CEMC Euclid, 2025

Since $4(x-2) = 2(x-4)$, then
$4x - 8 = 2x - 8$ which gives $2x = 0$ and so $x=0$.
Solution 1:

If $2x = 9$, then $6x = 3 \cdot 2x = 3 \cdot 9 = 27$.

This means that $2^{6x-23} = 2^{27 - 23} = 2^4 = 16$.

Solution 2:

If $2x = 9$, then $x = \frac{9}{2}$.

Therefore, $6x - 23 = 6 \cdot \tfrac{9}{2} - 23 = 27 - 23 = 4$.

Thus, $2^{6x - 23} = 2^4 = 16$.
Equating expressions for $y$,
we obtain $3x + 7 = 7x + 3$, which
gives $4 = 4x$ and so $x = 1$.

Since $x = 1$, then $y = 3x + 7 = 3 + 7 = 10$, and so the point
of intersection has coordinates $(1, 10)$.