To solve the system of equations by elimination, let's start by writing down the equations:

1. $3x - 4y = 11$
2. $2x + 3y = 13$

Our goal is to eliminate one of the variables by making the coefficients of either $x$ or $y$ the same in both equations.

Step 1: Make the coefficients of $x$ the same

To eliminate $x$, we can multiply the first equation by 2 and the second equation by 3, so that the coefficients of $x$ in both equations will be 6:

$2 \times (3x - 4y) = 2 \times 11 \Rightarrow 6x - 8y = 22$ $3 \times (2x + 3y) = 3 \times 13 \Rightarrow 6x + 9y = 39$

Now we have the following system:

1. $6x - 8y = 22$
2. $6x + 9y = 39$

Step 2: Eliminate $x$ by subtracting the equations

Subtract the first equation from the second to eliminate $x$:

$(6x + 9y) - (6x - 8y) = 39 - 22$ $6x + 9y - 6x + 8y = 17$ $17y = 17$

Now, divide by 17 to solve for $y$:

$y = 1$

Step 3: Substitute $y = 1$ back into one of the original equations

Let's substitute $y = 1$ into the first equation:

$3x - 4(1) = 11$ $3x - 4 = 11$ $3x = 15$ $x = 5$

Solution

The solution to the system is: $x = 5, \quad y = 1$

If you’d like further clarification or have any questions, please ask!

Here are some related questions to deepen your understanding:

1. How does the elimination method differ from the substitution method?
2. What happens if there is no unique solution in a system of linear equations?
3. Can you explain how to choose which variable to eliminate first?
4. How would you solve this system if you used substitution instead of elimination?
5. What is the geometric interpretation of solving a system of linear equations?

Tip: In elimination, always check if a multiplication is necessary to equalize coefficients, as sometimes coefficients might already be set up to cancel easily.