Finding The Right Multiplier To Solve Systems Of Equations

Systems of equations are a fundamental concept in algebra, representing a set of two or more equations with the same variables. Solving these systems involves finding values for the variables that satisfy all equations simultaneously. There are several methods to tackle this, including substitution, elimination, and graphing. In this article, we'll focus on the elimination method, specifically how to strategically manipulate equations to eliminate one variable and solve for the other. A crucial step in the elimination method involves identifying the appropriate multiplier to create opposite coefficients for one of the variables.

The elimination method shines when dealing with linear equations. The goal is to add or subtract the equations in such a way that one variable vanishes, leaving you with a single equation in one variable that is easily solved. This hinges on creating "like terms with opposite coefficients." For example, if one equation has +3x and another has -3x, adding the equations will eliminate x. If you have +2y in one equation and you want to eliminate y, you need -2y in the other equation. Finding the right multiplier is the key to achieving this balance.

To delve deeper, let's consider the given system of equations. The system presents a pair of linear equations: $\frac{1}{2}x + \frac{3}{2}y = -7$ and $-3x + 2y = -2$. Our objective is to determine the multiplier that, when applied to the first equation, will produce a term that cancels out with a term in the second equation. Specifically, we want to create like terms with opposite coefficients. Examining the x-terms, we see $\frac{1}{2}x$ in the first equation and $-3x$ in the second. To eliminate the x variable, we need to transform the $\frac{1}{2}x$ into a term that, when added to $-3x$, results in zero. This means we need to find a multiplier that, when multiplied by $\frac{1}{2}$, gives us the opposite of -3, which is 3. Therefore, we need to solve the equation: multiplier * $\frac{1}{2}$ = 3. Multiplying both sides by 2, we get multiplier = 6. So, multiplying the first equation by 6 will give us $3x$, which is the opposite of the -3x term in the second equation. Once we perform this multiplication, we can add the modified first equation to the second equation, effectively eliminating the x variable and allowing us to solve for y. This illustrates the core principle of using multipliers to set up the elimination method effectively. The next step would involve performing the multiplication and then proceeding with the elimination process to solve for the variables.

Identifying the Target Variable and its Coefficients

When confronted with a system of equations, the initial step in strategically selecting a multiplier involves pinpointing the variable you intend to eliminate. In other words, identifying the target variable is paramount. This decision often hinges on the coefficients of the variables within the equations. For example, if one variable already has coefficients that are multiples or factors of each other, it might be the easier one to eliminate. Additionally, consider the signs of the coefficients. If the coefficients of one variable have opposite signs, you can directly add the equations after applying the multiplier, simplifying the elimination process. If the signs are the same, you might need to subtract the equations, which can introduce more opportunities for errors. Understanding the structure of your equations is fundamental to efficient problem-solving.

Once you've decided which variable to target, the next crucial step is to carefully examine its coefficients in both equations. The coefficients are the numerical values that multiply the variables. These coefficients hold the key to determining the necessary multiplier. Look for relationships between the coefficients. Are they multiples of each other? Do they share common factors? For instance, if you have a '2x' in one equation and a '4x' in the other, it's clear that you can multiply the first equation by -2 to create opposite coefficients (-4x and 4x) for the x variable. Or, consider the fractions. The presence of fractions often means we will have to multiply one or both equations by a constant which is a multiple of the denominators of the fractions in the equation. This strategy will help us clear the fractions and work in the realm of integers and potentially simplify our work. The ultimate goal is to manipulate one or both equations so that the coefficients of your target variable become opposites – that is, they have the same numerical value but opposite signs. This precise understanding of the coefficients and their interplay is the bedrock of successful elimination.

