Mary Sony

2022-07-06

$If the system of inequalities 2-y< 2x and -x\le 4-y is graphed on the xy plane which quadrant contains no solution?$

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To solve the system of inequalities and determine the quadrant that contains no solution, let's first rewrite the given inequalities in a more standard form:
$2-y<2x$
$-x\le 4-y$
Now, let's graph these two inequalities on the xy-plane to visualize their solution regions.
Starting with the first inequality, $2-y<2x$, we can rearrange it to obtain:
$2x>2-y$
Subtracting 2 from both sides:
$2x-2>-y$
Multiplying both sides by -1 (which reverses the inequality):
$-2x+2
Now, let's plot the graph of $y=-2x+2$. This is a linear equation with a slope of -2 and a y-intercept of 2.
Next, let's graph the second inequality, $-x\le 4-y$. We can rearrange it as follows:
$y\le x+4$
The graph of $y=x+4$ is a straight line with a slope of 1 and a y-intercept of 4.
Now, we will shade the regions that satisfy each inequality.
For the inequality $2-y<2x$, we shade the region below the line $y=-2x+2$. We use a dashed line because the inequality is strict ($<$).
For the inequality $-x\le 4-y$, we shade the region below or on the line $y=x+4$. We use a solid line because the inequality is inclusive ($\le$).
After graphing the two inequalities, we can observe the regions where both shaded areas overlap.
Upon examining the graph, we can see that the shaded regions intersect in the second quadrant (top left) and the fourth quadrant (bottom right). However, the question asks for the quadrant that contains no solution.
Since there is an overlap in both the second and fourth quadrants, there is no quadrant that contains no solution. The system of inequalities has a solution in all quadrants.

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