Which Shows The Correct Solution Of The Equation When

Which Shows the Correct Solution of the Equation When
Determining whether a proposed value truly satisfies an equation is a fundamental skill in algebra, calculus, and beyond. The phrase “which shows the correct solution of the equation when” captures the moment when we test a candidate answer, substitute it back into the original statement, and verify that both sides balance. Mastering this verification process not only prevents careless errors but also builds confidence when dealing with more complex manipulations such as squaring both sides, clearing denominators, or applying logarithmic identities. In the following sections we break down the verification workflow, explain the underlying mathematical principles, highlight common pitfalls, and provide worked examples that illustrate exactly which shows the correct solution of the equation when each step is applied correctly.

Introduction to Solution Verification

When solving an equation, we often perform algebraic transformations that are not always reversible. For instance, multiplying both sides by a variable expression can introduce extraneous roots, while squaring both sides may lose sign information. Consequently, a solution derived purely from symbolic manipulation must be checked against the original equation. The act of substitution—plugging the candidate value into the original expression and simplifying—answers the question which shows the correct solution of the equation when the left‑hand side (LHS) equals the right‑hand side (RHS). If equality holds, the candidate is valid; if not, it must be discarded.

Step‑by‑Step Procedure for Verifying Solutions

1. Isolate the candidate solution
   After solving, you will have one or more potential values (e.g., (x = 3), (x = -2)). Write each candidate clearly.

2. Substitute into the original equation
   Replace every occurrence of the unknown variable with the candidate value, preserving parentheses and order of operations.

3. Simplify both sides independently
   Compute the LHS and RHS separately. Use exact arithmetic (fractions, radicals) whenever possible to avoid rounding errors.

4. Compare the results

   • If LHS = RHS (exactly), the candidate shows the correct solution of the equation when substituted.
   • If LHS ≠ RHS, the candidate is extraneous or incorrect.

5. State the conclusion
   List all values that passed the test as the solution set. If none pass, declare that the equation has no solution in the considered domain.

Why Substitution Works: The Mathematical Explanation

The validity of substitution rests on the substitution property of equality: if two expressions are equal, replacing a variable with the same value in both expressions preserves equality. Formally, for any real numbers (a), (b), and (c),

[ \text{If } a = b \text{ then } f(a) = f(b) ]

where (f) denotes any well‑defined function (addition, multiplication, exponentiation, etc.). When we solve an equation, we manipulate it using operations that are functions of the variable (e.g., adding 5, multiplying by (x), squaring). If each step is reversible, the final candidate must satisfy the original equation. However, when a step is not reversible (like squaring), the substitution property may fail for some candidates, producing extraneous solutions. Therefore, the substitution check is the safeguard that restores logical equivalence.

Common Pitfalls and How to Avoid Them

By staying vigilant about these issues, you ensure that the step “which shows the correct solution of the equation when” truly reflects a valid answer.

Worked Examples ### Example 1: Rational Equation

Solve (\displaystyle \frac{2}{x-1} + 3 = \frac{5}{x-1}).

Solution steps 1. Subtract (\frac{2}{x-1}) from both sides: (3 = \frac{3}{x-1}).
2. Multiply both sides by ((x-1)): (3(x-1) = 3).
3. Divide by 3: (x-1 = 1) → (x = 2).

Verification

• Substitute (x = 2) into the original equation:
  LHS = (\frac{2}{2-1} + 3 = 2 + 3 = 5).
  RHS = (\frac{5}{2-1} = 5).
• Since LHS = RHS, (x = 2) shows the correct solution of the equation when substituted.
• Note: (x = 1) would make denominators zero; it is excluded from the domain.

Example 2: Radical Equation

Solve (\sqrt{x+4} = x - 2).

Solution steps

1. Square both sides: (x+4 = (x-2)^2).
2. Expand RHS: (x+4 = x^2 -4x +4).
3. Bring all terms to one side: (0 = x^2 -5x).
4. Factor: (0 = x(x-5)) → candidates (x = 0) or (x = 5).

Verification

• For (x = 0): LHS = (\sqrt{0+4} = 2); RHS = (0-2 = -2). Not equal → extr

aneous solution.

• For (x = 5): LHS = (\sqrt{5+4} = 3); RHS = (5-2 = 3). Since LHS = RHS, (x = 5) shows the correct solution of the equation when substituted.

• Note: (x = 0) is an extraneous solution that arises from squaring the equation. Always verify candidates in the original equation to rule out such solutions.

Example 3: Logarithmic Equation

Solve (\ln(x) + \ln(x-3) = 1).

Solution steps

1. Combine logarithms: (\ln(x(x-3)) = 1).
2. Exponentiate both sides: (x(x-3) = e).
3. Rearrange into a quadratic equation: (x^2 - 3x - e = 0).
4. Solve using the quadratic formula: (x = \frac{3 \pm \sqrt{9 + 4e}}{2}).

Verification

• For (x = \frac{3 + \sqrt{9 + 4e}}{2}): Check if (x > 3) (since (\ln(x-3)) must be defined).

  • (x = \frac{3 + \sqrt{9 + 4e}}{2}) is greater than 3.
  • Substitute into the original equation to confirm it holds true.

• For (x = \frac{3 - \sqrt{9 + 4e}}{2}): Check if (x > 3).

  • (x = \frac{3 - \sqrt{9 + 4e}}{2}) is less than 3.
  • This value is outside the domain of (\ln(x-3)), so it is not a valid solution.

• Note: Only (x = \frac{3 + \sqrt{9 + 4e}}{2}) shows the correct solution of the equation when substituted.

Conclusion

Solving equations involves more than just manipulating symbols on paper; it requires a keen eye for potential pitfalls that can lead to incorrect or extraneous solutions. By recognizing common errors such as forgotten domain restrictions, sign loss after squaring, canceling zero factors, rounding errors, and misapplying inverse functions, one can avoid these traps. Always verify solutions in the original equation to ensure they are valid. Through careful attention to detail and rigorous verification, you can confidently determine the correct solutions to a wide range of equations.