To solve the system of linear equations using the elimination method, we will:

1. Eliminate one variable by multiplying the equations so that their coefficients align for one of the variables.
2. Subtract or add the equations to eliminate that variable.
3. Solve for the remaining variable.
4. Substitute the value of the found variable into one of the original equations to find the second variable.

The system of equations is: $2x + 3y = 16 \quad (1)$ $3x + 2y = 15 \quad (2)$

Step 1: Make the coefficients of $x$ or $y$ equal

Let's eliminate $x$. To do this, we'll multiply the first equation by 3 and the second equation by 2 to align the coefficients of $x$.

Multiply (1) by 3: $6x + 9y = 48 \quad (3)$

Multiply (2) by 2: $6x + 4y = 30 \quad (4)$

Step 2: Subtract the equations to eliminate $x$

Now subtract equation (4) from equation (3): $(6x + 9y) - (6x + 4y) = 48 - 30$ This simplifies to: $5y = 18$ Solve for $y$: $y = \frac{18}{5} = 3.6$

Step 3: Substitute the value of $y$ into one of the original equations

Substitute $y = 3.6$ into equation (1): $2x + 3(3.6) = 16$ $2x + 10.8 = 16$ Solve for $x$: $2x = 16 - 10.8$ $2x = 5.2$ $x = \frac{5.2}{2} = 2.6$

Final Solution:

$x = 2.6, \quad y = 3.6$

Do you want more details, or do you have any questions?

Here are 5 related questions:

1. How would you solve the system by substitution instead of elimination?
2. Can you apply elimination to non-linear systems of equations?
3. What happens if the system has no solutions or infinitely many solutions?
4. How do you interpret the geometric meaning of the solution for these linear equations?
5. How can you solve a system of three variables using elimination?

Tip: Always check your solution by plugging the values back into the original equations to ensure accuracy.