((9u+4)/5)+((3u-5)/2)=28

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Solution for ((9u+4)/5)+((3u-5)/2)=28 equation: 



((9u+4)/5)+((3u-5)/2)=28

We move all terms to the left:

((9u+4)/5)+((3u-5)/2)-(28)=0

We calculate fractions


(18u+8)/()+(15u-25)/()-28=0

We multiply all the terms by the denominator


(18u+8)+(15u-25)-28*()=0

We add all the numbers together, and all the variables

(18u+8)+(15u-25)=0

We get rid of parentheses

18u+15u+8-25=0

We add all the numbers together, and all the variables

33u-17=0

We move all terms containing u to the left, all other terms to the right

33u=17

u=17/33

u=17/33

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