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