Solving 4x + 3y = 16 And 3x + 2y = 11 Elimination And Substitution Methods

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Hey guys! Today, we're going to dive into a super important topic in math: solving systems of linear equations. Specifically, we'll be tackling the system:

  • 4x + 3y = 16
  • 3x + 2y = 11

We're going to use two powerful methods: elimination and substitution. These methods are like the dynamic duo of algebra, helping us crack problems that seem tough at first glance. So, grab your pencils, and let's get started!

Introduction to Systems of Linear Equations

Before we jump into the nitty-gritty, let's make sure we're all on the same page. What exactly is a system of linear equations? Well, it’s simply a set of two or more linear equations containing the same variables. Think of it like a puzzle where you need to find the values of the variables that make all the equations true at the same time. In our case, we have two equations with two variables, x and y. The solution to this system will be a pair of values (x, y) that satisfies both equations simultaneously. Systems of linear equations pop up everywhere in real-world scenarios, from calculating costs and quantities to modeling physical phenomena. Mastering how to solve them is a crucial skill in math and beyond.

Method 1: Elimination – The Art of Canceling

The elimination method, also known as the addition or subtraction method, is a clever technique where we manipulate the equations to eliminate one of the variables. The main idea here is to make the coefficients of either x or y the same (but with opposite signs) in both equations. That way, when we add the equations together, one variable cancels out, leaving us with a single equation in one variable. Let's see how it works with our system:

Step-by-Step Elimination

  1. Choose a variable to eliminate: Look at our system:

    • 4x + 3y = 16
    • 3x + 2y = 11 Notice that the coefficients of x are 4 and 3, and the coefficients of y are 3 and 2. We can choose to eliminate either x or y. For this example, let's eliminate x. To do this, we need to make the coefficients of x multiples of the same number. The least common multiple of 4 and 3 is 12. So, we'll aim to make the coefficients of x equal to 12 and -12.

  2. Multiply the equations:

    • Multiply the first equation by 3: 3 * (4x + 3y) = 3 * 16 --> 12x + 9y = 48
    • Multiply the second equation by -4: -4 * (3x + 2y) = -4 * 11 --> -12x - 8y = -44

    Now our system looks like this:

    • 12x + 9y = 48
    • -12x - 8y = -44

    See how the coefficients of x are now 12 and -12? Perfect!

  3. Add the equations: Now, we simply add the two equations together:

    (12x + 9y) + (-12x - 8y) = 48 + (-44)

    The 12x and -12x cancel each other out, leaving us with:

    y = 4

    Woo-hoo! We've found the value of y!

  4. Solve for the remaining variable: Now that we know y = 4, we can substitute this value into either of the original equations to solve for x. Let's use the first equation:

    4x + 3y = 16

    Substitute y = 4:

    4x + 3(4) = 16

    4x + 12 = 16

    Subtract 12 from both sides:

    4x = 4

    Divide by 4:

    x = 1

    Awesome! We've found the value of x as well.

Solution using Elimination

So, using the elimination method, we found that x = 1 and y = 4. This means the solution to the system of equations is the ordered pair (1, 4). This is the point where the two lines represented by the equations intersect on a graph. The elimination method is super handy because it systematically gets rid of one variable, making the problem much simpler to solve. It's like a strategic move in a game of algebra!

Method 2: Substitution – The Art of Replacing

The substitution method is another fantastic way to solve systems of linear equations. The main idea here is to solve one equation for one variable and then substitute that expression into the other equation. This turns the second equation into an equation with just one variable, which we can then solve. Once we find the value of that variable, we can substitute it back into either of the original equations to find the value of the other variable. Let's see how this works with our system:

Step-by-Step Substitution

  1. Solve one equation for one variable: Look at our system again:

    • 4x + 3y = 16
    • 3x + 2y = 11

    We need to choose one equation and solve it for either x or y. It's often easiest to choose the equation where a variable has a coefficient of 1 (if there is one), but in this case, none of the variables have a coefficient of 1. So, let's choose the second equation (3x + 2y = 11) and solve it for y. Why y? Just because! We could have chosen x, and the process would be similar.

    • 3x + 2y = 11

    Subtract 3x from both sides:

    • 2y = 11 - 3x

    Divide both sides by 2:

    • y = (11 - 3x) / 2

    We now have an expression for y in terms of x. This is the key to the substitution method.

  2. Substitute into the other equation: Now, we substitute this expression for y into the other equation (the one we didn't use in the previous step). That's the first equation, 4x + 3y = 16.

    • 4x + 3 * ((11 - 3x) / 2) = 16

    This looks a bit messy, but don't worry! We'll simplify it.

  3. Solve for the remaining variable: Let's simplify and solve for x.

    • 4x + (33 - 9x) / 2 = 16

    To get rid of the fraction, multiply the entire equation by 2:

    • 2 * (4x + (33 - 9x) / 2) = 2 * 16

    • 8x + 33 - 9x = 32

    Combine like terms:

    • -x + 33 = 32

    Subtract 33 from both sides:

    • -x = -1

    Multiply by -1:

    • x = 1

    Yes! We got x = 1, just like we did with the elimination method. This is a good sign that we're on the right track.

  4. Solve for the other variable: Now that we know x = 1, we can substitute this value back into the expression we found for y earlier:

    • y = (11 - 3x) / 2

    Substitute x = 1:

    • y = (11 - 3 * 1) / 2

    • y = (11 - 3) / 2

    • y = 8 / 2

    • y = 4

    Fantastic! We found y = 4. This matches our result from the elimination method, which is exactly what we want.

