for the first one it is everything to the left of -1/2 on your number line.
So put an open circle around -1/2 and an arrow to the left.
The open circle means everything to the left, but not the point itself.
For the second one, the point x = -3 is included so draw a solid, shaded in, circle around that point. Then your arrow points from there to the left.
Solve: 3+2(1-x) >6 or 2x+14 <=8. Graph the solution set on a number line.
When I solved the equations for the first one I got x<-1/2 and for the second one I got x<= -3.
How do I graph the solution set for these????
3 answers
Damon, since x ≤ -3 is already a subset of
x < -1/2, graphing x < -1/2 the way you just described it would suffice.
x < -1/2, graphing x < -1/2 the way you just described it would suffice.
Oh, I did not realize they were connected by a logical or.
1 answer