Question

\( 3x-\left[\begin{array}{cc}{\frac{1}{2}}&{-\left[\begin{array}{c}{x-\frac{1}{3}(1-\frac{x-2}{2})}\end{array}\right]}\\{}&{}\end{array}\right]-\frac{x-1}{3}=1-\frac{x}{4} \)

Ask by Reid Stuart. in Equatorial Guinea
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

La solución de la ecuación es \( x = \frac{22}{49} \).

Solution

Claro, resolvamos la ecuación paso a paso en español. **Ecuación dada:** \[ 3x - \left[\begin{array}{cc}\frac{1}{2} & -\left[\begin{array}{c}x - \frac{1}{3}\left(1 - \frac{x-2}{2}\right)\end{array}\right] \\ {} & {}\end{array}\right] - \frac{x-1}{3} = 1 - \frac{x}{4} \] Para simplificar, interpretaremos las matrices como agrupaciones de términos. La ecuación se puede reescribir de la siguiente manera: \[ 3x - \left( \frac{1}{2} - \left( x - \frac{1}{3}\left(1 - \frac{x-2}{2}\right) \right) \right) - \frac{x-1}{3} = 1 - \frac{x}{4} \] **Paso 1: Simplificar dentro de los paréntesis internos** Primero, simplificamos la expresión dentro del paréntesis más interno: \[ 1 - \frac{x-2}{2} = \frac{2 - (x - 2)}{2} = \frac{4 - x}{2} \] Luego, multiplicamos por \(\frac{1}{3}\): \[ \frac{1}{3} \left( \frac{4 - x}{2} \right) = \frac{4 - x}{6} \] Ahora, sustituimos de nuevo en la expresión: \[ x - \frac{4 - x}{6} = \frac{6x - (4 - x)}{6} = \frac{7x - 4}{6} \] **Paso 2: Sustituir y simplificar la expresión principal** Reemplazamos la expresión simplificada en la ecuación original: \[ 3x - \left( \frac{1}{2} - \frac{7x - 4}{6} \right) - \frac{x-1}{3} = 1 - \frac{x}{4} \] Simplificamos dentro del paréntesis: \[ \frac{1}{2} - \frac{7x - 4}{6} = \frac{3}{6} - \frac{7x - 4}{6} = \frac{3 - 7x + 4}{6} = \frac{7 - 7x}{6} = \frac{7(1 - x)}{6} \] Entonces, la ecuación queda: \[ 3x - \frac{7(1 - x)}{6} - \frac{x - 1}{3} = 1 - \frac{x}{4} \] **Paso 3: Eliminar denominadores multiplicando por 12** Multiplicamos cada término por 12 para eliminar los denominadores: \[ 12 \cdot 3x - 12 \cdot \frac{7(1 - x)}{6} - 12 \cdot \frac{x - 1}{3} = 12 \cdot \left(1 - \frac{x}{4}\right) \] Simplificando: \[ 36x - 14(1 - x) - 4(x - 1) = 12 - 3x \] **Paso 4: Expandir y combinar términos semejantes** Expandimos los términos: \[ 36x - 14 + 14x - 4x + 4 = 12 - 3x \] Combinamos términos semejantes: \[ (36x + 14x - 4x) + (-14 + 4) = 12 - 3x \] \[ 46x - 10 = 12 - 3x \] **Paso 5: Resolver para \(x\)** Sumamos \(3x\) a ambos lados y sumamos \(10\) a ambos lados: \[ 46x + 3x = 12 + 10 \] \[ 49x = 22 \] Finalmente, dividimos por 49: \[ x = \frac{22}{49} \] **Respuesta:** \[ x = \frac{22}{49} \]

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The Deep Dive

To solve the equation \( 3x - \left[\begin{array}{cc}{\frac{1}{2}}&{-\left[\begin{array}{c}{x-\frac{1}{3}(1-\frac{x-2}{2})}\end{array}\right]}\\{}&{}\end{array}\right] - \frac{x-1}{3} = 1 - \frac{x}{4} \), let's break this down step by step. 1. Start with simplifying the right side of the equation: \[ 1 - \frac{x}{4} = \frac{4 - x}{4} \] 2. Now, isolate the complex term on the left-hand side. We shift \( 3x \) and \(-\frac{x-1}{3}\) to the right: \[ \left[\begin{array}{cc}{\frac{1}{2}}&{-\left[\begin{array}{c}{x-\frac{1}{3}(1-\frac{x-2}{2})}\end{array}\right]}\\{}&{}\end{array}\right] = 3x - \frac{x-1}{3} - \frac{4-x}{4} \] Since the original expression has arrays, we need to understand them better if they signify anything mathematically. Here, it seems like the matrices or arrays might not be necessary if we are just focusing on variable isolation. The key is identifying and simplifying the variables step by step to solve for \( x \), but I would recommend checking to confirm any assumptions regarding notation. If it's matrices, isolation becomes trickier! Continuing from simplifying right-hand side and leaving arrays aside: 3. Start working out \( 3x - \frac{x-1}{3} \): By common denominator, the left side becomes \( \frac{9x - (x - 1)}{3} = \frac{9x - x + 1}{3} = \frac{8x + 1}{3} \). Thus, \[ \frac{8x + 1}{3} - \frac{4 - x}{4} = 0 \] 4. Now, simplifying gives us our way to standard forms; solve for \( x \) using equations! Continue to seek a numerical solution and clarify your interpretation through matrix representation if they represent equations or systems independently. Let me know if you need help solving further or clarifying the matrix terms!

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