Let's illustrate this principle with a concrete example. In our system of equations, we have $\frac{1}{2}x$ in the first equation and $-3x$ in the second. Focusing on the x variable, we want to find a multiplier for the first equation that will result in a coefficient that is the opposite of -3. That means we want to transform $\frac{1}{2}x$ into $3x$. To achieve this, we need to determine what number, when multiplied by $\frac{1}{2}$, equals 3. This can be represented by the equation $\frac{1}{2} * multiplier = 3$. Solving for the multiplier, we multiply both sides by 2, yielding multiplier = 6. This clearly demonstrates how a careful examination of the coefficients allows us to pinpoint the exact multiplier needed to set up the elimination process. Similarly, if we were to target the y variable with coefficients $\frac{3}{2}$ and 2, we'd need to find a multiplier that transforms $\frac{3}{2}$ into -2. This would involve solving the equation $\frac{3}{2} * multiplier = -2$. Multiplying both sides by $\frac{2}{3}$ gives us multiplier = $-\frac{4}{3}$. This process of analyzing coefficients and solving for the required multiplier is the cornerstone of effectively applying the elimination method in solving systems of equations.

Determining the Correct Multiplier

Once you've identified the target variable and examined its coefficients, the next step is to determine the correct multiplier. The multiplier is the number you'll multiply an equation by to create the desired opposite coefficient. This process often involves a bit of algebraic thinking. The objective is to find a number that, when multiplied by the coefficient of your target variable in one equation, yields the opposite of the coefficient of the same variable in the other equation. Essentially, we are transforming one of the equations in a way that facilitates the elimination of a variable when the equations are combined.

One common strategy for finding the multiplier involves setting up a simple equation. Let's say you want to eliminate the variable x. Identify the coefficients of x in both equations. Let's call them 'a' and 'b'. The goal is to find a multiplier, 'm', such that m times 'a' equals the opposite of 'b'. This can be represented by the equation: m * a = -b. Solving this equation for 'm' will give you the multiplier needed to create opposite coefficients for the x variable. This approach works effectively because it directly addresses the mathematical relationship required for elimination. If the coefficients are fractions, you can often clear the fractions by multiplying by the least common multiple of the denominators. This can simplify the arithmetic and make it easier to find the multiplier.

Consider our example equations: $\frac{1}{2}x + \frac{3}{2}y = -7$ and $-3x + 2y = -2$. We've already decided to target the x variable. The coefficient of x in the first equation is $\frac{1}{2}$, and the coefficient of x in the second equation is -3. We want to find a multiplier, 'm', such that m * $\frac{1}{2}$ = -(-3), which simplifies to m * $\frac{1}{2}$ = 3. To solve for 'm', we multiply both sides of the equation by 2, resulting in m = 6. This confirms that multiplying the first equation by 6 will produce a term that cancels out the -3x term in the second equation. If, instead, we had wanted to eliminate y, we would have set up a different equation based on the coefficients of y. This methodical approach ensures that you select the correct multiplier, setting the stage for a smooth and accurate elimination process.

Applying the Multiplier and Eliminating a Variable

Once the multiplier is determined, the next crucial step is to apply the multiplier to the entire equation. This means multiplying every term in the equation by the chosen multiplier, not just the term with the variable you're targeting for elimination. This ensures that the equation remains balanced and that you're creating an equivalent equation that still holds the same solutions as the original. It's a common mistake to only multiply one term, which can lead to incorrect results. Remember, an equation is like a balance scale – what you do to one side, you must do to the other to maintain equilibrium.

After multiplying the equation, the next step is to eliminate a variable by adding or subtracting the modified equation with the other equation in the system. This is the heart of the elimination method. If you've chosen your multiplier correctly, the coefficients of your target variable should now be opposites (e.g., +3x and -3x). When you add the equations together, these terms will cancel out, leaving you with an equation containing only one variable. If the coefficients of the target variable are the same (e.g., +3x and +3x), you'll need to subtract the equations instead of adding them. Careful attention to the signs is crucial here. A small mistake in addition or subtraction can throw off the entire solution.

Let's return to our example. We've established that the multiplier for the first equation, $\frac{1}{2}x + \frac{3}{2}y = -7$, is 6. Applying this multiplier to every term in the equation, we get: 6 * ($\frac{1}{2}x$) + 6 * ($\frac{3}{2}y$) = 6 * (-7), which simplifies to 3x + 9y = -42. Now, we have the modified first equation: 3x + 9y = -42, and the second equation remains: -3x + 2y = -2. Notice that the coefficients of x are now opposites (3x and -3x). Adding these two equations together, we get: (3x + 9y) + (-3x + 2y) = -42 + (-2). This simplifies to 11y = -44. The x variable has been successfully eliminated, and we're left with a simple equation in one variable. Now we can proceed to solve for y and then substitute the value of y back into one of the original equations to solve for x. This process illustrates the power and efficiency of the elimination method when the multiplier is correctly applied, and the equations are added or subtracted appropriately.

