1. **Problem statement:** We have the system of equations: $$y = 4x + 3$$ $$y - ax - b = 0$$ which can be rewritten as: $$y = ax + b$$ where $a$ and $b$ are constants. We want to find how many solutions (points of intersection) this system has for different values of $a$ and $b$. 2. **Formula and rules:** Two lines intersect where their $y$ values are equal: $$4x + 3 = ax + b$$ Rearranging: $$4x - ax = b - 3$$ $$x(4 - a) = b - 3$$ 3. **Cases:** - If $4 - a \neq 0$, then $$x = \frac{b - 3}{4 - a}$$ This gives exactly one solution (one intersection point). - If $4 - a = 0$, i.e., $a = 4$, then the equation becomes: $$0 \cdot x = b - 3$$ which means: - If $b - 3 = 0$, i.e., $b = 3$, then the two lines are identical ($y = 4x + 3$), so there are infinitely many solutions. - If $b - 3 \neq 0$, i.e., $b \neq 3$, then there is no solution because the lines are parallel but distinct. 4. **Summary:** - For $a \neq 4$, there is exactly one solution. - For $a = 4$ and $b = 3$, there are infinitely many solutions. - For $a = 4$ and $b \neq 3$, there are no solutions. This completes the analysis of the number of solutions depending on $a$ and $b$.