1. **State the problem:** We are given a triangle $\triangle A'B'C'$ with vertices $A'(-1, 2)$, $B'(-1, -1)$, and $C'(-4, -1)$ which is the image of $\triangle ABC$ after a translation by the rule $(x, y) \to (x + 1.5, y - 3.5)$. 2. **Goal:** Find the coordinates of the pre-image $\triangle ABC$. 3. **Translation rule:** The image coordinates are obtained by adding 1.5 to the $x$-coordinate and subtracting 3.5 from the $y$-coordinate: $$ (x', y') = (x + 1.5, y - 3.5) $$ To find the pre-image coordinates $(x, y)$, we reverse this translation: $$ (x, y) = (x' - 1.5, y' + 3.5) $$ 4. **Calculate each vertex of $\triangle ABC$:** - For $A'(-1, 2)$: $$ x_A = -1 - 1.5 = -2.5 $$ $$ y_A = 2 + 3.5 = 5.5 $$ - For $B'(-1, -1)$: $$ x_B = -1 - 1.5 = -2.5 $$ $$ y_B = -1 + 3.5 = 2.5 $$ - For $C'(-4, -1)$: $$ x_C = -4 - 1.5 = -5.5 $$ $$ y_C = -1 + 3.5 = 2.5 $$ 5. **Final answer:** The pre-image $\triangle ABC$ has vertices: $$ A(-2.5, 5.5), B(-2.5, 2.5), C(-5.5, 2.5) $$ This triangle is 1.5 units left and 3.5 units up from the image $\triangle A'B'C'$.