Question

A raiz de (y-3)/4+(y+1)/6=(y-1)/12

Answer 1

$\displaystyle \frac{\text y-3}{4}+\frac{\text y+1}{6}=\frac{\text y-1}{12} \\\\\\ \underline{\text{Multiplicando por 12 dos dois lados da igualdade}}: \\\\\\\frac{12(\text y-3)}{4}+\frac{12(\text y+1)}{6}=\frac{12.(\text y-1)}{12} \\\\\\ 3(\text y-3) + 2(\text y+1)=\text y-1 \\\\ 3\text y-9+2\text y+2=\text y-1 \\\\ 3\text y+2\text y -\text y = 9-2-1 \\\\ 4\text y = 6 \to \text y = \frac{6}{4} \\\\\\ \huge\boxed{\text y= \frac{3}{2} \ }\checkmark$