ĐK: x#0
Ta có:
(1) 1+2y/18 = 1+4y/24
=> 24 + 48y = 18 + 72y
<=> y=1/4
(2) 1+4y/24=1+6y/6x
Thay y=1/4 vào (2) ta tìm đc x=5 (TMĐK)
Theo đề bài, ta có;
\(\frac{1+2y}{18}=\frac{1+4y}{24}\)
\(\Rightarrow24\times\left(1+2y\right)=18\times\left(1+4y\right)\)
\(\Rightarrow24+48y=18+72y\)
\(\Rightarrow48y-72y=18-24\)
\(\Rightarrow-24y=-6\)
\(\Rightarrow y=\frac{-6}{-24}\)
\(\Rightarrow y=\frac{1}{4}\)
Với \(y=\frac{1}{4}\), ta có:
\(\frac{1+4y}{24}=\frac{1+6y}{6x}\) \(\left(x\ne0\right)\)
\(\Rightarrow\frac{1+4\times\frac{1}{4}}{24}=\frac{1+6\times\frac{1}{4}}{6x}\)
\(\Rightarrow\frac{1+1}{24}=\frac{1+\frac{3}{2}}{6x}\)
\(\Rightarrow\frac{2}{24}=\frac{\frac{5}{2}}{6x}\)
\(\Rightarrow\frac{1}{12}=\frac{\frac{5}{2}}{6x}\)
\(\Rightarrow6x=\frac{5}{2}\div\frac{1}{12}\)
\(\Rightarrow6x=\frac{5}{2}\times12\)
\(\Rightarrow6x=30\)
\(\Rightarrow x=30\div6\)
\(\Rightarrow x=5\)
Vậy \(x=5\) và \(y=\frac{1}{4}\) thì thỏa mãn đề bài.
