Давай решим оба уравнения по очереди.
\( \frac{6x + 1}{5} - \frac{3x - 8}{4} = 1 \)\[ 20 \left( \frac{6x + 1}{5} - \frac{3x - 8}{4} \right) = 20 \times 1 \]
\[ 4(6x + 1) - 5(3x - 8) = 20 \]
\[ 24x + 4 - 15x + 40 = 20 \]
\[ (24x - 15x) + (4 + 40) = 20 \]
\[ 9x + 44 = 20 \]
\[ 9x = 20 - 44 \]
\[ 9x = -24 \]
\[ x = \frac{-24}{9} \]
\[ x = -\frac{8}{3} \]
4x² + 5x = 0x за скобки:\[ x(4x + 5) = 0 \]
Значит, у нас два случая:
Случай 1: x = 0
Случай 2: 4x + 5 = 0
\[ 4x = -5 \]
\[ x = -\frac{5}{4} \]
Ответ:
x = -\frac{8}{3}x₁ = 0, x₂ = -\frac{5}{4}