Solve the following system of inequalities gr | vedclass

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(A) To solve the system $x+y < 2, x > 0, y > 1$ graphically,we consider the corresponding equations:
$1$. $x+y = 2$
$2$. $x = 0$
$3$. $y =

Vedclass · Accepted Answer

The region is in the first quadrant bounded by $x=0, y=1$ and $x+y=2$.

Vedclass · Answer

(A) To solve the system $x+y < 2, x > 0, y > 1$ graphically,we consider the corresponding equations: $1$. $x+y = 2$ $2$. $x = 0$ $3$. $y = 1$ Step $1$: Plot the line $x+y = 2$. It passes through $(2, 0)$ and $(0, 2)$. Since the inequality is $x+y < 2$,the region is below the line. Step $2$: The line $x = 0$ is the $y$-axis. The inequality $x > 0$ represents the region to the right of the $y$-axis. Step $3$: The line $y = 1$ is a horizontal line. The inequality $y > 1$ represents the region above this line. Combining these: We need $x > 0, y > 1$,and $x+y < 2$. If $y > 1$ and $x > 0$,then $x+y > 1$. For $x+y < 2$ to hold while $y > 1$ and $x > 0$,the region must be the triangle formed by the intersection of these lines. Intersection of $x=0$ and $y=1$ is $(0, 1)$. Intersection of $x+y=2$ and $y=1$ is $(1, 1)$. Intersection of $x+y=2$ and $x=0$ is $(0, 2)$. Since the inequalities are strict ($ < $ and $>$),the region is the interior of the triangle with vertices $(0, 1), (1, 1), (0, 2)$,excluding the boundaries.

Solve the following system of inequalities graphically: $x+y < 2, x > 0, y > 1$.

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