Represent solution set of each of the following inequations graphically in two dimensional plane:

Question 1: $x + 2y - 4 \leq 0$

To draw the equation, we calculate the intercepts:

 $x = 0$ $y = 2$ $y$ axis intercept $= (0,2)$ $x=4$ $y=0$ $x$ axis intercept $= ( 4, 0)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 0,0)$ in the equation we see that $-4 \leq 0$ which is true.

Hence,  $(0, 0)$ lies in the region that satisfies the inequation.$\\$

Question 2: $x + 2y \geq 6$

To draw the equation, we calculate the intercepts:

 $x = 0$ $y = 3$ $y$ axis intercept $= (0,3)$ $x=6$ $y=0$ $x$ axis intercept $= ( 6, 0)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 0,0)$ in the equation we see that $0 \geq 6$ which is not true.

Hence,  $(0, 0)$ does not lies in the region that satisfies the inequation. Hence the solution is on the other side of the line.$\\$

Question 3: $x + 2 \geq 0$

Answer:$\\$

Question 4: $x - 2y < 0$

To draw the equation, we calculate the intercepts:

 $x = 4$ $y = 2$ $y$ axis intercept $= (4, 2)$ $x=-6$ $y=-3$ $x$ axis intercept $= ( -6, -3)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 1,1)$ in the equation we see that $-1 < 0$ which is true.

Hence,  $(1, 1)$ lies in the region that satisfies the inequation. Hence the solution is on this side of the line.$\\$

Question 5: $-3 x + 2y \leq 6$

To draw the equation, we calculate the intercepts:

 $x = 0$ $y = 3$ $y$ axis intercept $= (0,3)$ $x=-2$ $y=0$ $x$ axis intercept $= ( -2, 0)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 0,0)$ in the equation we see that $0 \leq 6$ which is true.

Hence,  $(0, 0)$ lies in the region that satisfies the inequation.$\\$

Question 6: $x \leq 8 - 4y$

To draw the equation, we calculate the intercepts:

 $x = 0$ $y = 2$ $y$ axis intercept $= (0,2)$ $x=8$ $y=0$ $x$ axis intercept $= ( 8, 0)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 0,0)$ in the equation we see that $0 \leq 8$ which is true.

Hence,  $(0, 0)$ lies in the region that satisfies the inequation.$\\$

Question 7:  $0 \leq 2x-5y+10$

To draw the equation, we calculate the intercepts:

 $x = 0$ $y = 2$ $y$ axis intercept $= (0,2)$ $x=-5$ $y=0$ $x$ axis intercept $= ( -5, 0)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 0,0)$ in the equation we see that $0 \leq 10$ which is true.

Hence,  $(0, 0)$ lies in the region that satisfies the inequation.$\\$

Question 8: $3y > 6 - 2x$

To draw the equation, we calculate the intercepts:

 $x = 0$ $y = 2$ $y$ axis intercept $= (0, 2)$ $x=3$ $y=0$ $x$ axis intercept $= ( 3,0)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 0,0)$ in the equation we see that $0 > 6$ which is not true.

Hence,  $(0, 0)$ does not lies in the region that satisfies the inequation. Hence the solution is on the other side of the line.$\\$

Question 9: $y > 2x-8$

To draw the equation, we calculate the intercepts:

 $x = 0$ $y = -8$ $y$ axis intercept $= (0,-8)$ $x=4$ $y=0$ $x$ axis intercept $= ( 4, 0)$

To find out which side of the line the solution lies. We do a test of inequality.

If we substitute $( 0,0)$ in the equation we see that $0 > -8$ which is true.

Hence,  $(0, 0)$ lies in the region that satisfies the inequation.$\\$

Question 10: $3x-2y \leq x+y - 8$

 $x = 0$ $y = \frac{8}{3}$ $y$ axis intercept $= (0, \frac{8}{3})$ $x=-4$ $y=0$ $x$ axis intercept $= ( -4,0)$
If we substitute $( 0,0)$ in the equation we see that $0 \leq -8$ which is not true.
Hence,  $(0, 0)$ does not lies in the region that satisfies the inequation. Hence the solution is on the other side of the line.$\\$