Testing Solutions by Substitution - Sixth Grade Mathematics

Testing solutions by substitution determines whether a given candidate number satisfies an equation, providing an essential verification tool for checking algebraic solutions.

The Substitution Protocol

Before learning formal equation-solving algorithms, mathematicians verify solutions by substitution. This process checks whether a candidate value from a given replacement set creates equality between the left-hand side (LHS) and right-hand side (RHS).

+-------------------------------------------------------------------------+
|                  SOLUTION VERIFICATION PROTOCOL                         |
|                                                                         |
|   Step 1: Write down the original equation.                             |
|   Step 2: Replace the variable with the candidate value, enclosing the  |
|           substituted number in parentheses: (value).                   |
|   Step 3: Evaluate the arithmetic on the Left-Hand Side (LHS) using     |
|           the standard Order of Operations (PEMDAS).                    |
|   Step 4: Evaluate the arithmetic on the Right-Hand Side (RHS).         |
|   Step 5: Compare LHS and RHS:                                          |
|           - If LHS = RHS, the candidate value IS a valid solution.      |
|           - If LHS != RHS, the candidate value is NOT a solution.       |
+-------------------------------------------------------------------------+

Worked Example: Is x = 4 a solution to  5x - 3 = 17?

Step 1: Write equation:   5x - 3 = 17
Step 2: Substitute (4):   5(4) - 3 ?= 17
Step 3: Evaluate LHS:     20 - 3 = 17
Step 4: Compare sides:    17 = 17 (TRUE!)
Conclusion: Yes, x = 4 is a verified solution.

Testing Candidate Values from Replacement Sets

In many problem scenarios, you are given a finite replacement set of potential numbers and asked to identify which member is the true solution.

Problem: Which number from the replacement set { 3, 5, 7, 9 }
         is the solution to  4n + 6 = 34?

Test n = 3:  4(3) + 6 = 12 + 6 = 18  != 34  (False)
Test n = 5:  4(5) + 6 = 20 + 6 = 26  != 34  (False)
Test n = 7:  4(7) + 6 = 28 + 6 = 34  == 34  (TRUE!)
Test n = 9:  4(9) + 6 = 36 + 6 = 42  != 34  (False)

Conclusion: The only solution from the replacement set is n = 7.

Verifying Solutions in Multi-Step Equations

When solving equations algebraically, always build the habit of substituting your calculated answer back into the original equation. If an arithmetic error occurred during equation solving, substitution will immediately expose the mistake.

Exercises

Exercise 1. Determine whether x = 8 is a solution to each equation: (a) x + 15 = 23; (b) 4x - 10 = 24; (c) x / 2 + 5 = 9.

Exercise 2. Given the replacement set { 2, 4, 6, 8 }, determine which value satisfies the equation 3w - 5 = 13.

Exercise 3. A student solves the equation 25 - y = 9 and claims the solution is y = 14. Verify whether the student's solution is correct using substitution.

Solutions

Solution to Exercise 1. For (a) x + 15 = 23 with x = 8: 8 + 15 = 23; since 23 = 23, x = 8 is a solution. For (b) 4x - 10 = 24 with x = 8: 4(8) - 10 = 32 - 10 = 22; since 22 does not equal 24, x = 8 is not a solution. For (c) x / 2 + 5 = 9 with x = 8: 8 / 2 + 5 = 4 + 5 = 9; since 9 = 9, x = 8 is a solution.

Solution to Exercise 2. Test values from { 2, 4, 6, 8 } in 3w - 5 = 13: Test w = 2: 3(2) - 5 = 6 - 5 = 1 (False). Test w = 4: 3(4) - 5 = 12 - 5 = 7 (False). Test w = 6: 3(6) - 5 = 18 - 5 = 13 (True). Test w = 8: 3(8) - 5 = 24 - 5 = 19 (False). The solution from the set is w = 6.

Solution to Exercise 3. Substitute y = 14 into the original equation 25 - y = 9: Left-hand side: 25 - 14 = 11. Right-hand side: 9. Because 11 does not equal 9, the student's claim is false. (The correct solution is y = 16, since 25 - 16 = 9).