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Mathematics 10 Online
OpenStudy (anonymous):

solve x+3/5 = 2

OpenStudy (catlover5925):

is 3/5 a fraction or a division sign

OpenStudy (texaschic101):

multiply the entire equation by 5 to get rid of all fractions

OpenStudy (catlover5925):

so if it is a fraction it would be X + 3/5 = 2 ---> subtract 3/5 from the right side X = 1 and 2/5ths or 7/5

sammixboo (sammixboo):

Well to first solve this, we must subtract 3/5 on both sides of the equation, so we have x + \(\rm \frac{3}{5}\) - \(\rm \frac{3}{5}\) = 2 - \(\rm \frac{3}{5}\) x = 2 - \(\rm \frac{3}{5}\) Now we have x = 2 - \(\rm \frac{3}{5}\). We can make 2 into theimproper fraction \(\rm \frac{2}{1}\), because 2 and \(\rm \frac{2}{1}\) are the same, so now we have x = \(\rm \frac{2}{1}\) - \(\rm \frac{3}{5}\) Now let's find the lowest common multiple, so we can make the fractions have the same denominators to be able to subtract them.. Multiple of 5 - 5, 10, 15, 20 Multiples of 3 - 3, 6, 9, 12, 15, 18 What is the lowest multiple both 5 and 3 have.

sammixboo (sammixboo):

You can do it @texaschic101's way :P It might be easier

OpenStudy (catlover5925):

@texaschic101 so the answer would be 8 the way you are explaining it

OpenStudy (texaschic101):

listen...if you do not want to mess with fractions, multiply everything by 5 (the common denominator) x + 3/5 = 2 --- multiply by 5 5x + 3 = 10 -- subtract 3 5x = 10 - 3 5x = 7 x = 7/5 check.. 7/5 + 3/5 = 2 10/5 = 2 2 = 2 (correct) x = 7/5.....anytime you have fractions in an inequality or equation, if you multiply by the common denominator, this will get rid of the fractions....unless you just like to mess with fractions

sammixboo (sammixboo):

:) Yeah I was debating on which way to do it. With subtracting fractions or without

sammixboo (sammixboo):

Nice explanation too!*

OpenStudy (texaschic101):

well..either way will get you to the right answer...fractions or no fractions

sammixboo (sammixboo):

Yeppers =)

geerky42 (geerky42):

Problem could be \(\dfrac{x+3}{5}=2\) Is that what you mean or no? @patricia27

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