a) Graphically shade the feasible region for the constraints x+2y 7 , x y 4 , x, y 0 . [2] b) Convert hexadecimal number 81A into Octal number. [2] c) Using Gauss elimination method, solve the…

a) Feasible region for

$x+2y\le 7,x-y\le 4,x\ge 0,y\ge 0$

The inequalities are linear, so the feasible region is a convex polygon bounded by the lines
$x+2y=7,x-y=4,x=0,y=0$.

The intersection points of the boundary lines are:

These points together with the origin $(0,0)$ form the vertices of the feasible polygon.
The shaded region in the figure is the set of all $(x,y)$ that satisfy all four inequalities.

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b) Hexadecimal 81A → Octal

1. Convert 81A (hex) to decimal.
   $$8\cdot 16^2+1\cdot 16^1+10\cdot 16^0=8\cdot 256+1\cdot 16+10=2048+16+10=2074_{10}$$

2. Convert 2074 (decimal) to octal.
   Divide repeatedly by 8, keeping remainders.

Reading remainders from bottom to top gives $4072_8$.

Answer: $\boxed{{\text{4072}}_8}$

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c) Solve by Gauss‑Elimination

$$\{\begin{array}{l}x+2y=5\\ 5x-3y=-1\end{array}$$

Step 1: Write the augmented matrix

$$\left[\begin{array}{cccc}1& 2& |& 5\\ 5& -3& |& -1\end{array}\right]$$

Step 2: Eliminate $x$ from the second row

Multiply the first row by $5$ and subtract from the second row.

$$R_2\leftarrow R_2-5R_1$$

$$\left[\begin{array}{cccc}1& 2& |& 5\\ 0& -13& |& -26\end{array}\right]$$

Step 3: Solve for $y$

From the second row: $$-13y=-26\Rightarrow y=2$$

Step 4: Back‑substitute to find $x$

Use the first equation: $$x+2(2)=5\Rightarrow x+4=5\Rightarrow x=1$$

Solution: $\boxed{\text{x = 1, y = 2}}$

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