Solution using Substitution

So, using the substitution method, we also found that x = 1 and y = 4. Again, the solution to the system of equations is the ordered pair (1, 4). The substitution method is particularly useful when one of the equations is already solved for one variable or when it's easy to isolate a variable. It's like finding a clever shortcut to the solution!

Comparing Elimination and Substitution

Okay, so we've solved the same system of equations using two different methods: elimination and substitution. You might be wondering, which method is better? Well, the truth is, there's no single "best" method. Both elimination and substitution have their strengths and weaknesses, and the best method often depends on the specific system of equations you're dealing with.

When to Use Elimination

Elimination is often a good choice when:

  • The coefficients of one of the variables are already the same or easy to make the same (with opposite signs).
  • You want to avoid fractions (since substitution can sometimes lead to fractions).
  • The equations are in standard form (Ax + By = C).

When to Use Substitution

Substitution is often a good choice when:

  • One of the equations is already solved for one variable, or it's easy to solve for one variable.
  • You have a variable with a coefficient of 1 (making it easy to isolate).
  • You prefer working with expressions rather than manipulating entire equations.

The Bottom Line

The key is to be flexible and choose the method that seems most efficient for the problem at hand. And, of course, practice makes perfect! The more you work with these methods, the better you'll become at recognizing which one is the best fit for a given system of equations. Remember, both methods are just tools in your algebraic toolbox. The more tools you have, the more problems you can solve!

Verifying the Solution

Alright, we've found our solution: x = 1 and y = 4. But how do we know if we're right? It's always a good idea to verify your solution, especially in math. Luckily, it's super easy to do this with systems of equations. All we need to do is substitute our values for x and y back into the original equations and see if they hold true.

Step-by-Step Verification

  1. Original System:

    • 4x + 3y = 16
    • 3x + 2y = 11

  2. Substitute x = 1 and y = 4 into the first equation:

    • 4(1) + 3(4) = 16
    • 4 + 12 = 16
    • 16 = 16

    The first equation checks out! That's a good start.

  3. Substitute x = 1 and y = 4 into the second equation:

    • 3(1) + 2(4) = 11
    • 3 + 8 = 11
    • 11 = 11

    The second equation also checks out! 🎉

The Importance of Verification

Since our values for x and y satisfy both equations, we can confidently say that our solution (1, 4) is correct. Verifying your solution is like the final seal of approval on your work. It helps you catch any mistakes you might have made along the way and ensures that you're submitting the correct answer. It’s a simple step that can save you from unnecessary errors and boost your confidence in your problem-solving skills.

Real-World Applications of Systems of Linear Equations

So, we've mastered solving systems of linear equations using elimination and substitution. That's awesome! But you might be thinking, "Okay, this is cool math stuff, but where would I ever use this in real life?" Well, the truth is, systems of linear equations are everywhere! They pop up in all sorts of unexpected places. Let's explore a few real-world applications to see just how useful these skills can be.

1. Budgeting and Finance

Imagine you're trying to create a budget. You have a certain amount of money to spend each month, and you need to allocate it to different categories like rent, food, and entertainment. You can use systems of linear equations to figure out how much to spend in each category while staying within your budget. For example, you might have one equation representing your total income and another equation representing your total expenses. Solving the system will help you find the optimal balance between your spending categories.

2. Mixing Solutions

Chemists and pharmacists often use systems of linear equations to determine how much of different solutions to mix together to get a desired concentration. For instance, you might need to mix two solutions with different concentrations of acid to create a solution with a specific concentration. Each solution contributes to the total amount and the total concentration, which can be represented by a system of linear equations.

3. Supply Chain Management

Businesses use systems of linear equations to optimize their supply chains. They need to figure out how much of each product to order, how much to store, and how much to ship to different locations to minimize costs and meet demand. Each constraint, like storage capacity or transportation costs, can be represented by a linear equation. Solving the system helps businesses make efficient decisions about their logistics and operations.

4. Physics and Engineering

In physics and engineering, systems of linear equations are used to model a wide range of phenomena, from electrical circuits to structural mechanics. For example, when analyzing an electrical circuit, you might use Kirchhoff's laws to create a system of equations representing the currents and voltages in different parts of the circuit. Solving the system allows you to determine the behavior of the circuit and design it effectively.

5. Economics

Economists use systems of linear equations to model supply and demand, market equilibrium, and other economic phenomena. For example, you might have one equation representing the supply of a product and another equation representing the demand for that product. Solving the system will help you find the equilibrium price and quantity in the market.

The Big Picture

These are just a few examples, but the applications of systems of linear equations are vast and varied. From everyday budgeting to complex scientific modeling, these skills are essential for problem-solving in many different fields. By mastering elimination and substitution, you're not just learning math; you're learning a powerful tool that can help you make sense of the world around you. The ability to model real-world situations with equations and solve those equations is a valuable skill that will serve you well in your academic and professional life.

Conclusion

Alright, guys, we've covered a lot today! We've explored how to solve systems of linear equations using both the elimination and substitution methods. We've seen that the elimination method involves manipulating equations to cancel out one variable, while the substitution method involves solving for one variable and substituting that expression into the other equation. Both methods are powerful tools in your math arsenal, and the best one to use often depends on the specific problem you're facing.

We also emphasized the importance of verifying your solution by plugging your values back into the original equations. This step is crucial for catching errors and ensuring that you've arrived at the correct answer. And we took a tour of the real world to see how systems of linear equations are used in various fields, from budgeting to supply chain management. This shows just how practical and relevant these skills are.

So, what's the next step? Practice, practice, practice! The more you work with these methods, the more comfortable and confident you'll become. Try solving different systems of equations, and challenge yourself to choose the most efficient method for each problem. Remember, math is like any other skill: it gets easier with practice. Keep up the great work, and you'll be a system-solving pro in no time!