting it back into the original equation.

 

[Steve4Physics]

@chwala, neither of your answers are correct!
$-\left[\dfrac{3x+6}{x}\right] =20$ is wrong.

And even if the above equation were correct, your next line
$3x+6=20x$
would also be wrong; it should be
$-3x-6=20x$ or
$3x+6=-20x$

I would solve your original equation by multiplying both sidesby $x$. Or use @BvU's method.

Also, always check your answer by substituting it back into the original equation.

 

[chwala]

@chwala, neither of your answers are correct!
$-\left[\dfrac{3x+6}{x}\right] =20$ is wrong.

And even if the above equation were correct, your next line
$3x+6=20x$
would also be wrong; it should be
$-3x-6=20x$ or
$3x+6=-20x$

I would solve your original equation by multiplying both sidesby $x$. Or use @BvU's method.

Also, always check your answer by substituting it back into the original equation.

Interesting you're saying $x=\dfrac{-6}{17}$ is not correct? Note that in my approach i worked with improper fractions throughout.

Let me check my working again...

if i substitute;

$x=\dfrac{-6}{17}$ back into original equation i would have;

$2+5+\dfrac{2}{3}+1+11+\dfrac{1}{3}=20$

note that in my working i took;

$-2\dfrac{2}{x}= \dfrac{-2x-2}{x}$

Let me check my working again...back after break.

 

[pasmith]

I don't think I've ever seen juxtaposition used to indicate a mixed number with an unknown in the denominator, or indeed anywhere. I would naturally interpret <br /> 2\frac{2}{x} = 2 \times \frac{2}{x} = \frac{4}{x} rather than as 2 + \frac{2}{x}.

 

[BvU]

note that in my working i took;

$-2\dfrac{2}{x}= \dfrac{-2x-2}{x}$

Let me check my working again...back after break.

Huh ?
$-2-\dfrac{2}{x}= \dfrac{-2x-2}{x}$ !!

While you are checking anyway, please check if we are trying to solve

Find $x$ given:

$-2\dfrac{2}{x} - 1\dfrac{4}{x} =20$

or something else entirely !!

$\$

 

[chwala]

Huh ?
$-2-\dfrac{2}{x}= \dfrac{-2x-2}{x}$ !!

While you are checking anyway, please check if we are trying to solve

or something else entirely !!

$\$

My thinking on this is based on addition and subtraction of Mixed fractions...unless you may want to indicate if my way of thought is wrong on this.

 

[chwala]

Huh ?
$-2-\dfrac{2}{x}= \dfrac{-2x-2}{x}$ !!

While you are checking anyway, please check if we are trying to solve

or something else entirely !!

$\$

Yes we are trying to solve that...

 

[Steve4Physics]

@chwala, I think there is a basic communications problem here! To expand on what @pasmith said in Post #7....

An expression such as
$2\dfrac{2}{x}$
could be interpreted in two ways:

Interpretation-1 (the usual one): simple multiplication:
Take the product of $2$ and $\dfrac{2}{x}$

Interpretation-2: a mixed fraction:
Take the sum of $2$ and $\dfrac{2}{x}$

I’m guessing you are using interpretation-2 but I (and maybe others) have used interpretation-1.

Edit: typo'

 

[Mark44]

Homework Statement:: I normally set and create my own math problems ; This one was a bit interesting ...need your input.

Find $x$ given:

$-2\dfrac{2}{x} - 1\dfrac{4}{x} =20$

Based on comments along the way in this thread, it appears that you intend that the terms on the left side are to be interpreted as mixed fractions akin to 1 7/8; i.e., 1 + 7/8.

I taught algebra (and trig, calculus, linear algebra, and differential equations) for many years, but have never seen any algebra textbook that carried the notion of mixed fractions over to algebraic expressions.

As noted several times here, $-2\dfrac{2}{x}$ would be almost universally interpreted as $-2 \cdot \frac 2 x$. Making up your own problems is good, but you shouldn't make up your own notation that is in conflict with notation in common usage.

 

[chwala]

Based on comments along the way in this thread, it appears that you intend that the terms on the left side are to be interpreted as mixed fractions akin to 1 7/8; i.e., 1 + 7/8.

I taught algebra (and trig, calculus, linear algebra, and differential equations) for many years, but have never seen any algebra textbook that carried the notion of mixed fractions over to algebraic expressions.

As noted several times here, $-2\dfrac{2}{x}$ would be almost universally interpreted as $-2 \cdot \frac 2 x$. Making up your own problems is good, but you shouldn't make up your own notation that is in conflict with notation in common usage.

@Mark44 ...that's true maybe I should have indicated my intention...it depends on the context...when am solely looking at adding and subtraction Fractions, I do incorporate this ...cheers man!

 

[pbuk]

@Mark44 ...that's true maybe I should have indicated my intention...it depends on the context...when am solely looking at adding and subtraction Fractions, I do incorporate this ...cheers man!

Well you need to stop doing that. Just in case there is any doubt, you cannot simply state that "when I write $2 \frac 3 x$ I mean $2 + \frac 3 x$" and expect to communicate with anyone else. It is equally as invalid as saying "when I write $6 + 3 = 2$ I mean $6 \div 3 = 2$".

 

[chwala]

Well you need to stop doing that. Just in case there is any doubt, you cannot simply state that "when I write $2 \frac 3 x$ I mean $2 + \frac 3 x$" and expect to communicate with anyone else. It is equally as invalid as saying "when I write $6 + 3 = 2$ I mean $6 \div 3 = 2$".

I clearly get your point...
What I meant was when I am looking at fractions...I would occasionally add the mixed fractions as required (no variables involved) then later replace any of the numbers with a letter (variable)...then seek to find the unknown (letter)variable...

 

[chwala]

By the way do we have a general rule in Math on how to interprete the problem? it can either be looked at as multiplication ( accepted universally) or as addition of mixed fractions as I had indicated. In both contexts they follow math rules to the latter. Cheers.

 

[Mark44]

By the way do we have a general rule in Math on how to interprete the problem? it can either be looked at as multiplication ( accepted universally)

This -- based on the problem in post #1.

or as addition of mixed fractions as I had indicated. In both contexts they follow math rules to the latter.

Not this. In my experience with algebra and subsequent textbooks, mixed fractions play no role once we move from ordinary grade-school arithmetic to algebra. Terms that are juxtaposed in algebra expressions are interpreted as being multiplied.

$-2\dfrac{2}{x} - 1\dfrac{4}{x} =20$

This would have been much clearer:
$-2 - \frac 2 x - 1 - \frac 4 x = 20$

 

[pbuk]

By the way do we have a general rule in Math on how to interprete the problem? it can either be looked at as multiplication ( accepted universally) or as addition of mixed fractions as I had indicated. In both contexts they follow math rules to the latter. Cheers.

Yes. Mixed fractions are not used in mathematics or physics. They are used in the US in engineering (e.g. a 1⅜" diameter pipe). But you need to understand the difference between a mixed fraction, which is notation for a constant, a particular point on the number line, and an expression like $\frac 2 x$ which is notation saying "divide 2 by x".

Compare with the "-" sign which means something different in $a = -1$ and $a = 0 - 1$.