Question

como resolver esta ecuacion
\frac{3x-1}{4}+1=\frac{x-2}{5}+\frac{x+1}{2}

Answer 1

$\frac{3x-1}{4}+1=\frac{x-2}{5}+\frac{x+1}{2} \\ \frac{4 + 3x - 1}{4} = \frac{2(x - 2) + 5(x + 1)}{5 \times 2} \\ \frac{3 + 3x}{4} = \frac{2x - 4 + 5x + 5}{10} \\ \frac{3 + 3x}{4} = \frac{7x + 1}{10}$

$10(3 + 3x) = 4(7x + 1) \\ 30 + 30x = 28x + 4 \\ 30x - 28x = 4 - 30 \\ 2x = - 26 \\ x = - 26 \div 2 \\ \boxed{ x = - 13}$

COMPROBAMOS
$\frac{3x-1}{4}+1=\frac{x-2}{5}+\frac{x+1}{2} \\ \frac{3( - 13)-1}{4}+1=\frac{ - 13-2}{5}+\frac{ - 13+1}{2} \\ \frac{ - 39 - 1}{4} + 1 = \frac{ - 15}{5} + \frac{ - 12}{2} \\ \frac{ - 40}{4} + 1 = - 3 - 6 \\ - 10 + 1 = - 9 \\ - 9 = - 9$