\(2x+3y-14=186\)
\(\Rightarrow2x+3y=186+14=200\)
Từ \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{18}\) suy ra \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{18}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{18}=\dfrac{2x+3y}{30+60}=\dfrac{20}{9}\)
\(\Rightarrow x=15\cdot\dfrac{20}{9}=\dfrac{100}{3}\)
\(\Rightarrow y=20\cdot\dfrac{20}{9}=\dfrac{400}{9}\)
\(z=18\cdot\dfrac{20}{9}=40\)
2x + 3y - 14 = 186 => 2x + 3y = 186 + 14 = 200
\(\dfrac{x}{15}\) = \(\dfrac{y}{20}\) = \(\dfrac{z}{18}\) ⇒ \(\dfrac{2x}{30}\) = \(\dfrac{3y}{60}\) = \(\dfrac{2x+3y}{30+60}\) = \(\dfrac{200}{90}\) = \(\dfrac{20}{9}\)
=> x = \(\dfrac{20}{9}\) x 30 : 2 = \(\dfrac{100}{3}\); y = \(\dfrac{20}{9}\) x 60 : 3 = \(\dfrac{400}{9}\)
z = \(\dfrac{100}{3}\) : 15 x 18 = 40
Vậy (x, y, z) =( \(\dfrac{100}{3}\); \(\dfrac{400}{9}\); 40)
