1. **Problem statement:** Find the equation of the line parallel to the line $x - 3y = 4$ that passes through the point $P'(7, 10)$. 2. **Formula and rules:** The general form of a line is $Ax + By = C$. Lines are parallel if their coefficients $A$ and $B$ are the same. 3. **Given line:** $$x - 3y = 4$$ Here, $A=1$, $B=-3$. 4. **Equation of the parallel line:** It must have the form: $$1 \cdot x - 3 \cdot y = C$$ 5. **Find $C$ using point $P'(7, 10)$:** Substitute $x=7$, $y=10$: $$7 - 3 \times 10 = C$$ $$7 - 30 = C$$ $$C = -23$$ 6. **Final equation:** $$x - 3y = -23$$ This is the equation of the line parallel to $x - 3y = 4$ passing through $P'(7, 10)$.