How do you solve and graph #r+6< -8# and #r-3> -10#?

Let's solve the first one: #r+6 < -8# #r<-8-6# #r<-14# graph{x < -14 [-20.014, -7.527, -2.727, 3.52]}

The second one: #r-3> -10# #r>3 - 10# #r> -7# graph{x > -7 [-10.65, 0.456, -2.4, 3.154]}

To get the final answer you have to merge the 2 graphs: #r < -14# or #r> -7#

To solve and graph the inequalities r + 6 < -8 and r - 3 > -10, we first isolate r in each inequality: For r + 6 < -8: r < -8 - 6 r < -14 For r - 3 > -10: r > -10 + 3 r > -7 So, the solution for the system of inequalities is: -14 < r < -7. To graph this solution on a number line, we plot an open circle at -14 (since r is not equal to -14) and an open circle at -7 (since r is not equal to -7), then shade the region between these two points to represent the values of r that satisfy both inequalities.

To solve the inequalities r + 6 < -8 and r - 3 > -10, we need to isolate the variable r in each inequality. For the first inequality, r + 6 < -8: Subtract 6 from both sides: r < -8 - 6 r < -14 For the second inequality, r - 3 > -10: Add 3 to both sides: r > -10 + 3 r > -7 Now, we have two inequalities: r < -14 r > -7 To graph these inequalities on a number line, we'll represent each solution set separately: For r < -14, we'll shade the region to the left of -14 because r is less than -14. For r > -7, we'll shade the region to the right of -7 because r is greater than -7. So, on the number line, we'll have an open circle at -14 and shade to the left, indicating r < -14. Similarly, we'll have an open circle at -7 and shade to the right, indicating r > -7. The graph will show two separate shaded regions on the number line: one to the left of -14 and one to the right of -7. These regions represent the solutions to the given inequalities.