Question

Tìm x,y,z biết:

\(\dfrac{x}{3}=\dfrac{y}{5};3x=4z\) và \(x-y+2z=-20\)

Answer 1

Ta có \(x-y+2z=-20\)

\(\Leftrightarrow x-y+\dfrac{1}{2}\cdot4z=-20\Leftrightarrow x-y+\dfrac{1}{2}\cdot3x=-20\)

\(\Leftrightarrow\dfrac{5}{2}x-y=-20\Leftrightarrow5x-2y=-40\)

Theo đề bài: \(\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{5x}{15}=\dfrac{2y}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{5x}{15}=\dfrac{2y}{10}=\dfrac{5x-2y}{15-10}=-\dfrac{40}{5}=-8\)

\(\Rightarrow\dfrac{x}{3}=-8\Rightarrow x=-24\)

\(\Rightarrow\dfrac{y}{5}=-8\Rightarrow y=-40\)

\(\Rightarrow3\cdot\left(-24\right)=4z\Rightarrow z=-18\)