{2y-8x=16 {9x+4y=11

To solve the system of equations given by:

1. \( 2y - 8x = 16 \)
2. \( 9x + 4y = 11 \)

we can use either the substitution method or the elimination method. Here, we’ll use the elimination method.

First, let's rewrite the equations in standard form:

1. \( -8x + 2y = 16 \)
2. \( 9x + 4y = 11 \)

Next, we can multiply the entire first equation by 2 to align the \(y\) coefficients:

1. \( -16x + 4y = 32 \)
2. \( 9x + 4y = 11 \)

Now we can subtract the second equation from the first to eliminate \(y\):

\[ (-16x + 4y) - (9x + 4y) = 32 - 11 \]

This simplifies to:

\[ -16x - 9x + 4y - 4y = 21 \]

\[ -25x = 21 \]

Now, divide both sides by -25:

\[ x = -\frac{21}{25} \]

Now, substitute the value of \(x\) back into one of the original equations to find \(y\). We can substitute it into the first equation:

\[ 2y - 8(-\frac{21}{25}) = 16 \]

This simplifies to:

\[ 2y + \frac{168}{25} = 16 \]

Now, convert 16 into a fraction with the same denominator:

\[ 2y + \frac{168}{25} = \frac{400}{25} \]

Subtract \(\frac{168}{25}\) from both sides:

\[ 2y = \frac{400}{25} - \frac{168}{25} \]

\[ 2y = \frac{232}{25} \]

Now, divide both sides by 2:

\[ y = \frac{116}{25} \]

So the solution to the system of equations is:

\[ x = -\frac{21}{25}, \quad y = \frac{116}{25} \]

Thus, the solution is:

\[ \left(x, y\right) = \left(-\frac{21}{25}, \frac{116}{25}\right) \]