Solving Systems Of Equations Using Multiplication And Linear Combination

When faced with a system of equations, the linear combination method, also known as the elimination method, provides a powerful approach to finding solutions. This method is particularly effective when dealing with equations where variables have coefficients that can be easily manipulated to cancel each other out. In this comprehensive guide, we'll delve into the intricacies of using multiplication in conjunction with the linear combination method to solve systems of equations. We'll illustrate the process with a step-by-step example, ensuring you grasp the underlying concepts and can confidently apply them to various problems.

Understanding the Linear Combination Method

The linear combination method centers around the idea of manipulating equations to eliminate one variable, thereby allowing us to solve for the remaining variable. This manipulation often involves multiplying one or both equations by a constant factor. The goal is to create coefficients for one of the variables that are either additive inverses (e.g., 3 and -3) or the same (e.g., 2 and 2). By adding or subtracting the equations, we can eliminate the variable with matching or opposite coefficients, simplifying the system into a single equation with one unknown.

Key Steps in the Linear Combination Method

1. Align the Equations: Begin by arranging the equations in a standard form, typically with the variables on one side and the constants on the other. This alignment helps in identifying variables with potential for elimination.
2. Identify Target Variable: Determine which variable you want to eliminate first. Look for coefficients that are either the same or additive inverses, or can be easily made so through multiplication.
3. Multiply Equations (if needed): If necessary, multiply one or both equations by a constant factor to make the coefficients of the target variable additive inverses or the same. This is the crucial step where multiplication plays a pivotal role in the linear combination method.
4. Add or Subtract Equations: Add the equations together if the coefficients of the target variable are additive inverses. Subtract the equations if the coefficients are the same. This step eliminates the target variable, leaving you with a single equation in one variable.
5. Solve for Remaining Variable: Solve the resulting equation for the remaining variable. This will give you the value of one of the unknowns.
6. Substitute and Solve: Substitute the value obtained in the previous step back into either of the original equations. Solve this equation for the other variable. This will give you the value of the second unknown.
7. Check the Solution: Verify your solution by substituting both values into the original equations. If both equations hold true, your solution is correct.

Solving a System of Equations Using Multiplication

Let's consider the following system of equations:

6x - 3y = 3
-2x + 6y = 14

Our objective is to find the values of x and y that satisfy both equations simultaneously. We'll employ the linear combination method, utilizing multiplication to streamline the process.

Step 1: Align the Equations

The equations are already aligned in the standard form, with x and y terms on the left side and constants on the right.

Step 2: Identify Target Variable

Observing the equations, we notice that the coefficients of x (6 and -2) can be easily manipulated to become additive inverses. Specifically, if we multiply the second equation by 3, the coefficient of x will become -6, which is the additive inverse of 6.

Step 3: Multiply Equations

Multiply the second equation by 3:

3 * (-2x + 6y) = 3 * 14
-6x + 18y = 42

Now, our system of equations looks like this:

6x - 3y = 3
-6x + 18y = 42

Step 4: Add Equations

Add the two equations together. Notice that the x terms cancel out:

(6x - 3y) + (-6x + 18y) = 3 + 42
15y = 45

Step 5: Solve for Remaining Variable

Solve the resulting equation for y:

15y = 45
y = 45 / 15
y = 3

Step 6: Substitute and Solve

Substitute the value of y (3) back into either of the original equations. Let's use the first equation:

6x - 3(3) = 3
6x - 9 = 3
6x = 12
x = 12 / 6
x = 2

Step 7: Check the Solution

Verify the solution (x = 2, y = 3) by substituting these values into both original equations:

• Equation 1: 6(2) - 3(3) = 12 - 9 = 3 (True)
• Equation 2: -2(2) + 6(3) = -4 + 18 = 14 (True)

Since both equations hold true, our solution is correct.

Conclusion

The linear combination method, enhanced by the strategic use of multiplication, provides a robust technique for solving systems of equations. By carefully manipulating coefficients, we can eliminate variables and simplify the system, leading us to the solution. This method is especially valuable when dealing with equations where coefficients are not readily additive inverses or the same. Through practice and a clear understanding of the steps involved, you can master this powerful tool and confidently tackle a wide range of systems of equations.

In summary, solving systems of equations by multiplication with the linear combination method involves aligning equations, identifying target variables, strategically multiplying equations to create additive inverses or equal coefficients, adding or subtracting equations to eliminate a variable, solving for the remaining variable, substituting the value back to find the other variable, and finally, checking the solution for accuracy. This methodical approach ensures that you can effectively and efficiently solve systems of equations, even when they appear complex at first glance.

The key to success with the linear combination method lies in recognizing the potential for manipulation through multiplication and applying the steps systematically. By mastering this technique, you'll gain a valuable tool in your mathematical arsenal, enabling you to solve a wide variety of problems involving systems of equations.

Given the system of equations:

6x - 3y = 3
-2x + 6y = 14

We have already solved this system using the linear combination method with multiplication in the previous section. The solution we found was x = 2 and y = 3.

Now, let's match our solution with the provided options:

A. (2, 1) B. (2, -3) C. (2, -1) D. (2, 3)

Our solution (x = 2, y = 3) corresponds to option D. Therefore, the correct answer is:

D. (2, 3)

This confirms that the linear combination method, when applied correctly, leads to the accurate solution of the system of equations. By understanding the steps involved and practicing regularly, you can confidently solve similar problems and master the art of solving systems of equations.