The problem is to solve the following system of linear equations:

$2x + 3y = 9$ $y - 4x = -8$

Step 1: Rearrange the equations if necessary

Let’s rewrite the system:

1. $2x + 3y = 9$
2. $y - 4x = -8$

Step 2: Solve for $y$ in terms of $x$ from the second equation

From the second equation, solve for $y$: $y = 4x - 8$

Step 3: Substitute $y$ in the first equation

Now substitute $y = 4x - 8$ into the first equation: $2x + 3(4x - 8) = 9$

Step 4: Expand and simplify

$2x + 12x - 24 = 9$ Combine like terms: $14x - 24 = 9$ Add 24 to both sides: $14x = 33$ Divide by 14: $x = \frac{33}{14}$

Step 5: Substitute $x$ back to find $y$

Substitute $x = \frac{33}{14}$ into $y = 4x - 8$: $y = 4 \left(\frac{33}{14}\right) - 8$ $y = \frac{132}{14} - 8$ Convert 8 to a fraction with a denominator of 14: $y = \frac{132}{14} - \frac{112}{14}$ $y = \frac{20}{14} = \frac{10}{7}$

Solution

The solution to the system is: $x = \frac{33}{14}, \quad y = \frac{10}{7}$

Would you like further details on any specific step, or do you have questions?

Here are 5 additional questions for practice:

1. Can you solve a similar system with different coefficients?
2. How would you solve this system if it were presented in matrix form?
3. What happens if a system has no solution or infinitely many solutions?
4. How would you check your solution to verify its accuracy?
5. What methods can you use to solve systems of equations other than substitution?

Tip: When solving systems of equations, always check your solution by substituting the values back into the original equations.