To solve the system of two equations, we can follow these general steps:

1. Solve for one variable (algebraic expression):

• If you have two equations from Part A and Part B, let's represent them as:
  • Equation 1: $f(x, y) = 0$
  • Equation 2: $g(x, y) = 0$

You can solve one equation for one variable in terms of the other (e.g., solve for $y$ in terms of $x$ or $x$ in terms of $y$).

2. Substitute into the second equation:

• Once you have expressed one variable in terms of the other (e.g., $y = h(x)$), substitute this expression into the second equation.
• This will result in a new equation with only one variable (e.g., $x$).

3. Solve the single-variable equation:

• Solve the new equation to find the value of the single variable (e.g., $x$).

4. Substitute back to find the other variable:

• Once you have the value of one variable (e.g., $x$), substitute it back into one of the original equations (or the derived expression from Part A) to solve for the other variable (e.g., $y$).

5. Final solution:

• Now you have both $x$ and $y$, which is the solution to the system of equations.

Example:

If you had the system of equations:

1. $2x + y = 5$
2. $3x - y = 4$

Step 1: Solve the first equation for $y$:

$y = 5 - 2x$

Step 2: Substitute this into the second equation:

$3x - (5 - 2x) = 4$

Step 3: Simplify and solve for $x$:

5x - 5 = 4 \\ 5x = 9 \\ x = \frac{9}{5}$$ #### Step 4: Substitute $$x = \frac{9}{5}$$ into the first equation to find $$y$$: $$y = 5 - 2 \times \frac{9}{5} = 5 - \frac{18}{5} = \frac{25}{5} - \frac{18}{5} = \frac{7}{5}$$ #### Final Solution: $$x = \frac{9}{5}, \quad y = \frac{7}{5}$$ This process will yield the solution to the system of equations you have. --- Do you need more details or clarification on any part of this solution? Here are 5 questions related to this: 1. What methods can be used to solve systems of equations algebraically? 2. How can substitution be applied to more complex systems of equations? 3. What happens if both equations are linear and parallel? 4. Can elimination be more efficient than substitution in some cases? 5. How do systems of non-linear equations differ in their solutions? **Tip:** Always check your solution by substituting the values of both variables back into the original equations to ensure accuracy!