The Purplemath Forums

1/3 - (x+1)/5 = x/3 (only there are no parenthesis around the x+1 over 5... if that matters in this instance) ... and (y+5)/12 = 3/4 - (y+1)/8 (once again, no parenthesis around the fractions with variables in these equations...) My issue is figuring out whether if what I currently have is an accept...

... oh, solve. but i just wanted to make sure my work was accurate. finding a common denominator and then, one friend suggested, getting rid of the denominator and solving the problem as 5 - 3x + 3 = 5x, and 2y + 10 = 18 - 3y +3.

thank again for your help, btw:)