Question

Answer 1

\(-\dfrac{2}{3}x+\dfrac{1}{5}x=-\dfrac{14}{15}\)

=>\(x\left(-\dfrac{2}{3}+\dfrac{1}{5}\right)=-\dfrac{14}{15}\)

=>\(x\cdot\dfrac{-7}{15}=-\dfrac{14}{15}\)

=>x=2

Answer 2

\(c,\dfrac{-2}{3}x+\dfrac{1}{5}x=\dfrac{-14}{15}\)
\(x\left(\dfrac{-2}{3}+\dfrac{1}{5}\right)=\dfrac{-14}{15}\)
\(x\left(\dfrac{-10}{15}+\dfrac{3}{15}\right)=\dfrac{-14}{15}\)
\(x.\dfrac{-7}{15}=\dfrac{-14}{15}\)
\(x=\dfrac{-14}{15}:\dfrac{-7}{15}\)
\(x=\dfrac{-14}{15}.\dfrac{-15}{7}\)
\(x=\dfrac{\left(-2\right).\left(-1\right)}{1.1}\)
\(x=2\)
Vậy \(x=2\)