`-3x=2y `
`=> x/2 = -y/3 `
AD t/c của dãy tỉ số bằng nhau ta có
`x/2 =-y/3 = (x-y)/(2+3) = 6/5`
`=>{(x=2*6/5 = 12/5),(y=-3*6/5 =-18/5):}`
a) `6/x =-3/2`
`=>x =6 :(-3/2) = 6*(-2/3)=-4`
`b)`\(-3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)
Áp dụng t/c của DTSBN , ta đc :
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{x-y}{2+3}=\dfrac{6}{5}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{6}{5}\\\dfrac{y}{-3}=\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\y=-\dfrac{18}{5}\end{matrix}\right. \)
`a)`
`6/x=-3/2`
`x=6:(-3/2)`
`x=6*(-2/3)`
`x=-4`
a, \(\dfrac{6}{x}=-\dfrac{3}{2}\)
\(\Rightarrow3x=6.\left(-2\right)\)
\(\Rightarrow3x=-12\)
\(\Rightarrow x=-\dfrac{12}{3}\)
=>x=-4
b, \(+\left(-3\right)x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)
+x-y=6
+Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{x-y}{2-\left(-3\right)}=\dfrac{6}{5}\)
Suy ra \(\dfrac{x}{2}=\dfrac{6}{5}\Rightarrow x=\dfrac{6}{5}.2\Rightarrow x=\dfrac{12}{5}\)
\(\dfrac{y}{-3}=\dfrac{6}{5}\Rightarrow y=\dfrac{6}{5}.\left(-3\right)=-\dfrac{18}{5}\)
Vậy \(x=\dfrac{12}{5};y=-\dfrac{18}{5}\)