http://prntscr.com/iboc9u
your lines are good! so shade to the left of the vertical (solid) x graph and above the vertical line
(sorry I had to deal w/ some mod stuff) anyway let's check your lines again
2y + x = 6 if we plug in y = 0 we get x = 6 so (6,0) is a point on the graph if we plug in x = 0 we get y = 3, so (0,3) is a point on the graph so draw the dashed line that crosses these two points
and wheres the shaded ppart?
since it's 2y < something it needs to be below
for 3x - y > 4 we can re-arrange this to get y < 3x - 4 which also implies shading below so the second line needs to be a dashed line with slope 3, intercept -4, with shading below
anyway, that's a lot of info, just draw the two new lines and I'll let you know if they need fixing ok? ;;
i just need to know where to shade
below the first line below the second line
x + 1 > 3 x is greater than 2 so it needs to be to the right 2x - y < 4 y is less than ___ so it needs to be above/on the left side
|dw:1518057995201:dw|
|dw:1518058001132:dw|
and to the right of the blue line
x + y < 2 (y is less than, so needs to be below this line) 2x - y < 3 becomes y > 2x - 4 so needs to be above this line the orange lines x + y < 2 looks a bit off though, these need to be re-done
okay and the shaded part?
shading is below the orange line (after it's redone) and above the blue line
I know you're in a rush : x + y < 2 crosses (0,2) and (2,0) using the method from before
anyway, just for guidance: if you can re-arrange the equation into y > or y>= something, the shading needs to be above the line for y < or y<= something the shading needs to be below th eline
Thank You!!!
yeah the above rules will work, however you have to solve the inequality for y for this to work so it just has to be y by itself on one side of the inequality ^^ sorry if that's unclear
anyway, are you ready to move to the next one?
im done haha
thank you very much!!!!
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