Graph the system below and write its solution. \[ \begin{array}{l}x-2 y=-3 \\ y=-x+3\end{array} \]

1. Solve for \( x \) and \( y \): - We are given: \[ \begin{cases} x-2y=-3 \\ y=-x+3 \end{cases} \] - Substitute \( y=-x+3 \) into \( x-2y=-3 \): \[ x - 2(-x+3)=-3 \] - Simplify: \[ x + 2x - 6 = -3 \quad \Rightarrow \quad 3x-6=-3 \] - Add 6 to both sides: \[ 3x = 3 \] - Divide by 3: \[ x = 1 \] - Substitute \( x=1 \) back into \( y=-x+3 \): \[ y = -1+3 = 2 \] - Thus, the solution is: \[ (x,y)=(1,2) \] 2. Graph the lines: - **First Line: \( x-2y=-3 \)** - Rewrite in slope-intercept form: \[ x-2y=-3 \quad \Rightarrow \quad -2y=-x-3 \quad \Rightarrow \quad y=\frac{1}{2}x+\frac{3}{2} \] - Identify two points: - \( y \)-intercept: Set \( x=0 \) gives: \[ y=\frac{3}{2} \quad \Rightarrow \quad (0, \frac{3}{2}) \] - \( x \)-intercept: Set \( y=0 \) gives: \[ x-2(0)=-3 \quad \Rightarrow \quad x=-3 \quad \Rightarrow \quad (-3,0) \] - **Second Line: \( y=-x+3 \)** - This line is already in slope-intercept form. - Identify two points: - \( y \)-intercept: Set \( x=0 \) gives: \[ y=3 \quad \Rightarrow \quad (0,3) \] - \( x \)-intercept: Set \( y=0 \) gives: \[ 0=-x+3 \quad \Rightarrow \quad x=3 \quad \Rightarrow \quad (3,0) \] 3. Plot the lines on the coordinate plane: - Plot the points for each line and draw the lines through these points. - The point where the lines intersect is at: \[ (1,2) \] - This intersection point is the solution to the system. Final answer: \[ (x,y)=(1,2) \]