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I tried this(couldn't find latex code to cross over):

$$\frac{3}{5}-\frac{4}{5x}+\frac{1}{x}=\frac{3}{5}\cdot5x-\frac{4}{5x}\cdot5x+\frac{1}{x}\cdot5x\neq 3x-4+5$$

Why is this not possible?

My spesific problem is that I was under the impression that we could cross out (because everything can be adjustet to fit as long as we don't change the value) and else do this the same way we do fractions. I now see I did not have equal values. But why do we have to factorize an expression if it contains addition/subtracton in the numerator? Is this convention/definitions? For example we can't take (2x-4), we have to use 2(x-2) before we cross out? I'm working on my understanding. I have previously been very mechanic in mathematics. I did at least 40-50 different variants of this kind of problems before. Thats why I wonder how come I now suddenly don't know what I'm doing. I have forgotten alot/didn't work on in properly. So this time, before exam, - and because I know I have done this several times before - i try to only do problems whitout looking in the book before I really need it. The problem is that my book just says "do it the same way as we did with fractions, just remember parentheses around '''summation expressions(?)'''" I guess my question is more general than just my example, I will be more accurate in future questions. Thanks.

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    $\begingroup$ An expression is not equal to the same expression multiplied with $5x$. $$\frac{3x-4+5}{5x}$$ is correct assumng $x\ne 0$ $\endgroup$ – Peter Mar 26 '18 at 7:44
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Why would it be possible?

Why did you think that if $A = \frac{3}{5}-\frac{4}{5x}+\frac{1}{x}$ then $A = A*5x$???? Obviously you can't just multiply something be $5x$ and expect it to be the same value! That's just nutty.

But we can do

$\frac{3}{5}-\frac{4}{5x}+\frac{1}{x} = (\frac{3}{5}-\frac{4}{5x}+\frac{1}{x})*\frac {5x}{5x} = \frac {(\frac{3}{5}-\frac{4}{5x}+\frac{1}{x})5x}{5x}=\frac {3x -4 + 5}{5x}$. That is just fine.

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Note that two expressions are equal if they have the same value for all values of $x$

$$\frac{3}{5}-\frac{4}{5x}+\frac{1}{x}\ne 5x(\frac{3}{5}-\frac{4}{5x}x+\frac{1}{x}) = 3x-4+5$$

If you let $x=1$ you get $$\frac{3}{5}-\frac{4}{5x}+\frac{1}{x}= \frac{4}{5} $$

and $$3x-4+5 = 4$$

Obviously they are not the same.

Your confusion is that when we want to solve $$\frac{3}{5}-\frac{4}{5x}+\frac{1}{x}= 0$$ We drop the denominator because a fraction is zero when its numerator is zero and its denominator is not zero.

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  • $\begingroup$ I was thinking of $\frac{7z-7}{7z}=\frac{z-7}{z}$ but if we try z=2 in both we get different answers? $\endgroup$ – ReducedGosling Mar 26 '18 at 8:25
  • $\begingroup$ $\frac{7z-7}{7z}=\frac{z-1}{z}$ $\endgroup$ – Mohammad Riazi-Kermani Mar 26 '18 at 8:28
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    $\begingroup$ $\frac {7x - 7}{7z} \ne \frac {z - 7}z$. You must distribute and factor over sums and differences. $\frac {7z-7}{7z} = \frac {7*(z -1)}{7*z} = \frac 77\frac {z-1}{z} = \frac {z-1}z$ and for $z =2$, $\frac {14 - 7}{14}=\frac {7}{14} = \frac 12$ and $\frac {2 -1}2 = \frac 12$. Same answer. $\endgroup$ – fleablood Mar 26 '18 at 18:56

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