The image you provided shows a system of linear equations that we can solve. The system is:

1. $2x - 4y = 16$
2. $7x + 3y = 22$

To solve this, we can use either the substitution method or the elimination method. Let's proceed with the elimination method:

Step 1: Eliminate one variable

We need to manipulate these equations so that we can eliminate one of the variables. To do that, let's multiply the first equation by 3 and the second equation by 4. This will help us cancel out the $y$-terms when we add the equations.

$3(2x - 4y) = 3(16)$ $4(7x + 3y) = 4(22)$

This gives:

$6x - 12y = 48$ $28x + 12y = 88$

Step 2: Add the equations

Now, add the two equations together:

$(6x - 12y) + (28x + 12y) = 48 + 88$

This simplifies to:

$34x = 136$

Step 3: Solve for $x$

To find $x$, divide both sides of the equation by 34:

$x = \frac{136}{34} = 4$

Step 4: Substitute $x = 4$ into one of the original equations

Now that we know $x = 4$, we substitute this value into one of the original equations to find $y$. Let's use the first equation:

$2(4) - 4y = 16$ $8 - 4y = 16$ $-4y = 16 - 8$ $-4y = 8$ $y = \frac{8}{-4} = -2$

Final Solution:

The solution to the system of equations is: $x = 4, \quad y = -2$

Let me know if you would like any further clarification or details on how this solution was derived!

Here are some related questions you might find interesting:

1. How can we verify this solution by substituting the values of $x$ and $y$ into both original equations?
2. What are the advantages of using the elimination method compared to the substitution method?
3. Can you solve the same system using substitution instead of elimination?
4. How would the solution change if the coefficients of the variables were different, for example, $2x - 4y = 20$ and $7x + 3y = 25$?
5. What happens if the system of equations has no solution or an infinite number of solutions?

Tip: In elimination, it's often useful to manipulate the equations so that the coefficients of one variable match in both equations. Then, adding or subtracting the equations will eliminate that variable, making it easier to solve for the other.