Solve the Equation and Check Your Solution: A Step‑by‑Step Guide
The moment you encounter an algebraic problem, the most reliable way to be certain of your answer is to solve the equation and then check your solution. This two‑part process not only confirms correctness but also deepens your understanding of the underlying mathematics. Below you will find a detailed methodology, illustrative examples, common pitfalls, and a FAQ section that addresses typical questions learners have about verification.
Why Checking Matters
Even seasoned mathematicians make arithmetic slips. A misplaced sign, a forgotten distribution, or an accidental division by zero can turn a correct derivation into a false result. Because of that, by substituting the obtained value back into the original equation, you create an independent test that catches these errors. Also worth noting, the act of checking reinforces the logical flow: if the left‑hand side equals the right‑hand side after substitution, then the candidate value truly satisfies the equation.
Part 1: Solving the Equation
1. Identify the Type of Equation
| Equation Type | Typical Form | Primary Solving Technique |
|---|---|---|
| Linear | (ax + b = 0) | Isolate the variable (add/subtract, then divide/multiply) |
| Quadratic | (ax^2 + bx + c = 0) | Factoring, completing the square, or quadratic formula |
| Rational | (\frac{P(x)}{Q(x)} = 0) | Set numerator (P(x)=0) while noting restrictions from denominator |
| Radical | (\sqrt[n]{f(x)} = g(x)) | Raise both sides to the nth power, then solve the resulting polynomial |
| Exponential | (a^{f(x)} = b) | Take logarithms of both sides |
| Logarithmic | (\log_a f(x) = c) | Rewrite in exponential form: (f(x) = a^{c}) |
Recognizing the structure guides you toward the most efficient method and helps you avoid unnecessary algebraic gymnastics.
2. Perform Algebraic Manipulations
- Keep the equation balanced: Whatever you do to one side, do to the other.
- Combine like terms early to reduce clutter.
- Clear fractions by multiplying through by the least common denominator (LCD) when dealing with rational expressions.
- Isolate the variable step by step, watching for signs especially when multiplying or dividing by negative numbers.
3. Consider Domain Restrictions
Certain operations introduce implicit constraints:
- Denominators cannot be zero.
- Even‑root radicands must be non‑negative (for real numbers).
- Logarithmic arguments must be positive.
Jot these down before you begin; they will be crucial when you later check your solution No workaround needed..
4. Obtain Candidate Solution(s)
After simplification, you will have one or more potential values for the unknown. Because of that, write them clearly, e. g., (x = 3) or (x = -2, ; x = 5).
Part 2: Checking Your Solution
1. Substitute Back into the Original Equation
Replace every occurrence of the variable with the candidate value. Compute both the left‑hand side (LHS) and the right‑hand side (RHS) separately.
2. Simplify Each Side
Carry out the arithmetic exactly as the original expression dictates. Do not rearrange or cancel terms that were not present in the given equation; the goal is to see if the two sides match as they stand Which is the point..
3. Compare LHS and RHS
- If LHS = RHS (within the realm of real numbers, or exactly equal in symbolic form), the candidate satisfies the equation.
- If they differ, the candidate is extraneous—discard it and re‑examine your solving steps for algebraic mistakes or missed domain restrictions.
4. Verify Domain Conditions
Even if the substitution yields equality, ensure the value does not violate any restrictions identified earlier (e.g., making a denominator zero). A value that passes the substitution but breaks a domain rule is not a valid solution Worth knowing..
5. Document the Check
Write a brief verification statement, such as:
For (x = 3), LHS = (2(3) + 5 = 11) and RHS = (11); thus (x = 3) satisfies the original equation and respects the domain (denominator ≠ 0).
Worked Examples
Example 1: Linear Equation
Problem: Solve (4x - 7 = 9) and check your solution And that's really what it comes down to..
Solution Steps
- Add 7 to both sides: (4x = 16).
- Divide by 4: (x = 4).
Check
- LHS: (4(4) - 7 = 16 - 7 = 9).
- RHS: (9).
Since LHS = RHS and no domain issues exist, (x = 4) is correct.
Example 2: Quadratic Equation
Problem: Solve (x^2 - 5x + 6 = 0) and check your solution.
Solution Steps
- Factor: ((x - 2)(x - 3) = 0).
- Set each factor to zero: (x = 2) or (x = 3).
Check
- For (x = 2): LHS = (2^2 - 5(2) + 6 = 4 - 10 + 6 = 0); RHS = 0.
