Vì x,y tỉ lệ thuận nên \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)

a: \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)

nên \(\dfrac{x_1}{3}=\dfrac{-2}{\dfrac{3}{8}}=-2\cdot\dfrac{8}{3}=-\dfrac{16}{3}\)

=>\(x_1=-16\)

b: \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)

\(\Leftrightarrow\dfrac{x_2}{x_1}=\dfrac{y_2}{y_1}\)

\(\Leftrightarrow\dfrac{x_2}{-6}=\dfrac{y_2}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x_2}{-6}=\dfrac{y_2}{4}=\dfrac{y_2-x_2}{4-\left(-6\right)}=\dfrac{-5}{10}=-\dfrac{1}{2}\)

Do đó: \(x_2=3;y_2=-2\)

y : 1/8 +y : 1/2 = 140 + 1/4
y x 8 + y x 2 = 140 + 1/4
y x ( 8 + 2) = 140 + 1/4
y x 10 = 140 + 1/4 
Bạn tự giải nốt nhé 

a)y=5/6

b)y=11/4

\(a\left(\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{1}{3}\)

\(\frac{2}{5}.y=\frac{1}{3}\)

      \(y=\frac{1}{3}:\frac{2}{5}\)

     \(y=\frac{5}{6}\)

\(b,\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

     \(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

      \(\frac{10}{11}.y=\frac{2}{3}\)

              \(y=\frac{2}{3}:\frac{10}{11}\)

               \(y=\frac{22}{30}\)