Solving Systems Of Equations Efficiently A Step By Step Guide

Hey guys! Today, we're diving into the exciting world of solving systems of equations. Specifically, we're going to tackle the following system:

2x - 3y = -33
x + 2y = 15

We'll explore the most efficient methods to crack this code and express our answers as integers or simplified fractions. So, buckle up and let's get started!

Understanding Systems of Equations

Before we jump into solving, let's quickly recap what a system of equations actually is. A system of equations is simply a set of two or more equations that share the same variables. Our goal is to find the values for those variables that satisfy all the equations in the system simultaneously. Think of it as a puzzle where each equation is a piece, and we need to find the values that make all the pieces fit together perfectly. In the given system,

2x - 3y = -33
x + 2y = 15

we have two equations with two variables (x and y). This means we're looking for a specific pair of x and y values that will make both of these equations true at the same time. There are several methods we can use to solve systems of equations, and the most efficient method often depends on the specific equations we're dealing with. We'll explore two popular techniques: substitution and elimination. Each method has its strengths and weaknesses, and understanding them will empower you to choose the best approach for any given problem. It's like having different tools in your mathematical toolbox – knowing when to use each one is key!

When we talk about solving systems of equations, we're essentially looking for the point (or points) where the lines represented by these equations intersect on a graph. Each linear equation in the system represents a straight line, and the solution to the system is the coordinate pair (x, y) that lies on both lines. This point of intersection is the unique solution that satisfies both equations simultaneously. If the lines are parallel, they never intersect, indicating that the system has no solution. On the other hand, if the lines are the same, they overlap infinitely, meaning the system has infinitely many solutions. Visualizing the equations as lines can provide a helpful way to understand the nature of the solutions.

In practical terms, systems of equations pop up in a variety of real-world scenarios, from determining the break-even point in business to calculating the optimal mix of ingredients in a recipe. They're a fundamental tool in many fields, including science, engineering, economics, and computer science. Mastering the techniques for solving these systems is not only crucial for academic success in mathematics but also for applying mathematical concepts to solve practical problems. So, let's delve deeper into the methods that will help us tackle any system of equations with confidence and efficiency!

Method 1: Substitution – The Art of Variable Swapping

The substitution method is a powerful technique for solving systems of equations, and it's particularly effective when one of the equations is easily solved for one variable in terms of the other. The basic idea behind substitution is to isolate one variable in one equation and then substitute that expression into the other equation. This effectively reduces the system to a single equation with a single variable, which we can then solve. Once we've found the value of one variable, we can substitute it back into either of the original equations to find the value of the other variable. Think of it like replacing one piece of the puzzle with an equivalent expression to simplify the overall picture.

Let's apply the substitution method to our system:

2x - 3y = -33
x + 2y = 15

Notice that the second equation, x + 2y = 15, is relatively easy to solve for x. We can simply subtract 2y from both sides to get:

x = 15 - 2y

Now, we've isolated x in terms of y. This is the crucial first step in the substitution method. The next step is to substitute this expression for x into the other equation – the one we haven't used yet, which is 2x - 3y = -33. This is where the magic of substitution happens. We replace x with 15 - 2y in the first equation:

2(15 - 2y) - 3y = -33

Now we have a single equation with only one variable, y. We can solve this equation using basic algebraic techniques. First, distribute the 2:

30 - 4y - 3y = -33

Next, combine like terms:

30 - 7y = -33

Subtract 30 from both sides:

-7y = -63

Finally, divide both sides by -7:

y = 9

We've successfully found the value of y! Now that we know y = 9, we can substitute this value back into either of the original equations to find x. It's often easiest to use the equation where we already isolated x, which is x = 15 - 2y. Substituting y = 9 into this equation gives:

x = 15 - 2(9)
x = 15 - 18
x = -3

So, we've found that x = -3. Therefore, the solution to the system of equations is x = -3 and y = 9. We can write this as an ordered pair: (-3, 9). To verify our solution, we can substitute these values back into both original equations to make sure they hold true. This is a good practice to ensure we haven't made any errors along the way. In summary, the substitution method is a systematic approach that allows us to transform a system of equations into a single-variable equation, making it easier to solve for the unknowns.

