We are solving the system of equations:

8x + 3y = -7 \tag{1} 7x + 2y = -3 \tag{2}

Using Substitution:

1. Solve one equation for one variable.
   Let's solve equation (2) for $y$ in terms of $x$: $7x + 2y = -3$ Subtract $7x$ from both sides: $2y = -3 - 7x$ Divide by 2: y = \frac{-3 - 7x}{2} \tag{3}

2. Substitute $y$ from equation (3) into equation (1).
   Replace $y$ in $8x + 3y = -7$ with $\frac{-3 - 7x}{2}$: $8x + 3\left(\frac{-3 - 7x}{2}\right) = -7$ Distribute $3$ over the fraction: $8x + \frac{-9 - 21x}{2} = -7$ Rewrite with a common denominator (denominator of 2): $\frac{16x}{2} + \frac{-9 - 21x}{2} = -7$ Combine the terms in the numerator: $\frac{16x - 9 - 21x}{2} = -7$ Simplify the numerator: $\frac{-5x - 9}{2} = -7$

3. Solve for $x$: Multiply through by 2 to eliminate the fraction: $-5x - 9 = -14$ Add 9 to both sides: $-5x = -5$ Divide by $-5$: x = 1 \tag{4}

4. Substitute $x = 1$ into equation (3) to find $y$: Use $y = \frac{-3 - 7x}{2}$: $y = \frac{-3 - 7(1)}{2}$ Simplify: y = \frac{-3 - 7}{2} = \frac{-10}{2} = -5 \tag{5}

Solution:

$x = 1, \, y = -5$

Verification:

• Substitute $x = 1$ and $y = -5$ into equation (1): $8(1) + 3(-5) = 8 - 15 = -7 \quad \text{(True)}$
• Substitute $x = 1$ and $y = -5$ into equation (2): $7(1) + 2(-5) = 7 - 10 = -3 \quad \text{(True)}$

The solution satisfies both equations.

Would you like further details or explanations? Here are some related questions:

1. How does substitution work, and when should it be used?
2. What are alternative methods to solve systems of equations, such as elimination?
3. How do you check if a solution to a system is correct?
4. Can systems of equations have no solution or infinitely many solutions? How do you identify these cases?
5. How can graphing help visualize the solution to a system of equations?

Tip: Always verify your solution by substituting the values back into the original equations!