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????
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.
Damon, since x ≤ -3 is already a subset of
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.
To graph the solution sets of these inequalities on a number line, follow these steps:
1. Draw a horizontal line and label it with numbers.
2. Identify the solutions for each inequality separately and mark them on the number line.
For the first inequality, 3 + 2(1 - x) > 6, the solution is x < -1/2.
- To graph this, place an open circle at -1/2 on the number line.
- Then, draw an arrow to the left of the open circle to indicate that the solution set includes all the values less than -1/2.
For the second inequality, 2x + 14 ≤ 8, the solution is x ≤ -3.
- To graph this, place a filled-in circle at -3 on the number line.
- Then, draw a horizontal line from the filled-in circle to the left, covering all the values less than or equal to -3.
Lastly, combine the two solution sets on the number line:
- Place an open circle at -1/2 and draw an arrow to the left.
- Place a filled-in circle at -3 and draw a horizontal line to the left, covering all the values less than or equal to -3.
The final graph should show a line with a filled-in circle at -3 and an open circle at -1/2, with a line segment between them extending to the left.