Question

solve the equation of the following equation​

Answer 1

Answer:

(I) $\frac{x-1}{3}=\frac{x+2}{6}+3 \\ \\= \frac{x-1}{3}=\frac{x+2+18}{6}\\ \\=\frac{x-1}{3}=\frac{x+20}{6}\\ \\ = 6(x-1)=3(x+20)\\ \\= 6x-6=3x+60\\ \\= 6x-3x=60+6\\ \\ = 3x=66\\ \\=x=22$

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(II) $\frac{y+1}{3}-\frac{y-1}{2}=\frac{1+2y}{3}\\ \\= \frac{2y+2-3y+3}{6}=\frac{1+2y}{3}\\ \\= \frac{-y+5}{6}=\frac{1+2y}{3}\\ \\=3(-y+5)=6(1+2y)\\ \\ = - 3y+15=6+12y\\ \\=12y+3y= 6-15\\ \\= 15y= - 9\\ \\=y=\frac{-3}{5}$

Step-by-step explanation:

Answer 2

Answer:

1. x=22

2. y= 3/5

Step-by-step explanation: