-3y(x^2 + 4xy - 5y^2) = -3x^2y - 12xy^2 + 15y^3
Title: Solving the Equation –3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³ Step-by-Step
Introduction Understanding and simplifying algebraic equations is essential for students, researchers, and mathematics enthusiasts alike. One such equation—–3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³—can seem complex at first glance, but with a clear step-by-step breakdown, solving it becomes manageable. This article walks you through the full process of simplifying, expanding, and analyzing the equation to solve it clearly, alongside a practical example and key takeaways.
Understanding the Equation We begin with: –3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³ This is a bilinear equation in two variables x and y. The left-hand side (LHS) is a product involving y and a quadratic trinomial in x and y, while the right-hand side (RHS) is a cubic polynomial in x and y, including a pure cubic term in y³.
Rewriting the equation in standard form:
→ Move every term to the left-hand side: –3y(x² + 4xy − 5y²) + 3x²y + 12xy² − 15y³ = 0
Step 1: Expand the Left Side Distribute –3y across the parentheses: –3y·x² − 3y·4xy + 3y·5y² + 3x²y + 12xy² − 15y³ → –3x²y − 12xy² + 15y³ + 3x²y + 12xy² − 15y³
Step 2: Combine Like Terms Group similar terms:
- x²y: (–3x²y + 3x²y) = 0
- xy²: (–12xy² + 12xy²) = 0
- y³: (15y³ – 15y³) = 0
Thus, the entire expression simplifies to: 0 = 0
Step 3: Interpret the Result Since both sides reduce to zero, the equation –3y(x² + 4xy − 5y²) = −3x²y − 12xy² + 15y³ is identically true for all real numbers x and y. In other words, the equation represents a trivial identity — it holds universally.
This means every ordered pair (x, y) satisfies the equation, and thus the expression is an algebraic identity.
Example Substitution (for clarity) Choose x = 1, y = 2:
- LHS: –3(2)(1² + 4(1)(2) – 5(2)²) = –6(1 + 8 – 20) = –6(–11) = 66
- RHS: –3(1)²(2) – 12(1)(2)² + 15(2)³ = –6 – 48 + 120 = 66 ✅ Both sides equal 66 — confirming the identity.
Why This Matters (Mathematical Insight) Recognizing identities like this streamlines problem-solving. It avoids unnecessary algebraic steps and clarifies relationships between variables. Identities simplify calculus operations, parametric modeling, and error-checking in applied mathematics.
Key Takeaways
- Expand and simplify expressions carefully.
- Combine like terms accurately.
- Recognizing identical forms (like 0 = 0) indicates an identity.
- Substituting sample values verifies correctness.
- This equation is universally true for all real x, y.
Conclusion While seemingly cryptic, –3y(x² + 4xy − 5y²) = –3x²y − 12xy² + 15y³ simplifies elegantly to 0 = 0, revealing it as a masterful algebraic identity. Mastery of such manipulations strengthens foundational algebra skills—essential in advanced mathematics, engineering, and computational modeling.
If you’re tackling similar equations, remember: always simplify fully, look for patterns, and verify with substitution.
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Further Reading:
- How to verify algebraic identities
- Common trinomial expansions and factorizations
- Linear vs. nonlinear equations in two variables
FAQ Q: Is this equation solvable for specific x and y? A: Yes, it’s true for all real x and y—it’s an identity, not a solvable equation with unique solutions. Q: Why does simplifying lead to 0 = 0? A: Because the LHS expands exactly into the RHS, canceling terms completely. Q: Can I apply this method to other polynomials? A: Absolutely—this technique applies broadly to simplify and verify algebraic expressions.
Stay algebraically sharp—understanding identities unlocks deeper mathematical mastery.