Method 2: Elimination – The Art of Strategic Cancellation

The elimination method, also known as the addition method, is another powerful technique for solving systems of equations. This method shines when the coefficients of one of the variables in the two equations are either the same or easily made the same (or opposites). The core idea behind elimination is to manipulate the equations so that when you add them together, one of the variables cancels out, leaving you with a single equation in one variable. This simplifies the system and allows you to solve for the remaining variable. It's like strategically neutralizing one variable to reveal the value of the other.

Let's revisit our system of equations:

2x - 3y = -33
x + 2y = 15

Looking at the coefficients of x and y, we can see that it would be relatively easy to eliminate x. To do this, we can multiply the second equation by -2. This will give us a -2x term, which will cancel out the 2x term in the first equation. Let's perform this multiplication:

-2(x + 2y) = -2(15)
-2x - 4y = -30

Now we have a new second equation: -2x - 4y = -30. Our system now looks like this:

2x - 3y = -33
-2x - 4y = -30

Notice that the coefficients of x are now opposites (2 and -2). This is exactly what we wanted! Now, we can add the two equations together. When we do this, the x terms will cancel out:

(2x - 3y) + (-2x - 4y) = -33 + (-30)
2x - 3y - 2x - 4y = -63
-7y = -63

As you can see, the x terms have disappeared, and we're left with a simple equation in y. Now we can solve for y by dividing both sides by -7:

y = 9

Great! We've found y = 9, just like we did with the substitution method. Now we need to find x. To do this, we can substitute the value of y back into either of the original equations. Let's use the second original equation, x + 2y = 15:

x + 2(9) = 15
x + 18 = 15

Subtract 18 from both sides:

x = -3

Again, we find that x = -3. So, the solution to the system is x = -3 and y = 9, or (-3, 9) as an ordered pair. We can verify this solution by plugging these values back into both original equations to ensure they hold true. The elimination method provides an alternative pathway to solving systems of equations, especially when the coefficients are conducive to cancellation. It's a strategic approach that streamlines the process and helps us efficiently find the solution.

Choosing the Most Efficient Method: A Strategic Decision

So, we've explored two powerful methods for solving systems of equations: substitution and elimination. But how do you decide which method is the most efficient for a particular problem? The answer, like many things in mathematics, often depends on the specific characteristics of the equations you're dealing with. Think of it as choosing the right tool for the job – a screwdriver is great for screws, but you wouldn't use it to hammer a nail!

When to use Substitution:

• Substitution shines when one of the equations is already solved for one variable or can be easily solved for one variable. For example, if you have an equation like y = 3x + 2, substitution is likely the way to go. You can simply substitute the expression 3x + 2 for y in the other equation. This avoids the need for further manipulation to isolate a variable.
• Substitution is also a good choice when one of the variables has a coefficient of 1 or -1. This makes isolating that variable straightforward, as you don't need to divide by any other number. This simplifies the substitution process and reduces the risk of fractions.

When to use Elimination:

• Elimination excels when the coefficients of one of the variables in the two equations are either the same, opposites, or easily made the same or opposites by multiplying one or both equations by a constant. This allows you to eliminate a variable by simply adding or subtracting the equations.
• Elimination is particularly efficient when the equations are in standard form (Ax + By = C). In this form, it's easy to identify the coefficients and determine the best way to manipulate the equations for elimination.

Analyzing our Example:

Let's look back at our original system:

2x - 3y = -33
x + 2y = 15

In this case, both substitution and elimination are viable options. However, we can argue that substitution might be slightly more efficient. Notice that the second equation, x + 2y = 15, can be easily solved for x by subtracting 2y from both sides (x = 15 - 2y). This makes substitution a relatively quick and direct approach. While elimination would also work (as we demonstrated), it requires an extra step of multiplying the second equation by -2 before adding the equations. This additional step adds a bit more complexity compared to the straightforward substitution.

General Tips for Choosing a Method:

• Scan the equations: Take a quick look at the equations and identify any obvious opportunities for substitution or elimination.
• Look for easy isolations: If one of the equations can be easily solved for a variable, substitution might be the best choice.
• Check for matching coefficients: If the coefficients of one variable are the same or opposites (or easily made so), elimination is often the most efficient method.
• Don't be afraid to try both: If you're unsure which method is best, try starting with one. If it becomes cumbersome, you can always switch to the other method. The key is to develop a flexible approach and be comfortable with both techniques.

Ultimately, the