The image shows a system of two linear equations:

1. $6x + 8y = 16$
2. $7x + 4y = 3$

I'll solve this system of equations using the elimination or substitution method. Let’s proceed with the elimination method here:

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

To eliminate $y$, we can multiply the first equation by 2 and the second equation by -4:

• First equation: $2 \times (6x + 8y) = 2 \times 16$ becomes $12x + 16y = 32$
• Second equation: $-4 \times (7x + 4y) = -4 \times 3$ becomes $-28x - 16y = -12$

Step 2: Add the two equations

Now we add the transformed equations:

$12x + 16y + (-28x - 16y) = 32 - 12$

This simplifies to:

$-16x = 20$

Then, divide by -16:

$x = -\frac{20}{16} = -\frac{5}{4}$

Step 3: Substitute $x = -\frac{5}{4}$ into one of the original equations

Using the first equation:

$6\left(-\frac{5}{4}\right) + 8y = 16$

This simplifies to:

$-\frac{30}{4} + 8y = 16$

$-7.5 + 8y = 16$

Add 7.5 to both sides:

$8y = 23.5$

Now, divide by 8:

$y = \frac{23.5}{8} = 2.9375$

Solution

The solution to the system is:

$x = -\frac{5}{4}, \quad y = 2.9375$

Let me know if you would like further details on any step!

Would you like more clarification on the steps, or do you have any questions? Here are some related questions for practice:

1. How would the solution change if the constants in both equations were different?
2. Can this system of equations be solved using the substitution method, and if so, how?
3. What is the graphical interpretation of this solution?
4. How can we verify if the solution is correct?
5. How would you solve the system if the coefficients of $x$ and $y$ were fractions?

Tip: Always check your solution by substituting the values back into the original equations. This helps ensure accuracy.