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Solve for \( x \) \( x+\frac{6}{x+10}=\frac{5 x}{x+10} \) \( x= \) One or more solutions: No solution

Ask by Haynes Bolton. in the United States
Feb 03,2025

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Tutor-Verified Answer

Answer

The solutions are \( x = -3 \) and \( x = -2 \).

Solution

Solve the equation \( x+\frac{6}{x+10}=\frac{5x}{x+10} \). Solve the equation by following steps: - step0: Solve for \(x\): \(x+\frac{6}{x+10}=\frac{5x}{x+10}\) - step1: Find the domain: \(x+\frac{6}{x+10}=\frac{5x}{x+10},x\neq -10\) - step2: Multiply both sides of the equation by LCD: \(\left(x+\frac{6}{x+10}\right)\left(x+10\right)=\frac{5x}{x+10}\times \left(x+10\right)\) - step3: Simplify the equation: \(x^{2}+10x+6=5x\) - step4: Move the expression to the left side: \(x^{2}+10x+6-5x=0\) - step5: Subtract the terms: \(x^{2}+5x+6=0\) - step6: Factor the expression: \(\left(x+2\right)\left(x+3\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x+2=0\\&x+3=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=-2\\&x=-3\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&x=-2\\&x=-3\end{align},x\neq -10\) - step10: Find the intersection: \(\begin{align}&x=-2\\&x=-3\end{align}\) - step11: Rewrite: \(x_{1}=-3,x_{2}=-2\) The solutions to the equation \(x+\frac{6}{x+10}=\frac{5x}{x+10}\) are \(x=-3\) and \(x=-2\).

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Mind Expander

To solve the equation \( x + \frac{6}{x+10} = \frac{5x}{x+10} \), first eliminate the fraction by multiplying both sides by \( x + 10 \) (noting that \( x \neq -10 \)): \[ (x + 10)\left(x + \frac{6}{x + 10}\right) = (x + 10)\left(\frac{5x}{x + 10}\right) \] This simplifies to: \[ x(x + 10) + 6 = 5x \] \[ x^2 + 10x + 6 = 5x \] Rearranging gives: \[ x^2 + 5x + 6 = 0 \] Now we can factor this quadratic: \[ (x + 2)(x + 3) = 0 \] Setting each factor to zero, we find the solutions: \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] Thus, the solutions are \( x = -2 \) and \( x = -3 \). To assure these solutions are valid, check if they cause division by zero in the original equation: neither \( -2 \) nor \( -3 \) equals \( -10 \), so both solutions are valid. In summary, \( x = -2 \) and \( x = -3 \).

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