Question

Tìm x,y,z thỏa mãn

3x=4y=5z-3x-4y và 2x+y=z-38

Answer 1

Ta có : 3x=4y=>y=\(\frac{3x}{4}\) (1)

3x=5z-3x-4y,mà 4y=3x

\(\rightarrow\) 3x=5z-3x-3x

\(\rightarrow\) 3x+3x+3x=5z

\(\rightarrow\) 9x=5z

\(\rightarrow\) z=\(\frac{9x}{5}\) (2)

Thay(1),(2) vào 2x+y=z-38

=>2x+\(\frac{3x}{4}\) =\(\frac{9x}{5}\) -38

=>2x+\(\frac{3x}{4}\)-\(\frac{9x}{5}\)=-38

=>x(2+\(\frac{3}{4}\)-\(\frac{9}{5}\))=-38

=>x.\(\frac{19}{20}\)=-38

=>x=-38:\(\frac{19}{20}\)

=>x=-40

=>y=\(\frac{3\left(-40\right)}{4}\)=-30

=>z=\(\frac{9\left(-40\right)}{5}\)=72

Vậy (x,y,z)=(-40,-30,72)

Chúc bạn học tốt nha !! :v