To solve the given system of equations, we'll first simplify and rewrite the equations in a more manageable form.
The equations are:
[tex]\[
\left\{\begin{array}{l}
\frac{2(x+1)+3=y}{3} \\
3(x+5-y)+3x=12
\end{array}\right.
\][/tex]
### First Equation
Starting with the first equation:
[tex]\[
\frac{2(x+1) + 3 = y}{3}
\][/tex]
We can get rid of the fraction by multiplying both sides by 3:
[tex]\[
2(x + 1) + 3 = y
\][/tex]
Now, simplify inside the parentheses first:
[tex]\[
2x + 2 + 3 = y
\][/tex]
Combine like terms:
[tex]\[
2x + 5 = y
\][/tex]
This simplifies the first equation to:
[tex]\[
y = 2x + 5
\][/tex]
### Second Equation
Now, consider the second equation:
[tex]\[
3(x + 5 - y) + 3x = 12
\][/tex]
First, distribute the 3:
[tex]\[
3x + 15 - 3y + 3x = 12
\][/tex]
Combine like terms:
[tex]\[
6x + 15 - 3y = 12
\][/tex]
Subtract 15 from both sides to isolate the terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
6x - 3y = -3
\][/tex]
Divide through by 3 to simplify:
[tex]\[
2x - y = -1
\][/tex]
### Solving the System
We now have a system of two equations:
[tex]\[
\left\{\begin{array}{l}
y = 2x + 5 \\
2x - y = -1
\end{array}\right.
\][/tex]
Substitute [tex]\(y = 2x + 5\)[/tex] into the second equation:
[tex]\[
2x - (2x + 5) = -1
\][/tex]
Simplify inside the parentheses:
[tex]\[
2x - 2x - 5 = -1
\][/tex]
This reduces to:
[tex]\[
-5 = -1
\][/tex]
Since [tex]\(-5\)[/tex] does not equal [tex]\(-1\)[/tex], this is a contradiction. Therefore, the system of equations has no solution. There are no values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
Hence, the solution to this system of equations is:
[tex]\[
\boxed{\text{No solution}}
\][/tex]
