It looks like you have a system of equations, but there seems to be a formatting issue. Let's rewrite the equations clearly:

1. \(\frac{3x}{4} + \frac{2y}{5} = 2\)
2. \(X = \frac{3y}{5} = 0.8\)

The second equation seems to have an error because it equates two different expressions to 0.8. Let's assume it should be two separate equations:

2. \(X = \frac{3y}{5}\)
3. \(X = 0.8\)

Now we have three equations:

1. \(\frac{3x}{4} + \frac{2y}{5} = 2\)
2. \(X = \frac{3y}{5}\)
3. \(X = 0.8\)

Since \(X = 0.8\), we can substitute \(X\) in the second equation:

\[0.8 = \frac{3y}{5}\]

Solving for \(y\):

\[0.8 \times 5 = 3y\]
\[4 = 3y\]
\[y = \frac{4}{3}\]

Now we have \(y = \frac{4}{3}\). We can substitute \(y\) into the first equation to find \(x\):

\[\frac{3x}{4} + \frac{2 \times \frac{4}{3}}{5} = 2\]

Simplify the second term:

\[\frac{3x}{4} + \frac{8}{15} = 2\]

To solve for \(x\), first isolate \(\frac{3x}{4}\):

\[\frac{3x}{4} = 2 - \frac{8}{15}\]

Find a common denominator to subtract the fractions:

\[2 = \frac{30}{15}\]
\[\frac{3x}{4} = \frac{30}{15} - \frac{8}{15}\]
\[\frac{3x}{4} = \frac{22}{15}\]

Now solve for \(x\):

\[3x = \frac{22 \times 4}{15}\]
\[3x = \frac{88}{15}\]
\[x = \frac{88}{45}\]

Simplify the fraction:

\[x = \frac{88}{45} = \frac{8}{5}\]

So the solution to the system of equations is:

\[x = \frac{8}{5}\]
\[y = \frac{4}{3}\]