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

To graph the system, first rearrange each equation into slope-intercept form (y = mx + b). For the first equation \( x - 2y = -3 \): 1. Move \( x \) to the right: \( -2y = -x - 3 \) 2. Divide by -2: \( y = \frac{1}{2}x + \frac{3}{2} \) For the second equation \( y = -x + 3 \), it's already in slope-intercept form. Now, let's plot the lines: 1. For \( y = \frac{1}{2}x + \frac{3}{2} \): - When \( x = 0 \), \( y = \frac{3}{2} = 1.5 \) (point (0, 1.5)). - When \( x = 2 \), \( y = \frac{1}{2}(2) + \frac{3}{2} = 2.5 \) (point (2, 2.5)). 2. For \( y = -x + 3 \): - When \( x = 0 \), \( y = 3 \) (point (0, 3)). - When \( x = 3 \), \( y = 0 \) (point (3, 0)). Next, draw the lines through these points. The intersection of these two lines is the solution to the system. To find the intersection algebraically, set the equations equal: \[ \frac{1}{2}x + \frac{3}{2} = -x + 3 \] Multiply through by 2 to eliminate the fraction: \[ x + 3 = -2x + 6 \] Combine like terms: \[ 3x = 3 \] Thus, \( x = 1 \). Substituting \( x = 1 \) back into one of the original equations to find \( y \): \[ y = -1 + 3 = 2 \] The solution to the system is \( (1, 2) \).