Let's go through the questions step by step. The first question (#15) involves solving a system of linear equations. Let’s start by addressing this one.

Problem 15:

The system of linear equations:

$x + 3y = 6 \quad \text{(Equation 1)}$ $2x - 3y = 12 \quad \text{(Equation 2)}$

We need to solve this system of equations, and we can do this by either substitution or elimination. Let’s use elimination.

1. Add Equation 1 and Equation 2: $(x + 3y) + (2x - 3y) = 6 + 12$ This simplifies to: $3x = 18 \implies x = 6$

2. Substitute $x = 6$ into Equation 1: $6 + 3y = 6 \implies 3y = 0 \implies y = 0$

Thus, the solution is $x = 6$ and $y = 0$.

Next Parts:

• (a) The point of intersection is the solution of the system, which is $(6, 0)$.

• (b) The number of solutions of the equation $x + 24 = 8$ is straightforward: $x + 24 = 8 \implies x = 8 - 24 = -16$ Hence, the solution is $x = -16$, and since it's a linear equation, there is one solution.

  For $2x + 4y = 16$, the number of solutions depends on whether the system is consistent. This is also a line, so if paired with another independent equation, there is exactly one solution unless they are parallel, in which case no solution, or coincident, in which case infinitely many solutions exist.

• (c) If $(1, 2)$ is the solution to the system $ax + by = 3$ and $2x + by = 12$, we substitute $x = 1$ and $y = 2$ into the equations to find $a$ and $b$.

  1. From $a(1) + b(2) = 3$, we get: $a + 2b = 3 \quad \text{(Equation 1)}$
  2. From $2(1) + b(2) = 12$, we get: $2 + 2b = 12 \implies 2b = 10 \implies b = 5$
  3. Substitute $b = 5$ into Equation 1: $a + 2(5) = 3 \implies a + 10 = 3 \implies a = -7$ So, $a = -7$ and $b = 5$.

Would you like to continue solving the other questions, or do you need more details for this solution?