Sign In
Rewrite this inequality as a compound inequality.
Solution Set: {h | - 5 23
We can split this compound inequality into two cases, one where 3h + 12 is greater than -8 and one where 3h + 12 is less than 8. 3h+1/2>- 8 and 3h+1/2< 8 Let's isolate h in both of these cases before graphing the solution set.
LHS* 2 > RHS * 2
LHS-1>RHS-1
.LHS /3.>.RHS /3.
Put minus sign in front of fraction
Rewrite 17 as 15+2
Write as a sum of fractions
Calculate quotient
Add terms
Put minus sign in front of fraction
The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality. First Solution Set: & - 5 23< h [0.5em] Second Solution Set: & h < 5 [0.5em] Intersecting Solution Set:& - 5 23< h< 5
The graph of this inequality includes all values from - 5 23 to 5, not inclusive. We show this by using open circles on the endpoints.