What is the equation of the line parallel to the line $3x-4y+7=0$ and passing through the point (2, -3)?

Step 1 ：Step 1: Rewrite the equation $3x-4y+7=0$ in the slope-intercept form y = mx + b, where m represents the slope and b represents the y-intercept. This yields $y=\frac{3}{4}x+\frac{7}{4}$. The slope of this line is $\frac{3}{4}$.

Step 2 ：Step 2: Since parallel lines have equal slopes, the slope of the line we are looking for is also $\frac{3}{4}$.

Step 3 ：Step 3: Insert the coordinates of the given point (2, -3) into the equation y = mx + b, replacing y with -3, m with $\frac{3}{4}$, and x with 2. This yields $-3=\frac{3}{4}\ast 2+b$. Solving for b gives $b=-3-\frac{3}{4}\ast 2=-\frac{9}{2}$.

Step 4 ：Step 4: Substitute m = $\frac{3}{4}$ and b = $-\frac{9}{2}$ into the equation y = mx + b to get the equation of the line we are looking for, which is $y=\frac{3}{4}x-\frac{9}{2}$.