How will the solution of the system $y>2 x+\frac{2}{3}$ and $y<2 x+\frac{1}{3}$ change if the inequality sign on both inequalities is reversed to $y<2 x+\frac{2}{3}$ and $y>2 x+\frac{1}{3}$?
Answer (2)
The solution changes from no solution in the original system to a viable region between two lines in the modified system. The first system has inequalities that cannot be satisfied simultaneously, while the reversed inequalities allow for an area where solutions exist. Thus, the reversal creates a feasible range of values for y .
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The original system 2x + \frac{2}{3}"> y > 2 x + 3 2 and y < 2 x + 3 1 has no solution because there is no y that can be simultaneously greater than 2 x + 3 2 and less than 2 x + 3 1 .
The modified system y < 2 x + 3 2 and 2x + \frac{1}{3}"> y > 2 x + 3 1 represents the region between the two parallel lines y = 2 x + 3 1 and y = 2 x + 3 2 .
Reversing the inequality signs changes the solution from no solution to the region between the two parallel lines.
The solution changes from no solution to a region between two parallel lines.
Explanation
Analyzing the Original System Let's analyze the original system of inequalities: 2x + \\frac{2}{3}"> y > 2 x + f r a c 2 3 and y < 2 x + f r a c 1 3 . We want to find the region where y is simultaneously greater than 2 x + f r a c 2 3 and less than 2 x + f r a c 1 3 .
Comparing the Lines Notice that the lines y = 2 x + f r a c 2 3 and y = 2 x + f r a c 1 3 are parallel because they have the same slope, which is 2. The y-intercept of the first line is f r a c 2 3 , and the y-intercept of the second line is f r a c 1 3 . Since \\frac{1}{3}"> f r a c 2 3 > f r a c 1 3 , the line y = 2 x + f r a c 2 3 is above the line y = 2 x + f r a c 1 3 .
Determining the Solution of the Original System For the original system, we need y to be greater than 2 x + f r a c 2 3 and less than 2 x + f r a c 1 3 . However, since 2x + \\frac{1}{3}"> 2 x + f r a c 2 3 > 2 x + f r a c 1 3 , there is no value of y that can satisfy both inequalities simultaneously. Therefore, the original system has no solution.
Analyzing the Modified System Now let's analyze the modified system of inequalities: y < 2 x + f r a c 2 3 and 2x + \\frac{1}{3}"> y > 2 x + f r a c 1 3 . We want to find the region where y is simultaneously less than 2 x + f r a c 2 3 and greater than 2 x + f r a c 1 3 .
Determining the Solution of the Modified System In this case, we need y to be between the two parallel lines y = 2 x + f r a c 1 3 and y = 2 x + f r a c 2 3 . Since 2 x + f r a c 1 3 < 2 x + f r a c 2 3 , there are values of y that satisfy both inequalities. The solution to the modified system is the region between the two parallel lines.
Comparing the Solutions Therefore, the solution changes from no solution to the region between the two parallel lines.
Examples
Imagine you're trying to define a safe zone for a robot moving along a conveyor belt. The original inequalities might represent conflicting safety requirements, making it impossible for the robot to operate safely. Reversing the inequalities creates a feasible safety zone between two defined boundaries, allowing the robot to function within those limits. This concept applies to various scenarios, such as setting acceptable ranges for temperature control, defining permissible levels of noise pollution, or establishing boundaries for financial investments.
