Hello, DooBeeDoo!
Consider the following linear programming problem:
Maximise: .$\displaystyle Z \:= \:100x + 200y$
Subject to: .$\displaystyle \begin{Bmatrix}3x + 2y & \leq &120 & [1]\\ x + 4y & \leq & 80 & [2]\\ 2x + y & \leq & 100 & [3]\\ x, y & \geq & 0 & [4]\end{Bmatrix}$
Solution is: .$\displaystyle Z=5600,\;x =32,\;y=12,\;s3=24$
If a new constraint $\displaystyle 2x+ 2y \leq 100$ was included in the original problem,
how would this affect the solution?
My answer is that there is no change. . . . . I agree
Is there a quicker way of getting the answer?
. . You weren't shown how to graph these problems?
Constraint [4] puts us in Quandrant 1.
For $\displaystyle [1]\;3x + 2y \:\leq \:120$, graph the line: .$\displaystyle 3x + 2y \:=\:120$
It has intercepts: (40, 0), (0, 60)
Draw the line and shade the region below it.
For $\displaystyle [2]\;x + 4y \:\leq \:80$ graph the line: .$\displaystyle x + 4y \:=\:80$
It has intercepts: (80, 0), (0, 20)
Draw the line and shade the region below it.
The shaded region looks like this . . . Code:

60 *
*
 *
 *
 *
 *
20 o *
:::* *
:::::::o
::::::::* *
:::::::::* *
::::::::::* *
  o      o      *  
 40 80
Constraint [3] is useless.
The line $\displaystyle 2x + y \:=\:100$ has intercepts (50, 0), (0, 100)
. . and falls outside of the shaded region.
We are concerned with the vertices of the shaded region.
We can see three of them: .(0, 0), (40, 0), (0, 20)
We find the fourth by locating the intersection of the two slanted lines.
We solve the system: .$\displaystyle \begin{array}{ccc}3x + 2y &=&120 \\ x + 4y &=& 80\end{array}\;\text{ and get: }\32,\:12)$
We substitute these coordinates into the Zfunction
. . to determine which one maximizes Z.