Question
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Solve the equation \( \frac{x+1}{x-1}=\frac{-11}{x+3}+\frac{8}{x^{2}+2 x-3} \) List all valid solutions, separated by commas. \( x=\square \)

Ask by Stephens Floyd. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

\( x=-16 \)

Solution

Solve the equation \( \frac{x+1}{x-1}=\frac{-11}{x+3}+\frac{8}{x^{2}+2x-3} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{x+1}{x-1}=\frac{-11}{x+3}+\frac{8}{x^{2}+2x-3}\) - step1: Find the domain: \(\frac{x+1}{x-1}=\frac{-11}{x+3}+\frac{8}{x^{2}+2x-3},x \in \left(-\infty,-3\right)\cup \left(-3,1\right)\cup \left(1,+\infty\right)\) - step2: Rewrite the fraction: \(\frac{x+1}{x-1}=-\frac{11}{x+3}+\frac{8}{x^{2}+2x-3}\) - step3: Multiply both sides of the equation by LCD: \(\frac{x+1}{x-1}\times \left(x-1\right)\left(x+3\right)=\left(-\frac{11}{x+3}+\frac{8}{x^{2}+2x-3}\right)\left(x-1\right)\left(x+3\right)\) - step4: Simplify the equation: \(x\left(x+3\right)+x+3=-11x+19\) - step5: Calculate: \(x^{2}+3x+x+3=-11x+19\) - step6: Simplify: \(x^{2}+4x+3=-11x+19\) - step7: Move the expression to the left side: \(x^{2}+4x+3-\left(-11x+19\right)=0\) - step8: Calculate: \(x^{2}+15x-16=0\) - step9: Factor the expression: \(\left(x-1\right)\left(x+16\right)=0\) - step10: Separate into possible cases: \(\begin{align}&x-1=0\\&x+16=0\end{align}\) - step11: Solve the equation: \(\begin{align}&x=1\\&x=-16\end{align}\) - step12: Check if the solution is in the defined range: \(\begin{align}&x=1\\&x=-16\end{align},x \in \left(-\infty,-3\right)\cup \left(-3,1\right)\cup \left(1,+\infty\right)\) - step13: Find the intersection: \(x=-16\) The valid solution to the equation \( \frac{x+1}{x-1}=\frac{-11}{x+3}+\frac{8}{x^{2}+2x-3} \) is \( x=-16 \).

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The Deep Dive

To solve the equation \[ \frac{x+1}{x-1}=\frac{-11}{x+3}+\frac{8}{x^{2}+2x-3}, \] we first factor the denominator of the third term on the right side. The expression \( x^2 + 2x - 3 \) can be factored as \( (x+3)(x-1) \). Thus, we can rewrite the equation as: \[ \frac{x+1}{x-1} = \frac{-11}{x+3} + \frac{8}{(x+3)(x-1)}. \] Next, we need to combine the terms on the right-hand side: \[ \frac{-11(x-1) + 8}{(x+3)(x-1)} = \frac{-11x + 11 + 8}{(x+3)(x-1)} = \frac{-11x + 19}{(x+3)(x-1)}. \] Now we have: \[ \frac{x+1}{x-1} = \frac{-11x + 19}{(x+3)(x-1)}. \] Cross-multiply to eliminate the fractions: \[ (x + 1)(x + 3) = (-11x + 19)(x - 1). \] Expanding both sides gives: \[ x^2 + 4x + 3 = -11x^2 + 11x + 19. \] Now, combine like terms: \[ 12x^2 - 7x - 16 = 0. \] Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 12, b = -7, c = -16 \). Calculate the discriminant: \[ b^2 - 4ac = (-7)^2 - 4(12)(-16) = 49 + 768 = 817. \] Then substituting into the formula gives: \[ x = \frac{7 \pm \sqrt{817}}{24}. \] We have two potential solutions: \[ x = \frac{7 + \sqrt{817}}{24}, \quad x = \frac{7 - \sqrt{817}}{24}. \] We should also verify that these solutions do not violate any restrictions from the original equation. The values that would make the denominator zero are: - \(x - 1 = 0 \Rightarrow x = 1\) - \(x + 3 = 0 \Rightarrow x = -3\) Neither of the founding solutions equal 1 or -3, so they are valid. Thus, the solutions are: \[ x = \frac{7 + \sqrt{817}}{24}, \frac{7 - \sqrt{817}}{24}. \] Final answer: \( x = \frac{7 + \sqrt{817}}{24}, \frac{7 - \sqrt{817}}{24} \)

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