2 Answers | Add Yours
(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<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.
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.
We’ve answered 395,978 questions. We can answer yours, too.Ask a question