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Solve for \( x \) : \[ \frac{3}{(x+2)(x-1)}=-3+\frac{1}{x-1} \] If there is more than one solution, separate them with commas. If there is no solution, click on "No solution". \( x=\frac{-1+\sqrt{34}}{3}, \frac{-1-\sqrt{34}}{3} \)

Ask by Bolton Estrada. in the United Arab Emirates
Jan 21,2025

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There is no solution.

Solución

Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3}{\left(x+2\right)\left(x-1\right)}=-3+\frac{1}{x-1}\) - step1: Find the domain: \(\frac{3}{\left(x+2\right)\left(x-1\right)}=-3+\frac{1}{x-1},x \in \left(-\infty,-2\right)\cup \left(-2,1\right)\cup \left(1,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\frac{3}{\left(x+2\right)\left(x-1\right)}\times \left(x-1\right)\left(x+2\right)=\left(-3+\frac{1}{x-1}\right)\left(x-1\right)\left(x+2\right)\) - step3: Simplify the equation: \(3=-3x^{2}-2x+8\) - step4: Swap the sides: \(-3x^{2}-2x+8=3\) - step5: Move the expression to the left side: \(-3x^{2}-2x+8-3=0\) - step6: Subtract the numbers: \(-3x^{2}-2x+5=0\) - step7: Factor the expression: \(\left(-x+1\right)\left(3x+5\right)=0\) - step8: Separate into possible cases: \(\begin{align}&-x+1=0\\&3x+5=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=1\\&x=-\frac{5}{3}\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=1\\&x=-\frac{5}{3}\end{align},x \in \left(-\infty,-2\right)\cup \left(-2,1\right)\cup \left(1,+\infty\right)\) - step11: Find the intersection: \(x=-\frac{5}{3}\) To solve the equation \[ \frac{3}{(x+2)(x-1)} = -3 + \frac{1}{x-1}, \] we can start by simplifying the right side. First, we can rewrite \(-3\) with a common denominator of \(x-1\): \[ -3 = \frac{-3(x-1)}{x-1} = \frac{-3x + 3}{x-1}. \] Now, substituting this back into the equation gives us: \[ \frac{3}{(x+2)(x-1)} = \frac{-3x + 3 + 1}{x-1} = \frac{-3x + 4}{x-1}. \] Next, we can cross-multiply to eliminate the fractions: \[ 3(x-1) = (x+2)(-3x + 4). \] Expanding both sides: \[ 3x - 3 = -3x^2 + 4x - 6x - 8. \] This simplifies to: \[ 3x - 3 = -3x^2 - 2x - 8. \] Rearranging all terms to one side gives us: \[ 3x + 3x^2 + 2x - 3 + 8 = 0, \] which simplifies to: \[ 3x^2 + 5x + 5 = 0. \] Now, we can use the quadratic formula to find the values of \(x\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3} = \frac{-5 \pm \sqrt{25 - 60}}{6} = \frac{-5 \pm \sqrt{-35}}{6}. \] Since the discriminant is negative, this indicates that there are no real solutions. Thus, the final answer is: **No solution.**

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The given equation is \[ \frac{3}{(x+2)(x-1)} = -3 + \frac{1}{x-1}. \] To solve for \( x \), first, we can simplify the right-hand side. Getting a common denominator gives us \[ -3 + \frac{1}{x-1} = \frac{-3(x-1) + 1}{x-1} = \frac{-3x + 3 + 1}{x-1} = \frac{-3x + 4}{x-1}. \] Now, we can rewrite the equation as: \[ \frac{3}{(x+2)(x-1)} = \frac{-3x + 4}{x-1}. \] Cross-multiplying yields: \[ 3(x-1) = (x+2)(-3x + 4). \] Expanding both sides gives: \[ 3x - 3 = (-3x^2 + 4x - 6x - 8), \] which simplifies to: \[ 3x - 3 = -3x^2 - 2x - 8. \] Rearranging everything to one side results in: \[ 3x^2 + 5x + 5 = 0. \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = 5, c = 5 \): \[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3} = \frac{-5 \pm \sqrt{25 - 60}}{6} = \frac{-5 \pm \sqrt{-35}}{6} = \frac{-5 \pm i\sqrt{35}}{6}. \] Thus, the solutions are: \[ x = \frac{-5 + i\sqrt{35}}{6}, \, \frac{-5 - i\sqrt{35}}{6}. \]

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