Solve for y: (30y/20+3)/15=(y+2)/10

Question

Confused about how to solve this equation with a fraction that also has variable as the numerator. I'm trying to cross multiply but am not sure if that's right. 

Answer 1

I'm assuming this is the problem...

[(30y/20)+3]/15=(y+2)/10

[(3y/2)+3]/15=(y+2)/10

you can cross multiply if you like

10[(3y/2)+3]=15(y+2)

distribute

15y+30=15y+30 (this is a true statement)

y=all Real numbers (any real number makes this equation true)

you could have multiplied (y+2) by (1 1/2) and multiplied 10 by 1 1/2 to make the 10 a 15

now because both denominators are equal, the numerators must be equal

[(3y/2+3)]=(1 1/2)(y+2)

[(3y/2+3)]=(3/2)y+3 which again is a true statement (3y/2=(3/2)y)