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[(2x+1)/3]+[(5x-2)/2]-[(3x-3)/5]=2 

 Feb 8, 2016

Best Answer 

 #2
avatar+95288 
+15

Hi Rarinstraw,

 

This is the BEST way to do it :)

 

\(\frac{(2x+1)}{3}+\frac{(5x-2)}{2}-\frac{(3x-3)}{5}=2 \\ \mbox{Multiply through by the lowest common multiple - that is 30}\\ \frac{30(2x+1)}{3}+\frac{30(5x-2)}{2}-\frac{30(3x-3)}{5}=60 \\ \frac{10(2x+1)}{1}+\frac{15(5x-2)}{1}-\frac{6(3x-3)}{1}=60 \\ 10(2x+1)+15(5x-2)-6(3x-3)=60 \\ \mbox{See that got rid of all the fractions :)}\\ 20x+10+75x-30-18x+18=60\\ 20x+75x-18x+18+10-30=60\\ 77x-2=60\\ 77x=62\\ x=\frac{62}{77}\\ \)

.
 Feb 8, 2016
 #1
avatar+5258 
+3

2/3x + 1/3 + 2 1/2x - 1 - 3/5x - 3/5 =2

 

Combine like terms.

 

The decimals are approximate, the numbers would have repeated for ages otherwise.

 

2.57x - 1.26 = 2

 

Add 1.26 to both sides.

 

2.57x=3.26

 

2.57x=3.26 = x=(326/257)

 Feb 8, 2016
 #2
avatar+95288 
+15
Best Answer

Hi Rarinstraw,

 

This is the BEST way to do it :)

 

\(\frac{(2x+1)}{3}+\frac{(5x-2)}{2}-\frac{(3x-3)}{5}=2 \\ \mbox{Multiply through by the lowest common multiple - that is 30}\\ \frac{30(2x+1)}{3}+\frac{30(5x-2)}{2}-\frac{30(3x-3)}{5}=60 \\ \frac{10(2x+1)}{1}+\frac{15(5x-2)}{1}-\frac{6(3x-3)}{1}=60 \\ 10(2x+1)+15(5x-2)-6(3x-3)=60 \\ \mbox{See that got rid of all the fractions :)}\\ 20x+10+75x-30-18x+18=60\\ 20x+75x-18x+18+10-30=60\\ 77x-2=60\\ 77x=62\\ x=\frac{62}{77}\\ \)

Melody Feb 8, 2016
 #3
avatar+5258 
0

Looks a whole lot easier! Good job Melody!

rarinstraw1195  Feb 8, 2016
 #4
avatar+95288 
0

Thanks rarinstraw :)

 Feb 9, 2016

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