2y+((3y-2)/2)=10

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Solution for 2y+((3y-2)/2)=10 equation: 


y in (-oo:+oo)

(3*y-2)/2+2*y = 10 // - 10

(3*y-2)/2+2*y-10 = 0

(3*y-2)/2+(2*2*y)/2+(-10*2)/2 = 0

3*y+2*2*y-10*2-2 = 0

7*y-20-2 = 0

7*y-22 = 0

(7*y-22)/2 = 0

(7*y-22)/2 = 0 // * 2

7*y-22 = 0

7*y-22 = 0 // + 22

7*y = 22 // : 7

y = 22/7

y = 22/7

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