- For (x = 3): LHS = (3^2 - 5(3) + 6 = 9 - 15 + 6 = 0); RHS = 0.
Both values satisfy the equation; no restrictions apply, so both are valid Not complicated — just consistent..
Example 3: Rational Equation
Problem: Solve (\frac{2}{x+1} = \frac{3}{x-2}) and check your solution.
Solution Steps
- Cross‑multiply: (2(x-2) = 3(x+1)).
- Expand: (2x - 4 = 3x + 3).
- Bring terms: (-4 - 3 = 3x - 2x) → (-7 = x).
Check
- LHS with (x = -7): (\frac{2}{-7+1} = \frac{2}{-6} = -\frac{1}{3}).
- RHS with (x = -7): (\frac{3}{-7-2} = \frac{3}{-9} = -\frac{1}{3}).
LHS = RHS.
Domain Check: Denominators (x+1) and (x-2) become (-6) and (-9), neither zero. Hence (x = -7) is acceptable Simple, but easy to overlook..
Example 4: Radical Equation (Extraneous Root)
Problem: Solve (\sqrt{x+3} = x - 1) and check your solution.
Solution Steps
-
Square both sides: (x + 3 = (x-1)^2).
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Expand RHS
-
Expand RHS: ((x-1)^2 = x^2 - 2x + 1).
The equation becomes
[ x + 3 = x^2 - 2x + 1. ] -
Bring all terms to one side:
[ 0 = x^2 - 2x + 1 - x - 3 = x^2 - 3x - 2. ]
Thus we need to solve the quadratic (x^2 - 3x - 2 = 0) Took long enough.. -
Apply the quadratic formula:
[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2} = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}. ]
The two algebraic candidates are
[ x_1 = \frac{3 + \sqrt{17}}{2}, \qquad x_2 = \frac{3 - \sqrt{17}}{2}. ] -
Check each candidate in the original radical equation (\sqrt{x+3}=x-1).
For (x_1 = \frac{3 + \sqrt{17}}{2}):
[ \text{LHS} = \sqrt{\frac{3 + \sqrt{17}}{2} + 3} = \sqrt{\frac{3 + \sqrt{17} + 6}{2}} = \sqrt{\frac{9 + \sqrt{17}}{2}}. ]
[ \text{RHS} = \frac{3 + \sqrt{17}}{2} - 1 = \frac{3 + \sqrt{17} - 2}{2} = \frac{1 + \sqrt{17}}{2}. ]
Squaring both sides shows equality:
[ \left(\frac{1 + \sqrt{17}}{2}\right)^2 = \frac{1 + 2\sqrt{17} + 17}{4} = \frac{18 + 2\sqrt{17}}{4} = \frac{9 + \sqrt{17}}{2}, ]
which is exactly the radicand of the LHS. Since both sides are non‑negative for this value, (x_1) satisfies the original equation Small thing, real impact..For (x_2 = \frac{3 - \sqrt{17}}{2}):
[ \text{LHS} = \sqrt{\frac{3 - \sqrt{17}}{2} + 3} = \sqrt{\frac{9 - \sqrt{17}}{2}}. ]
[ \text{RHS} = \frac{3 - \sqrt{17}}{2} - 1 = \frac{1 - \sqrt{17}}{2}. ]
Here the RHS is negative because (\sqrt{17}>1), whereas the LHS (a principal square root) is non‑negative. Squaring both sides would give equality, but the sign mismatch means the original equation fails. On top of that, substituting (x_2) into the radicand yields a positive number, so the domain restriction (x+3\ge0) is satisfied, but the equation (\sqrt{x+3}=x-1) is not. Hence (x_2) is an extraneous root introduced by squaring. -
Domain check: The original radical requires (x+3\ge0). Both candidates satisfy this ((x_1\approx3.56), (x_2\approx-0.56)). The extraneous root is rejected solely because it violates the equality after accounting for the sign of the square root Nothing fancy..
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Verification statement:
For (x = \frac{3+\sqrt{17}}{2}), LHS = (\sqrt{\frac{9+\sqrt{17}}{2}}) and RHS = (\frac{1+\sqrt{17}}{2}); squaring shows LHS² = RHS² and both sides are non‑negative, so the equality holds. The other algebraic root (\frac{3-\sqrt{17}}{2}) makes RHS negative while LHS remains non‑negative, thus it is extraneous.
Additional Example: Logarithmic Equation
Problem: Solve (\log_{2}(x-1) + \log_{2}(x+1) = 3) and check your solution Turns out it matters..
