Turnbull Bryant
11/21/2023 · Middle School

59. \( \frac{1 x}{2}+2=x+4 \) 61. \( x \cdot \frac{3 x}{4}=\frac{x}{7}+3 \) 63. \( \frac{9 x}{4} \cdot 6=\frac{2 x}{3}+\frac{1}{3} \) 65. \( \frac{3 x}{5} \cdot 7=\frac{2 x}{6}+1 \) 67. \( \frac{x}{3}+x=10+\frac{2 x}{9} \) 69. \( \frac{x}{5}+\frac{x}{2}=x \cdot 3 \)

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Claro, aquí tienes las soluciones a las ecuaciones que proporcionaste: 59. \( \frac{1 x}{2}+2=x+4 \) Multiplicamos todo por 2 para eliminar el denominador: \[ x + 4 = 2x + 8 \] Reorganizamos: \[ 4 - 8 = 2x - x \] \[ -4 = x \] 60. \( x \cdot \frac{3 x}{4}=\frac{x}{7}+3 \) Multiplicamos todo por 28 para eliminar los denominadores: \[ 21x^2 = 4x + 84 \] Reorganizamos: \[ 21x^2 - 4x - 84 = 0 \] Usamos la fórmula cuadrática: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 21 \cdot (-84)}}{2 \cdot 21} \] Calculamos el discriminante y resolvemos. 63. \( \frac{9 x}{4} \cdot 6=\frac{2 x}{3}+\frac{1}{3} \) Multiplicamos todo por 12 para eliminar los denominadores: \[ 9x \cdot 18 = 8x + 4 \] Reorganizamos: \[ 162x - 8x = 4 \] \[ 154x = 4 \implies x = \frac{4}{154} = \frac{2}{77} \] 65. \( \frac{3 x}{5} \cdot 7=\frac{2 x}{6}+1 \) Multiplicamos todo por 30 para eliminar los denominadores: \[ 42x = 10x + 30 \] Reorganizamos: \[ 42x - 10x = 30 \] \[ 32x = 30 \implies x = \frac{30}{32} = \frac{15}{16} \] 67. \( \frac{x}{3}+x=10+\frac{2 x}{9} \) Multiplicamos todo por 9 para eliminar los denominadores: \[ 3x + 9x = 90 + 2x \] Reorganizamos: \[ 12x - 2x = 90 \] \[ 10x = 90 \implies x = 9 \] 69. \( \frac{x}{5}+\frac{x}{2}=x \cdot 3 \) Multiplicamos todo por 10 para eliminar los denominadores: \[ 2x + 5x = 30x \] Reorganizamos: \[ 7x - 30x = 0 \] \[ -23x = 0 \implies x = 0 \] Si necesitas más ayuda o aclaraciones, ¡hazmelo saber!

Quick Answer

59. \( x = -4 \) 60. Solución usando la fórmula cuadrática. 63. \( x = \frac{2}{77} \) 65. \( x = \frac{15}{16} \) 67. \( x = 9 \) 69. \( x = 0 \)
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