Multiply both sides by \( x + 5 \) (assuming \( x
["# Multiply Both Sides by ( x + 5 ): A Step-by-Step Guide to Solving Equations", "Solving linear equations often requires manipulating both sides to isolate the variable. One common strategy is multiplying both sides of an equation by an expression—such as ( x + 5 )—to simplify and solve for ( x ). But what happens if ( x + 5 = 0 )? This article explores the process of multiplying both sides by ( x + 5 ), especially when ( x ) could yield values that make this expression zero, and how to handle such cases carefully.", "## Why Multiply Both Sides by ( x + 5 )", "Multiplying both sides of an equation by ( x + 5 ) is useful because it eliminates parentheses and combines like terms, transforming the equation into a simpler linear form. This technique preserves equality as long as ( x + 5 <br/>\neq 0 ), allowing straightforward solving. However, when ( x + 5 = 0 ), multiplying introduces a potential extraneous solution that must be verified.", "---", "## The Equation: Multiplying by ( x + 5 )", "Consider a basic equation:", "[\n2(x + 5) = 3x\n]", "To eliminate the parentheses, multiply both sides by ( x + 5 ), provided ( x + 5 <br/>\neq 0 ):", "[\n2(x + 5) \cdot (x + 5) = 3x(x + 5)\n]", "Simplify both sides:", "[\n2(x + 5)^2 = 3x(x + 5)\n]", "Expand both sides:", "- Left: ( 2(x^2 + 10x + 25) = 2x^2 + 20x + 50 )\n- Right: ( 3x^2 + 15x )", "Bring all terms to one side:", "[\n2x^2 + 20x + 50 - 3x^2 - 15x = 0\n]", "Simplify:", "[\n- x^2 + 5x + 50 = 0\n]", "Multiply through by –1 to simplify:", "[\nx^2 - 5x - 50 = 0\n]", "This quadratic equation now solves easily using factoring, the quadratic formula, or completing the square.", "---", "## Handling the Zero Case: When ( x + 5 = 0 )", "A critical step is checking if ( x + 5 = 0 ) creates a valid solution. If ( x = -5 ), then multiplying both sides of any original equation by ( x + 5 ) results in:", "[\n\ ext{LHS} = \ ext{RHS} \ imes 0 = 0\n]", "So the equation becomes ( 0 = 0 ), which is always true but does not yield any specific value for ( x ). Thus, ( x = -5 ) is not a valid solution of the original equation—it is an extraneous solution introduced by multiplication.", "Hence, always verify solutions when multiplying both sides by an expression containing the variable.", "---", "## Practical Example", "Solve:", "[\n\frac{2}{x + 5} = \frac{4}{x}\n]", "Multiply both sides by ( x(x + 5) ), assuming ( x <br/>\neq 0 ) and ( x <br/>\neq -5 ):", "[\n2x = 4(x + 5)\n]", "Expand:", "[\n2x = 4x + 20\n]", "Simplify:", "[\n-2x = 20 \Rightarrow x = -10\n]", "Check:\n( x = -10 ) does not make any denominator zero.\nSubstitute back into original:", "LHS: ( \frac{2}{-10 + 5} = \frac{2}{-5} = -\frac{2}{5} )\nRHS: ( \frac{4}{-10} = -\frac{2}{5} ) ✓", "Valid solution. Note: ( x = 0 ) and ( x = -5 ) would make denominators zero and are excluded.", "---", "## Key Takeaways", "- Multiplying both sides by ( x + 5 ) simplifies equations but requires checking assumptions.\n- When ( x + 5 = 0 ), the equation loses meaning—avoid ansatz without verification.\n- Always isolate and check potential solutions when using this method.\n- This technique is powerful but must be applied carefully to prevent errors.", "---", "## Summary", "Multiplying both sides by ( x + 5 ) is a valid algebraic step when ( x <br/>\neq -5 ), streamlining equation solving. However, awareness of critical values ensures correctness—never accept solutions that violate the domain restrictions imposed by the original equation. Mastering this skill helps solve more complex algebraic expressions confidently.", "---", "Keywords: Multiply both sides by ( x + 5 ), solve linear equations, algebra technique, avoid extraneous solutions, domain restrictions, equation solving, quadratic equation, verify solutions, algebraic manipulation."]
