When a=5 and b=-2, x≤ -3. Now we can complete the statement.
When a=5 and b=-2, x≤-3.
b Since we want the solution of the inequality to be all real numbers, we have to eliminate x from the inequality and obtain an statement that is always true. This can be done if we have the same x-terms on both sides of the inequality.
ax+4≤ 3x+b
Since we have 3x on the right-hand side, we will substitute a= 3.
If we want 4 ≤ b to be always true, we can substitute any number for b that makes this inequality true. Let's choose b= 5.
4 ≤ 5
This statement is always true and the solution to the original inequality will be all real numbers.
Now we can complete the statement.
When a=3 and b=5, the solution of the inequality is all real numbers.
c Since we want the inequality to have no solution, we have to eliminate x from the inequality and obtain an statement that is always false. We found in Part B that we can eliminate x if we let a=3.
If we want 4 ≤ b to be a contradiction, we have to substitute a number for b that makes this inequality false. Let's choose b= 3.
4 ≰ 3
This is always false and the original inequality will have no solution. Therefore, when a=3 and b=3, the inequality has no solution. Now we can complete the statement.
