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
\[
w=\square
\]

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 \(\boxed{w=5}\)

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 \(\boxed{\text{No solution}}\)