[tex]\frac{4y + 2}{3} + \frac{2y + 1}{4} = \frac{6y + 3}{4} [/tex] y =...(dengan cara)

[tex] \frac{4y+2}{3} + \frac{2y+1}{4} = \frac{6y+3}{4} \\ \frac{4y+2}{3} = \frac{6y+3-(2y+1)}{4} \\ \frac{4y+2}{3} = \frac{4y+2}{4} \\ 16y + 8 = 12y + 6 \\ 16y - 12y = 6 - 8 \\ 4y = -2 \\ y = - \frac{1}{2} [/tex]