Solving Systems of Equations by Graphing Solve each swstem by graphing. \( \begin{array}{ll}\text { 1) } y=-\frac{5}{3} x+3-\frac{6}{3} x+3=\frac{1}{3} x-3 & \text { 2) } y=4 x+3 \\ y=\frac{1}{3} x-3 & x=1\end{array} \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=4x+3\\x=1\end{array}\right.\) - step1: Substitute the value of \(x:\) \(y=4\times 1+3\) - step2: Simplify: \(y=7\) - step3: Calculate: \(\left\{ \begin{array}{l}x=1\\y=7\end{array}\right.\) - step4: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=7\end{array}\right.\) - step5: Rewrite: \(\left(x,y\right) = \left(1,7\right)\) Solve the system of equations \( y=-\frac{5}{3} x+3; y=-\frac{1}{3} x+3 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-\frac{5}{3}x+3\\y=-\frac{1}{3}x+3\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-\frac{5}{3}x+3=-\frac{1}{3}x+3\) - step2: Cancel equal terms: \(-\frac{5}{3}x=-\frac{1}{3}x\) - step3: Add or subtract both sides: \(-\frac{5}{3}x-\left(-\frac{1}{3}x\right)=0\) - step4: Subtract the terms: \(-\frac{4}{3}x=0\) - step5: Change the signs: \(\frac{4}{3}x=0\) - step6: Rewrite the expression: \(x=0\) - step7: Substitute the value of \(x:\) \(y=-\frac{5}{3}\times 0+3\) - step8: Substitute back: \(y=3\) - step9: Calculate: \(\left\{ \begin{array}{l}x=0\\y=3\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=3\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right) = \left(0,3\right)\) Let's solve the systems of equations step by step by graphing. ### System 1: The equations are: 1. \( y = -\frac{5}{3} x + 3 \) 2. \( y = -\frac{1}{3} x + 3 \) To find the intersection point, we can set the two equations equal to each other: \[ -\frac{5}{3} x + 3 = -\frac{1}{3} x + 3 \] By simplifying, we can see that both equations intersect at the point where \( x = 0 \). Substituting \( x = 0 \) into either equation gives: \[ y = -\frac{5}{3}(0) + 3 = 3 \] Thus, the solution for the first system is: \[ (x, y) = (0, 3) \] ### System 2: The equations are: 1. \( y = 4x + 3 \) 2. \( x = 1 \) Since the second equation is a vertical line at \( x = 1 \), we can substitute \( x = 1 \) into the first equation to find \( y \): \[ y = 4(1) + 3 = 4 + 3 = 7 \] Thus, the solution for the second system is: \[ (x, y) = (1, 7) \] ### Summary of Solutions: 1. For the first system: \( (x, y) = (0, 3) \) 2. For the second system: \( (x, y) = (1, 7) \)