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find all solutions to the equations $\frac{15}{x(x+4)} = \frac{x+2}{x+4}$

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Solve multi-step equations with fractional coefficients

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Q. find all solutions to the equations

\frac{15}{x(x+4)} = \frac{x+2}{x+4}

Write Equation: First, let's write down the equation clearly and make sure we understand it correctly. $\newline$ The equation is $\frac{15}{x(x+4)} = \frac{x+2}{x+4}$ .
Eliminate Denominators: Next, we will multiply both sides of the equation by $x(x+4)$ to eliminate the denominators. $\newline$ $\frac{15}{x(x+4)} \cdot x(x+4) = \frac{x+2}{x+4} \cdot x(x+4)$ $\newline$ This simplifies to: $\newline$ $15 = (x+2)x$
Distribute $x$ : Now, let's distribute $x$ on the right side of the equation. $15 = x^2 + 2x$
Set to Zero: We will move all terms to one side to set the equation to zero and solve for $x$ as a quadratic equation. $x^2 + 2x - 15 = 0$
Factor Quadratic: Now, we factor the quadratic equation. $\newline$ $(x + 5)(x - 3) = 0$
Find Values of x: Next, we find the values of $x$ that make each factor equal to zero. $x + 5 = 0$ or $x - 3 = 0$
Check Solutions: Solving each equation for $x$ gives us the solutions. $x = -5$ or $x = 3$
Check Solutions: Solving each equation for $x$ gives us the solutions. $x = -5$ or $x = 3$ We must check these solutions in the original equation to ensure they do not make the denominator zero, as that would be undefined. $\newline$ For $x = -5$ : $\newline$ The original denominator $x(x+4)$ becomes $-5(-5+4)$ which is $-5(-1) = 5$ , not zero. $\newline$ For $x = 3$ : $\newline$ The original denominator $x(x+4)$ becomes $3(3+4)$ which is $x = -5$ $0$ , not zero. $\newline$ Both solutions are valid.

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