To solve a system of equations using the elimination method, we generally follow the steps outlined below, with specific choices for the missing dropdowns. Let me walk you through the process to fill in the blanks.

Given:

Assume we are solving this system of equations:

1. Equation 1: $a_1x + b_1y = c_1$
2. Equation 2: $a_2x + b_2y = c_2$

Step-by-Step Elimination Process

1. Multiply one or both equations to eliminate one variable.

   • For example, multiply Equation 2 by a constant such that the coefficient of $x$ or $y$ becomes the same (or opposite) in both equations. This will give us a new equation. So:
   • Drop Down 1: Multiply Equation 2 by a constant. (This is typically the coefficient needed to match one of the variables with the corresponding term in Equation 1.)

2. New Equation 2: After multiplying, we have a new version of Equation 2.

3. Add Equation 1 and New Equation 2:

   • This step eliminates one variable.

   • Drop Down 2: After adding, you'll have a new equation in terms of only one variable (typically $y$ or $x$).

4. Divide by: After simplifying the equation with the remaining variable, we divide by the coefficient of that variable.

   • Drop Down 3: This would be the coefficient of the variable you're solving for.

5. Resulting Equation: After division, the equation will give the value of one variable, say $y$.

   • Drop Down 4: This is the resulting value of the variable you solved for.

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Example:

Let’s solve the following system:

3x + 2y = 16 \tag{Equation 1} 5x - 4y = -2 \tag{Equation 2}

Step 1: Multiply Equation 1 or Equation 2 to eliminate one variable.

• To eliminate $y$, we can multiply Equation 1 by 2 to match the coefficient of $y$ in Equation 2 (with opposite signs):

  • $2 \times (3x + 2y) = 2 \times 16$ gives $6x + 4y = 32$ (New Equation 1)

  • Equation 2 is $5x - 4y = -2$ (remains unchanged)

• Drop Down 1: Multiply Equation 1 by 2.

Step 2: Add the two equations.

• Adding the two equations results in:

$(6x + 4y) + (5x - 4y) = 32 + (-2)$ $11x = 30$

• Drop Down 2: Resulting equation is $11x = 30$.

Step 3: Solve for $x$.

• Divide by 11:

$x = \frac{30}{11}$

• Drop Down 3: Divide by 11.
• Drop Down 4: $x = \frac{30}{11}$.

Step 4: Solve for $y$.

• Substitute $x = \frac{30}{11}$ back into one of the original equations, for example, Equation 1:

$3\left(\frac{30}{11}\right) + 2y = 16$

Simplify and solve for $y$:

$\frac{90}{11} + 2y = 16$ $2y = 16 - \frac{90}{11} = \frac{176}{11} - \frac{90}{11} = \frac{86}{11}$ $y = \frac{43}{11}$

Summary for Dropdowns:

• Drop Down 1: Multiply Equation 1 by 2.
• Drop Down 2: Resulting equation is $11x = 30$.
• Drop Down 3: Divide by 11.
• Drop Down 4: $x = \frac{30}{11}$.

Would you like to explore further details of this example or go over additional elimination cases?

Five Related Questions:

1. How would the steps change if we eliminated $x$ instead of $y$?
2. Can you solve the same system of equations using substitution instead of elimination?
3. What happens if the system of equations has no solution or infinite solutions?
4. How does elimination work with three-variable systems?
5. Can we use matrices to solve the same system?

Tip:

When multiplying equations for elimination, always look for the smallest coefficients that will eliminate a variable efficiently.