\( 6 x-y-4=0 \) and \( 9 x+1=-2 y \)

Solve the system of equations \( 6x-y-4=0;9x+1=-2y \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}6x-y-4=0\\9x+1=-2y\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=6x-4\\9x+1=-2y\end{array}\right.\) - step2: Substitute the value of \(y:\) \(9x+1=-2\left(6x-4\right)\) - step3: Expand the expression: \(9x+1=-12x+8\) - step4: Move the expression to the left side: \(9x+12x=8-1\) - step5: Add and subtract: \(21x=8-1\) - step6: Add and subtract: \(21x=7\) - step7: Divide both sides: \(\frac{21x}{21}=\frac{7}{21}\) - step8: Divide the numbers: \(x=\frac{1}{3}\) - step9: Substitute the value of \(x:\) \(y=6\times \frac{1}{3}-4\) - step10: Substitute back: \(y=6\times 3^{-1}-4\) - step11: Calculate: \(y=-2\) - step12: Calculate: \(\left\{ \begin{array}{l}x=\frac{1}{3}\\y=-2\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=\frac{1}{3}\\y=-2\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(\frac{1}{3},-2\right)\) The solution to the system of equations \(6x-y-4=0\) and \(9x+1=-2y\) is \(x=\frac{1}{3}\) and \(y=-2\).