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Question 1195571: Can I please have help with this, Thank you!
On the grid, sketch the inequalities below and determine the solution region.
y is greater than or equal to 0
x is greater than or equal to 0
y is less than or equal to x+2
y is less than or equal to 14-2x
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
We'll have these boundary equations
y = 0
x = 0
y = x+2
y = -2x+14
The first two equations represent the x axis and y axis in that exact order.
The last equation is equivalent to y = 14 - 2x
Each equation mentioned is a solid line due to the "or equal to" as part of the inequality sign.
We're told that x ≥ 0 and y ≥ 0, which places us in the upper right quadrant.
The shaded region is a subset of this quadrant. Meaning that the shaded region will not be found in any other quadrant.
To graph out y ≤ x+2, we will shade below the boundary equation y = x+2. This trick only works when y is fully isolated.
Similarly, the graph of y ≤ -2x+14 will also be shaded below its corresponding boundary line.
This is what the final shaded region looks like when we overlap the four inequalities together
I used GeoGebra to make this graph.
The input command I typed in was: ( x >= 0) && (y >= 0) && (y <= x+2) && (y <= -2x+14)
Desmos is another tool you can use, and here's the link to the interactive graph:
https://www.desmos.com/calculator/mkjjbdalpl
This is what you would type in x >= 0 { y >= 0 } {y <= x+2} {y <= -2x+14}
The curly braces are important so Desmos knows where one inequality ends and another begins. The first inequality doesn't need curly braces.
Points inside the blue shaded region are solutions to the system of inequalities.
The point (4,1) is one such example. It makes x ≥ 0 and y ≥ 0 true; it also makes y ≤ x+2 and y ≤ -2x+14 true as well.
Confirmation for y ≤ x+2
y ≤ x+2
1 ≤ 4+2
1 ≤ 6
which is true
Confirmation for y ≤ -2x+14
y ≤ -2x+14
1 ≤ -2(4)+14
1 ≤ 6
which is also true
Therefore, the point (4,1) has been confirmed to work for the last two inequalities given.
I'll let you try other points in the blue region.
Points on the boundary directly adjacent to the interior region are solutions as well
However, we cannot include boundary points that aren't next to the blue interior.
Eg: (2,4) is a solution but (-1,1) is not a solution even though it resides on a boundary line
The point (-1,1) doesn't work because it doesn't satisfy x ≥ 0
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