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4x/5 -3x/4=3x/10-1

 May 3, 2014

Best Answer 

 #1
avatar+118677 
+8

$${\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{5}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{4}}}} = {\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{10}}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{x}} = {\mathtt{4}}$$

That is the answer straight from the site calculator. I putit in using Math(Input=result) tab.

Now I'll do it by hand - When you do it you should line up all the equal signs.

I multiplied by 20 because 20 is  the LCD Lowest common denominator and that got rid of all the fractions.

$$\frac{4x}{5}-\frac{3x}{4}=\frac{3x}{10}-1\\\\
20\times \left(\frac{4x}{5}-\frac{3x}{4}\right)=20\times
\left(\frac{3x}{10}-1\right)\\\\
16x-15x=6x-20\\\\
x=6x-20\\\\
-5x=-20\\\\
x=4$$

.
 May 3, 2014
 #1
avatar+118677 
+8
Best Answer

$${\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{5}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{4}}}} = {\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{10}}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{x}} = {\mathtt{4}}$$

That is the answer straight from the site calculator. I putit in using Math(Input=result) tab.

Now I'll do it by hand - When you do it you should line up all the equal signs.

I multiplied by 20 because 20 is  the LCD Lowest common denominator and that got rid of all the fractions.

$$\frac{4x}{5}-\frac{3x}{4}=\frac{3x}{10}-1\\\\
20\times \left(\frac{4x}{5}-\frac{3x}{4}\right)=20\times
\left(\frac{3x}{10}-1\right)\\\\
16x-15x=6x-20\\\\
x=6x-20\\\\
-5x=-20\\\\
x=4$$

Melody May 3, 2014
 #2
avatar+893 
0

I truly hate the result from the site calculator.

The result, x=4, is correct, but the presentation, saying that both sides of the original equation  are equal to x and equal to 4 is most certainly not correct.

 May 3, 2014
 #3
avatar+118677 
0

Yes, thank you for your input Bertie,

I don't like that either, maybe Andre could put an arrow in the middle - that would be an improvement.

Andre is very busy with a great many problems and I don't want to pester him all of the time but I do hope some of these 'annoying issues' will eventually be improved.

I probably should set up a list of these more minor issues somewhere so that eventually we can talk to Andre about them.

 May 4, 2014
 #4
avatar+33661 
0

I'm in complete agreement with Bertie's comment here too!

 May 4, 2014

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