Ta có:\(2x+y=z-38\Rightarrow2x+y-z=-38\)
Vì \(3x=4y=5x-3x-4y\) nên \(3x=5z-3x-3x\)
\(\Rightarrow3x-5z-6x\)
\(\Rightarrow9x=5z\)
\(\Rightarrow\frac{x}{5}=\frac{z}{9}\Rightarrow\frac{x}{20}=\frac{z}{36}\)(1)
Vì \(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}\Rightarrow\frac{x}{20}=\frac{z}{15}\) (2)
Từ (1) và (2)\(\Rightarrow\frac{x}{20}=\frac{y}{15}=\frac{z}{36}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{20}=\frac{y}{15}=\frac{z}{36}=\frac{2x+y-z}{2.20+15-36}=\frac{-38}{19}=-2\)
\(\frac{x}{20}=-2\Rightarrow x=20.\left(-2\right)=-40\)
\(\frac{y}{15}=-2\Rightarrow y=15.\left(-2\right)=-30\)
\(\frac{z}{36}=-2\Rightarrow z=36.\left(-2\right)=-72\)
Vậy \(x=-40;y=-30;z=-72\)
Nhớ tick cho mình nha!
Thay 4y=3x, ta được:
3x=5z-3x-3x
⇒9x=5z
Thay y=\(\frac{3}{4}\)x và z=\(\frac{9}{5}\)x
Ta được:
x(2+\(\frac{3}{4}\))=\(\frac{9}{5}\)x-38
\(\frac{11}{4}\)x=\(\frac{9}{5}\)x-38
\(\frac{19}{20}\)x=-38
x=\(\frac{\left(-38\right).20}{19}=20.\left(-2\right)=-40\)
\(y=\left(-40\right).\frac{3}{4}=-30\)
\(z=\frac{9}{5}x=\frac{9}{5}\left(-40\right)=-72\) Vậy,\(\left(x,y,z\right)=\left(-40,-30,-72\right)\)
