, 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)$.