No solution to a system of equations

When a system of two lines has no solution, it means the lines never intersect.
The only way two straight lines never meet is if they are parallel.
Parallel lines have the same slope but different $y$-intercepts.

The figure shows two parallel lines.
They rise at the same rate, so they stay the same distance apart and never cross.
With no crossing point, no pair $(x, y)$ works in both equations.

The algebra shows this as a contradiction.
Setting $\frac{3}{4}x + 15 = \frac{3}{4}x + 6$ cancels the $x$ terms and leaves $15 = 6$, which is false.
A false statement like that means the system has no solution.

You can also check the coefficients.
For $ax + by = c$ and $px + qy = r$, there is no solution when $\frac{a}{p} = \frac{b}{q} \ne \frac{c}{r}$.
The slopes match but the constants do not, so the lines are parallel.

How many solutions does the system $y = 2x + 1$ and $y = 2x + 5$ have?
Both lines have slope $2$ but different $y$-intercepts, so they are parallel.
The system has no solution.

Two lines have slopes $\frac{k}{3}$ and $\frac{2}{3}$ and different $y$-intercepts. What value of $k$ gives no solution?
For parallel lines the slopes must be equal: $\frac{k}{3} = \frac{2}{3}$, so $k = 2$.
With different intercepts, the lines are parallel and the system has no solution.

Solve the system $y = -2x + 3$ and $y = -2x + 7$.
Set them equal: $-2x + 3 = -2x + 7$, which gives $3 = 7$.
That is impossible, so the system has no solution.