2y+x<6
3x-y>4
x+y>7
y<0
where do i shade in? where do i plot the points im really confused help plzz..
3x-y>4
x+y>7
y<0
where do i shade in? where do i plot the points im really confused help plzz..
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It's best to make all of these into slope-intercept form first (solve for y)
So 2y+x<6 becomes y<-x/2+3. Start at (0,3) then move down 1, right 2 per the slope. This is a dashed line because it does not have the "equal to". Because it is less than, shade below this line.
3x-y>4 becomes y<3x-4. Start at (0,4) then move up 3 right 1 (or down 3, left 1) per the slope. This, too, is a dashed line and will be shaded below.
Are these two inequalities one problem? If so, it's a system and it's solution is where the two shaded areas overlap when both lines are graphed on the same coordinate plane.
The second set of inequalities:
x+y>7 becomes y>-x+7. Start at (0,7) and move up one, right one per the slope. Dashed line with shading above the line since it's greater than.
y<0 is the x-axis. It should be a dashed line and shaded below.
So 2y+x<6 becomes y<-x/2+3. Start at (0,3) then move down 1, right 2 per the slope. This is a dashed line because it does not have the "equal to". Because it is less than, shade below this line.
3x-y>4 becomes y<3x-4. Start at (0,4) then move up 3 right 1 (or down 3, left 1) per the slope. This, too, is a dashed line and will be shaded below.
Are these two inequalities one problem? If so, it's a system and it's solution is where the two shaded areas overlap when both lines are graphed on the same coordinate plane.
The second set of inequalities:
x+y>7 becomes y>-x+7. Start at (0,7) and move up one, right one per the slope. Dashed line with shading above the line since it's greater than.
y<0 is the x-axis. It should be a dashed line and shaded below.