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How would you describe and correct the error in graphing this sytem of inequalities?

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anya4one | Student, Undergraduate | Honors

Posted July 14, 2013 at 5:25 PM via web

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How would you describe and correct the error in graphing this sytem of inequalities?

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aruv | High School Teacher | Valedictorian

Posted July 14, 2013 at 5:55 PM (Answer #1)

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`y<=(3/2)x+3`     (i)

`-y<2x`       (ii)

(i) is inequality where as (ii) is strict inequality.

(i) represented by solid line  which is correctly repn.

(ii)  represented by doted line again true repn.

Pivotal point point in case first may be (0,0) work but in case of second it will nt work . So let us choose  (0,-1) to determine region.

So whole pane divided in four parts. Among them which is common and satisfy both will solution of system of inequality.

So when we choose (0,-1)  in second inequality ,we find it does not satify i.e.

`-(-1)<2xx0`

`1<0`  not true.

Thus a points in  plane opposite to point (0,-1)  will satisfy (ii).

Thus correct solution are points in quadrant I and IV ,the common region between solid and doted line

The region in which blue line , are solution set of the inequality.

Green lne please assume doted line.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted July 15, 2013 at 1:42 AM (Answer #2)

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The incorrect shading is for the inequality -y<2x. What the student did was to graph the line y=2x; dotted since the inequality is strict. This was done correctly.

However, they then shaded as if it was y<-2x. But when you divide/multiply by a negative number, you must reverse the inequality. Thus -y<2x is equivalent to y>-2x.

So the correct region to shade is below `y=3/2x+3` and above y=-2x.

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