\( \left. \begin{array} { l } { 11 , y = - \frac { 5 } { 3 } x + 3 } \\ { y = \frac { 1 } { 3 } x - 3 } \end{array} \right. \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}-\frac{5}{3}x+3=y\\\frac{1}{3}x-3=y\end{array}\right.\) - step1: Rewrite the expression: \(\left\{ \begin{array}{l}y=-\frac{5}{3}x+3\\\frac{1}{3}x-3=y\end{array}\right.\) - step2: Substitute the value of \(y:\) \(\frac{1}{3}x-3=-\frac{5}{3}x+3\) - step3: Multiply both sides of the equation by LCM: \(\left(\frac{1}{3}x-3\right)\times 3=\left(-\frac{5}{3}x+3\right)\times 3\) - step4: Calculate: \(x-9=\left(-\frac{5}{3}x+3\right)\times 3\) - step5: Calculate: \(x-9=-5x+9\) - step6: Move the expression to the left side: \(x+5x=9+9\) - step7: Add and subtract: \(6x=9+9\) - step8: Add and subtract: \(6x=18\) - step9: Divide both sides: \(\frac{6x}{6}=\frac{18}{6}\) - step10: Divide the numbers: \(x=3\) - step11: Substitute the value of \(x:\) \(y=-\frac{5}{3}\times 3+3\) - step12: Calculate: \(y=-2\) - step13: Calculate: \(\left\{ \begin{array}{l}x=3\\y=-2\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=3\\y=-2\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(3,-2\right)\) To solve the system of equations given by: 1. \( y = -\frac{5}{3}x + 3 \) 2. \( y = \frac{1}{3}x - 3 \) we can set the two equations equal to each other since they both equal \( y \): \[ -\frac{5}{3}x + 3 = \frac{1}{3}x - 3 \] Now, let's solve for \( x \): 1. First, add \(\frac{5}{3}x\) to both sides: \[ 3 = \frac{1}{3}x + \frac{5}{3}x - 3 \] 2. Combine the \( x \) terms: \[ 3 = \left(\frac{1}{3} + \frac{5}{3}\right)x - 3 \] \[ 3 = \frac{6}{3}x - 3 \] 3. Add 3 to both sides: \[ 3 + 3 = 2x \] \[ 6 = 2x \] 4. Divide both sides by 2: \[ x = 3 \] Now that we have \( x \), we can substitute it back into either equation to find \( y \). Let's use the second equation: \[ y = \frac{1}{3}(3) - 3 \] \[ y = 1 - 3 \] \[ y = -2 \] Thus, the solution to the system of equations is: \[ (x, y) = (3, -2) \]