For each equation, choose the statement that describes its solution. If applicable, give the solution.
$$4(w+1)+5=6(w-1)+w$$

No solution
$$w=◻$$

All real numbers are solutions
$$2(v+2)+v=3(v-1)+6$$

No solution
$v=$

All real numbers are solutions

Step 1 ：Distribute the 4 and 6 to the terms inside the parentheses for the first equation: $4(w+1)+5=6(w-1)+w\Rightarrow 4w+4+5=6w-6+w$

Step 2 ：Combine like terms for the first equation: $4w+4+5=6w-6+w\Rightarrow 4w+9=7w-6$

Step 3 ：Isolate the variable $w$ on one side for the first equation: $4w+9=7w-6\Rightarrow 9+6=7w-4w\Rightarrow 15=3w$

Step 4 ：Solve for $w$ in the first equation: $15=3w\Rightarrow w=\frac{15}{3}\Rightarrow w=5$

Step 5 ：The solution for the first equation is $\boxed{{\displaystyle w=5}}$

Step 6 ：Distribute the 2 and 3 to the terms inside the parentheses for the second equation: $2(v+2)+v=3(v-1)+6\Rightarrow 2v+4+v=3v-3+6$

Step 7 ：Combine like terms for the second equation: $2v+4+v=3v-3+6\Rightarrow 3v+4=3v+3$

Step 8 ：Attempt to isolate the variable $v$ on one side for the second equation: $3v+4=3v+3\Rightarrow 4=3$

Step 9 ：Since $4$ does not equal $3$, there is no solution for the second equation

Step 10 ：The solution for the second equation is $\boxed{{\displaystyle \text{No solution}}}$