Solve the system of equations: -3y + 10x - 0.1 = 0 5x + 4y = 2.7

Solution:

We have the system of equations: \[\begin{cases} -3y + 10x - 0.1 = 0 \\ 5x + 4y = 2.7 \end{cases}\] We can rewrite the first equation as: \[10x - 3y = 0.1\] \[5x + 4y = 2.7\] Multiply the first equation by 4 and the second by 3 to eliminate \(y\): \[\begin{cases} 4(10x - 3y) = 4(0.1) \\ 3(5x + 4y) = 3(2.7) \end{cases}\] \[\begin{cases} 40x - 12y = 0.4 \\ 15x + 12y = 8.1 \end{cases}\] Add the two equations: \[40x - 12y + 15x + 12y = 0.4 + 8.1\] \[55x = 8.5\] \[x = \frac{8.5}{55} = \frac{85}{550} = \frac{17}{110}\] Now substitute the value of \(x\) into the second equation to find \(y\): \[5(\frac{17}{110}) + 4y = 2.7\] \[\frac{85}{110} + 4y = 2.7\] \[4y = 2.7 - \frac{85}{110}\] \[4y = 2.7 - \frac{17}{22}\] \[4y = \frac{2.7 \cdot 22 - 17}{22} = \frac{59.4 - 17}{22} = \frac{42.4}{22} = \frac{424}{220} = \frac{106}{55}\] \[y = \frac{106}{55 \cdot 4} = \frac{106}{220} = \frac{53}{110}\] Therefore, the solution is: \[x = \frac{17}{110}\] \[y = \frac{53}{110}\] Let's verify the solution: \[-3(\frac{53}{110}) + 10(\frac{17}{110}) - 0.1 = \frac{-159 + 170}{110} - \frac{1}{10} = \frac{11}{110} - \frac{11}{110} = 0\] \[5(\frac{17}{110}) + 4(\frac{53}{110}) = \frac{85 + 212}{110} = \frac{297}{110} = \frac{27}{10} = 2.7\]

Answer: x = 17/110, y = 53/110