graphical inequalities: Pick system of linear inequalities that best fits picture

jamie bean

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May 31, 2018
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5
. From the systems of inequalities that follow, select the system that best defines the shaded area in the graph below.


  • [FONT=STIXMathJax_Main-italic]y[/FONT][FONT=STIXMathJax_Main]≤[/FONT][FONT=STIXMathJax_Main]4[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]11[/FONT][FONT=STIXMathJax_Main],[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]≤[/FONT][FONT=STIXMathJax_Main]4[/FONT]y≤4x11,x≤4
  • [FONT=STIXMathJax_Main-italic]y[/FONT][FONT=STIXMathJax_Main]≥[/FONT][FONT=STIXMathJax_Main]4[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]11[/FONT][FONT=STIXMathJax_Main],[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]>[/FONT][FONT=STIXMathJax_Main]4[/FONT]y≥4x11,x>4
  • [FONT=STIXMathJax_Main-italic]y[/FONT][FONT=STIXMathJax_Main]≥[/FONT][FONT=STIXMathJax_Main]11[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]4[/FONT][FONT=STIXMathJax_Main],[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]≥[/FONT][FONT=STIXMathJax_Main]4[/FONT]y≥11x4,x≥4
  • [FONT=STIXMathJax_Main-italic]y[/FONT][FONT=STIXMathJax_Main]≤[/FONT][FONT=STIXMathJax_Main]11[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]4[/FONT][FONT=STIXMathJax_Main],[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]=[/FONT][FONT=STIXMathJax_Main]4[/FONT]y≤11x4,x=4
  • [FONT=STIXMathJax_Main-italic]y[/FONT][FONT=STIXMathJax_Main]≥[/FONT][FONT=STIXMathJax_Main]4[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]11[/FONT][FONT=STIXMathJax_Main],[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]=[/FONT][FONT=STIXMathJax_Main]4[/FONT]
  • Why is y calculated in this way?????? isnt y larger than or equal to 11?
  • Solution: The correct answer is C. The shaded area is to the right of the vertical line x = 4, so x ≥ 4 must be in the correct system. Fortunately, only one of the answer choices has x ≥ 4—choice C.
  • To find the slope, find the Δy: Δy = 11 − 0 = 11, then divide by Δx: Δx = 4 − 0 = 4.
    The slope = [FONT=STIXMathJax_Main]11[/FONT][FONT=STIXMathJax_Main-italic]4[/FONT][FONT=STIXMathJax_Main].[/FONT]114. This can be verified, since the shading is above the line and the inequality is [FONT=STIXMathJax_Main-italic]y[/FONT][FONT=STIXMathJax_Main]≥[/FONT][FONT=STIXMathJax_Main]11[/FONT][FONT=STIXMathJax_Main-italic]x[/FONT][FONT=STIXMathJax_Main]4[/FONT][FONT=STIXMathJax_Main].[/FONT]
 

stapel

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The formatting of your post is difficult to read. I'm guessing that you didn't notice that your copy-n-paste from your online assessment didn't work quite right...? The below is what I think is the question:

From the systems of inequalities that follow, select the system that best defines the shaded area in the graph below.


(a) y ≤ (4/11)x with x ≤ 4

(b)
y ≥ (4/11x with x > 4

(c)
y ≥ (11/4)x with x ≥ 4

(d)
y ≤ (11/4)x with x = 4

(e)
y ≥ (4/11)x with x = 4



Solution: The correct answer is C.

The shaded area is to the right of the vertical line x = 4, so x ≥ 4 must be in the correct system.
To find the slope, find the Δy: Δy = 11 − 0 = 11, then divide by Δx: Δx = 4 − 0 = 4.
The slope is m = 11/4. This can be verified, since the shading is above the line and the inequality is y ≥ (11/4)x.




Why is y calculated in this way?????? isnt y larger than or equal to 11?
Yes, but only under a specific condition on x. For instance, the point (2, 12) is not in the solution region.

To learn how this process works, please try here. Thank you! ;)
 

HallsofIvy

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Jan 27, 2012
Messages
4,805
The vertical line is x= 4. Any line through (0, 0) has equation y= ax. Since it goes through (4, 11) 11= 4a so a= 11/4. The equation of that line is y= (11/4)x. The shaded portion is to the right of x= 4 so has x> 4 and to the above y= (11/4)x so y> (11/4)x.

"y> 11" would be the entire half plane (x, y) with x anything and y> 11. The point (5, 11.01), for example, satisfies y> 11 but is NOT above that line.
 
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