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Solve the rational equation: \( \frac{x-8}{x}=\frac{63}{x+8} \)

Ask by Schmidt Crawford. in the United States
Feb 27,2025

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Answer

The solutions are \( x=-1 \) and \( x=64 \).

Solution

Solve the equation \( \frac{x-8}{x}=\frac{63}{x+8} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{x-8}{x}=\frac{63}{x+8}\) - step1: Find the domain: \(\frac{x-8}{x}=\frac{63}{x+8},x \in \left(-\infty,-8\right)\cup \left(-8,0\right)\cup \left(0,+\infty\right)\) - step2: Cross multiply: \(\left(x-8\right)\left(x+8\right)=x\times 63\) - step3: Simplify the equation: \(\left(x-8\right)\left(x+8\right)=63x\) - step4: Expand the expression: \(x^{2}-64=63x\) - step5: Move the expression to the left side: \(x^{2}-64-63x=0\) - step6: Factor the expression: \(\left(x-64\right)\left(x+1\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x-64=0\\&x+1=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=64\\&x=-1\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&x=64\\&x=-1\end{align},x \in \left(-\infty,-8\right)\cup \left(-8,0\right)\cup \left(0,+\infty\right)\) - step10: Find the intersection: \(\begin{align}&x=64\\&x=-1\end{align}\) - step11: Rewrite: \(x_{1}=-1,x_{2}=64\) The solutions to the rational equation \( \frac{x-8}{x}=\frac{63}{x+8} \) are \( x=-1 \) and \( x=64 \).

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To solve the equation \( \frac{x-8}{x} = \frac{63}{x+8} \), we can start by cross-multiplying to eliminate the fractions. This gives us: \[ (x-8)(x+8) = 63x \] Next, we expand the left-hand side: \[ x^2 - 64 = 63x \] Now, rearranging the equation to set it to zero results in: \[ x^2 - 63x - 64 = 0 \] We can factor this quadratic equation. We are looking for two numbers that multiply to \(-64\) and add to \(-63\). These numbers are \(-64\) and \(1\). Therefore, we can factor the equation as follows: \[ (x - 64)(x + 1) = 0 \] Setting each factor equal to zero gives us the possible solutions: 1. \(x - 64 = 0 \Rightarrow x = 64\) 2. \(x + 1 = 0 \Rightarrow x = -1\) However, we must check for extraneous solutions by substituting back into the original equation. For \(x = 64\): \[ \frac{64-8}{64} = \frac{56}{64} = \frac{7}{8} \] \[ \frac{63}{64+8} = \frac{63}{72} = \frac{7}{8} \] So, \(x = 64\) is valid. For \(x = -1\): \[ \frac{-1-8}{-1} = \frac{-9}{-1} = 9 \] \[ \frac{63}{-1+8} = \frac{63}{7} = 9 \] So, \(x = -1\) is also valid. Thus, the solutions to the equation are: \[ \boxed{64 \text{ and } -1} \]

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