((4y+2)/5)+((3y-1)/2)=16

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Solution for ((4y+2)/5)+((3y-1)/2)=16 equation: 



((4y+2)/5)+((3y-1)/2)=16

We move all terms to the left:

((4y+2)/5)+((3y-1)/2)-(16)=0

We calculate fractions


(8y+4)/()+(15y-5)/()-16=0

We multiply all the terms by the denominator


(8y+4)+(15y-5)-16*()=0

We add all the numbers together, and all the variables

(8y+4)+(15y-5)=0

We get rid of parentheses

8y+15y+4-5=0

We add all the numbers together, and all the variables

23y-1=0

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

23y=1

y=1/23

y=1/23

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