Sealed Solution Eric

Problem 1: Envelope and Card Combinations
Known Cards: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Envelope Rules:
Two cards are in each envelope.
The sum of the two cards is written on the envelope.
Envelope Sums: 3, 7, 8, 13, 14
Goal: Determine the possible pairs of cards in the envelope marked "8".
Step 1: List Possible Pairs Summing to 8
Possible combinations (duplicates excluded):
0 + 8
1 + 7
2 + 6
3 + 5
4 + 4 (not possible, only one "4" card is available)
Step 2: Eliminate Impossible Pairs
Investigating other envelope sums, valid pairs for "8" are:
0 and 8
1 and 7
2 and 6
3 and 5
Problem 2: Solving the Inequality n+10>2n+3n+10>2n+3
Step 1: Set Up the Inequality
Two expressions are being compared:
n+10n+10
2n+32n+3
Goal: Find values of nn where n+10>2n+3n+10>2n+3.
Step 2: Solve Algebraically
Subtract 2n2n from both sides:
n+10−2n>3n+10−2n>3
−n+10>3−n+10>3
Subtract 10 from both sides:
−n>−7−n>−7
Multiply by −1−1 (reverse the inequality sign):
n<7n<7
Important Note: Multiplying/dividing by a negative number reverses the inequality.
Step 3: Check the Solution
For n<7n<7: n+10>2n+3n+10>2n+3 (e.g., n=5n=5: 15>1315>13).
For n>7n>7: 2n+3>n+102n+3>n+10 (e.g., n=8n=8: 19>1819>18).
At n=7n=7: Both expressions equal 17.
Graphical Interpretation
The lines y=n+10y=n+10 and y=2n+3y=2n+3 intersect at n=7n=7.
n+10n+10 is above 2n+32n+3 when n<7n<7.
General Strategy for Comparing Functions
Set up f(n)>g(n)f(n)>g(n).
Rearrange to solve for nn.
Solve, maintaining inequality rules.
Interpret the solution to determine dominance regions.