SOLUTION: These conditions were given: AC + BC > AB 1st Triangle Inequality condition AB + BC > AC 2nd Triangle Inequality condition AB + AC > BC 3rd Triangle Inequality condition

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Question 1119574: These conditions were given:
AC + BC > AB 1st Triangle Inequality condition
AB + BC > AC 2nd Triangle Inequality condition
AB + AC > BC 3rd Triangle Inequality condition

Activity: The sum of the lengths of any two sides of a triangle is greater
than the length of the third side. What can you conclude about AC in
∆ABC? (Hint: Use the above example as a guideline.)
1. AB = 6; BC + 2AC = 18 (5 pts)
BC + 2AC = 18 > 6 (I know it seems very wrong but I couldn't understand it)

2. BC = 7; AC = 2 + AB (5 pts)
2 + AB > 7

3. AB + AC = 5; AC + BC = 4 (5 pts)
AB + AC = 5 > AB + BC = 4

4. BC + AC = 22; AB + BC = 12 (5 pts)
BC + AC = 22 > AB + BC =12
Please show your work so I can understand it better, thank you.

Did I write my inequalities correctly?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
I will only do the first one.

These conditions were given:
AC + BC > AB 1st Triangle Inequality condition
AB + BC > AC 2nd Triangle Inequality condition
AB + AC > BC 3rd Triangle Inequality condition

Activity: The sum of the lengths of any two sides of a triangle is greater
than the length of the third side. What can you conclude about AC in
∆ABC? (Hint: Use the above example as a guideline.)
1. AB = 6; BC + 2AC = 18 (5 pts) 
BC + 2AC = 18

Solve for BC

BC = 18 - 2AC

Substitute AB = 6 and BC = 18 - 2AC in:

AC + BC > AB 1st Triangle Inequality condition

AC + (18 - 2AC) > 6
  AC + 18 - 2AC > 6
       -AC + 18 > 6
            -AC > -12
             AC < 12

Next, substitute AB = 6 and BC = 18 - 2AC in
 
AB + BC > AC 2nd Triangle Inequality condition
6 + (18 - 2AC) > AC
  6 + 18 - 2AC > AC
      24 - 2AC > AC
            24 > 3AC
             8 > AC
            AC < 8

Finally, substitute AB = 6 and BC = 18 - 2AC in

AB + AC > BC 3rd Triangle Inequality condition 
 6 + AC > 18 - 2AC
    -12 > -3AC
      4 < AC
     AC > 4

So putting all three results together

AC < 12, AC < 8, and AC > 4

Since AC is less than 8, it's automatically less than 12,
so we can dispense with AC < 12.  So we have:

AC < 8 and AC > 4  which can be combined and written as

    4 < AC < 8    [that is, AC is between 4 and 8]

--------------
Do the others the same way.  
1.  Solve the given equation for the side other than AC.  
2.  Substitute in each of the Triangle Inequality conditions.
3.  Solve for AC in each.
4.  Consider the three inequalities together. 
5.  Give the inequality interval for AC.

Edwin