Solving for the Remaining Variable and Finding the Solution

With one variable eliminated, you're left with a simplified equation containing only one unknown. The next step is to solve for the remaining variable. This usually involves basic algebraic manipulation, such as adding or subtracting constants from both sides of the equation, or dividing both sides by the coefficient of the variable. The goal is to isolate the variable on one side of the equation, giving you its numerical value. This value is one part of the solution to the system of equations.

Once you've found the value of one variable, the final step is to find the solution by substituting the known value back into one of the original equations. It doesn't matter which original equation you choose; the solution should be the same. This substitution will create a new equation with only one unknown – the variable you haven't solved for yet. Solve this equation using the same algebraic techniques as before to find the value of the second variable. Now you have the values for both variables, which together form the solution to the system of equations. It's always a good practice to check your solution by substituting both values back into both original equations to ensure they are satisfied. This confirms that your solution is correct.

Continuing our example, we had the equation 11y = -44 after eliminating x. To solve for y, we divide both sides by 11, which gives us y = -4. Now that we know y = -4, we can substitute this value back into either of the original equations. Let's use the first original equation: $\frac{1}{2}x + \frac{3}{2}y = -7$. Substituting y = -4, we get: $\frac{1}{2}x + \frac{3}{2}(-4) = -7$. This simplifies to $\frac{1}{2}x - 6 = -7$. Adding 6 to both sides, we get $\frac{1}{2}x = -1$. Multiplying both sides by 2, we find x = -2. Therefore, the solution to the system of equations is x = -2 and y = -4. To verify this solution, we can substitute these values back into both original equations. For the first equation: $\frac{1}{2}(-2) + \frac{3}{2}(-4) = -1 - 6 = -7$, which is correct. For the second equation: -3(-2) + 2(-4) = 6 - 8 = -2, which is also correct. This confirms that our solution (x = -2, y = -4) is accurate and satisfies both equations in the system. This comprehensive process of solving for the remaining variable and verifying the solution is essential for mastering the elimination method and accurately solving systems of equations.

Applying the Concept to the Given Problem

Now, let's apply these principles to the specific problem presented: $\frac{1}{2}x + \frac{3}{2}y = -7$ and $-3x + 2y = -2$. The question asks: "You could produce a pair of like terms with opposite coefficients by multiplying the first equation by what number?" This is directly related to the concept of finding the multiplier for the elimination method.

As we discussed earlier, the key is to identify the target variable and its coefficients. Looking at the x terms, we have $\frac{1}{2}x$ in the first equation and -3x in the second equation. To eliminate x, we need to transform $\frac{1}{2}x$ into 3x (the opposite of -3x). To do this, we need to find the multiplier that, when multiplied by $\frac{1}{2}$, equals 3. We set up the equation: multiplier * $\frac{1}{2}$ = 3.

Solving for the multiplier, we multiply both sides of the equation by 2, which gives us: multiplier = 6. Therefore, multiplying the first equation by 6 will create the term 3x, which has the opposite coefficient of the -3x term in the second equation. This allows us to eliminate x when we add the equations together. Based on this analysis, the correct answer is C. 6.

This problem perfectly illustrates the practical application of understanding multipliers in the context of solving systems of equations. By carefully examining the coefficients and setting up a simple equation, we can quickly determine the correct multiplier to create opposite coefficients and facilitate the elimination process. This skill is fundamental for efficiently solving systems of equations using the elimination method.

In conclusion, mastering the art of finding the right multiplier is crucial for efficiently solving systems of equations using the elimination method. By strategically analyzing the coefficients of the variables, setting up appropriate equations, and applying the multiplier correctly, you can transform equations to create opposite coefficients and eliminate variables. This process leads you to a simplified equation that can be easily solved, ultimately providing the solution to the system. The ability to confidently identify and apply multipliers not only streamlines the solving process but also demonstrates a deeper understanding of algebraic principles.