Question

(ECUACIONES DE PRIMER GRADO Y FUNCIÓN LINEAL)

Resuelva los ejercicios propuestos.

Solo necesito el e, g, h, i, j, k, l

Ayuda porfavor, es para mañana .​​

Answer 1

Respuesta:

Usa Simbolab :v

Explicación paso a paso:

es una calculadora online

Answer 2

Respuesta:

g) $x=\frac{11}{10}$

h)  $x=-\frac{29}{60}$

i)  $x=\frac{11}{5}$

j)  $x=\frac{79}{33}$

k)  $x=-11$

l) $x=\frac{17}{7}$

Explicación paso a paso:

g) Procedimiento:

$\frac{2}{3}x-\frac{1}{2}=\frac{4}{3}-x\\\\\frac{2}{3}x-\frac{1}{2}+\frac{1}{2}=\frac{4}{3}-x+\frac{1}{2}\\\\\frac{2}{3}x=-x+\frac{11}{6}\\\\\frac{2}{3}x+x=-x+\frac{11}{6}+x\\\\\frac{5}{3}x=\frac{11}{6}\\\\3\cdot \frac{5}{3}x=\frac{11\cdot \:3}{6}\\\\5x=\frac{11}{2}\\\\\frac{5x}{5}=\frac{\frac{11}{2}}{5}\\\\x=\frac{11}{10}$

h)Procedimiento:

$\frac{2}{5}-\frac{7}{2}x+\frac{1}{2}\left(x+\frac{1}{3}\right)=\frac{5}{2}+x\\\\-3x+\frac{2}{5}+\frac{1}{6}=\frac{5}{2}+x\\\\-3x+\frac{2}{5}+\frac{1}{6}-\left(\frac{2}{5}+\frac{1}{6}\right)=\frac{5}{2}+x-\left(\frac{2}{5}+\frac{1}{6}\right)\\\\-3x=x+\frac{29}{15}\\\\-3x-x=x+\frac{29}{15}-x\\\\-4x=\frac{29}{15}\\\\\frac{-4x}{-4}=\frac{\frac{29}{15}}{-4}\\\\x=-\frac{29}{60}$

i)Procedimiento:

$4x-\frac{1}{3}\left\{15x-3\left(-1+2x\right)\right\}=\frac{3}{2}\left(9-x-6\right)\\\\4x-\frac{15x-3\left(-1+2x\right)}{3}=\frac{3}{2}\left(9-x-6\right)\\\\4x\cdot \:6-\frac{15x-3\left(-1+2x\right)}{3}\cdot \:6=\frac{3}{2}\left(9-x-6\right)\cdot \:6\\\\24x-6\left(3x+1\right)=9\left(-x+3\right)\\\\6x-6=9\left(-x+3\right)\\\\6x-6=-9x+27\\\\6x-6+6=-9x+27+6\\\\6x=-9x+33\\\\6x+9x=-9x+33+9x\\\\15x=33\\\\\frac{15x}{15}=\frac{33}{15}\\\\x=\frac{11}{5}$

j) Procedimiento: $\frac{1}{2}+\frac{1}{3}+\frac{1}{2}x-\frac{12}{4}x=-2\left\{-1+x-3\left(2x+6-x-2\left(3+x-1\right)\right)\right\}\\\\\frac{1}{2}+\frac{1}{3}+\frac{1}{2}x-3x=-2\left\{-1+x-3\left(2x+6-x-2\left(3+x-1\right)\right)\right\}\\\\-3x+\frac{5}{6}+\frac{1}{2}x=-2\left\{-1+x-3\left(2x+6-x-2\left(3+x-1\right)\right)\right\}\\\\\frac{1}{2}x-3x+\frac{5}{6}=-2\left\{-1+x-3\left(2x+6-x-2\left(3+x-1\right)\right)\right\}\\\\-\frac{5}{2}x+\frac{5}{6}=-8x+14\\\\-\frac{5}{2}x+8x=-8x+\frac{79}{6}+8x$

$\frac{11}{2}x=\frac{79}{6}\\\\2\cdot \frac{11}{2}x=\frac{79\cdot \:2}{6}\\\\11x=\frac{79}{3}\\\\\frac{11x}{11}=\frac{\frac{79}{3}}{11}\\\\x=\frac{79}{33}$

k) Procedimiento;

$x-3\cdot \frac{2x+1}{2}=-3\left(-\frac{3x+9}{3}-2+x\right)-\frac{x}{2}\\\\x-\frac{3\left(2x+1\right)}{2}=-3\left(-\frac{3x+9}{3}-2+x\right)-\frac{x}{2}\\\\x\cdot \:2-\frac{3\left(2x+1\right)}{2}\cdot \:2=-3\left(-\frac{3x+9}{3}-2+x\right)\cdot \:2-\frac{x}{2}\cdot \:2\\\\x\cdot \:2-3\left(2x+1\right)=-6\left(-\left(x+3\right)-2+x\right)-x\\\\-4x-3=30-x\\\\-4x-3+3=30-x+3\\\\-4x=-x+33\\\\-4x+x=-x+33+x\\\\-3x=33\\\\\frac{-3x}{-3}=\frac{33}{-3}\\\\x=-11$

l) Procedimiento:

$\frac{2x+5-x-1}{2}-\frac{9-x-5+3x-2}{3}=\frac{x-3\left(2x-4\right)+2}{2}\\\\\frac{x+4}{2}-\frac{3x-x+9-5-2}{3}=\frac{x-3\left(2x-4\right)+2}{2}\\\\\frac{x+4}{2}-\frac{2x+2}{3}=\frac{x-3\left(2x-4\right)+2}{2}\\\\\frac{x+4}{2}\cdot \:6-\frac{2x+2}{3}\cdot \:6=\frac{x-3\left(2x-4\right)+2}{2}\cdot \:6\\\\3\left(x+4\right)-2\left(2x+2\right)=3\left(x-3\left(2x-4\right)+2\right)\\\\-x+8=-15x+42\\\\-x+8-8=-15x+42-8\\\\-x=-15x+34\\\\-x+15x=-15x+34+15x\\\\14x=34\\\\\frac{14x}{14}=\frac{34}{14}$

$x=\frac{17}{7}$