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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