what to do to get rid of a fraction in the denominator?

[eddy2017]

&& Hi, I was wondering if what i have done here is correct[math][/math] 2(5/3 - 1) + 2(7/5-1) / 5/3 -7/5[math][/math]2(2/3) + 2(2/5)/ 4/15[math][/math] my question is , can I do this to get trid of the 4/15 i nthe denominator, that is, multiply the numerator by the multiplicative inverse of the fraction in the denominator?[math][/math] 2(2/3) + 2(2/5) * 15/4

 

[BigBeachBanana]

Wrong: 2(2/3) + 2(2/5)/ 4/15
Eddy, I can't stress this enough. Please use parenthesis to indicate the correct order of operation!
What you wrote mean [math]\frac{2(2/5)}{4*15}[/math]What you really meant is [math]\frac{{2(2/5)}}{\frac{4}{15}}[/math]It's 2 different expression.
Correct: 2(2/3) + 2(2/5)/ (4/15)
But yes, this is correct 2(2/3) + 2(2/5) * 15/4 if you meant to write the second expression.

 

[eddy2017]

Wrong: 2(2/3) + 2(2/5)/ 4/15
Eddy, I can't stress this enough. Please use parenthesis to indicate the correct order of operation!
What you wrote mean [math]\frac{2(2/5)}{4*15}[/math]What you really meant is [math]\frac{{2(2/5)}}{\frac{4}{15}}[/math]It's 2 different expression.
Correct: 2(2/3) + 2(2/5)/ (4/15)
But yes, this is correct 2(2/3) + 2(2/5) * 15/4 if you meant to write the second expression.

okay, thanks for the heads-up about the grouping symbols. yes that is what I wanted to know. so, Taking the multiplicative inverse of the fraction in the denominator and multiplying it by the numerator I can get rid it. Great!
thanks