3x+y>8x+y<4

Split the inequalities first and do them separately:

1. 3x+y>8x+y ( when you subtract y on both sides it will cancel out)

   3x >8x      

3x-8x >0

-5x>0 ( when we divide by a negative number we need to "flip" inequality sign)

x<0 ( this will be your inequality- when graphed it would be dashed vertical line on y-axis with shading to the left)

2. 8x+y<4 (subtract 8x on both sides)

 y<-8x+4  ( you can graph it on the coordinate plane: line going to the left with y-int=4 and line needs to be dashed and shaded below)

your solution will be when shading intersects (overlaps)

3x+y>8x+y<4 we have to solve first part 3x+y>8x+y subtract y from each side 3x>8x  we can solve it for y or it is true for all y values.  from here we can say x is all negative numbers.    second part since we know x is negative 8x+y < 4  => y<4-8x