Solving Systems Of Equations: A Step-by-Step Guide

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

{3x+2y=125x−y=7{ \begin{cases} 3x+2y = 12\\ 5x-y=7 \end{cases} }

Don't worry if it looks intimidating at first glance. We'll break it down step by step, making it super easy to understand. Whether you're a student grappling with algebra or just someone who loves a good mathematical puzzle, this guide is for you. So, grab your pencils, and let's get started!

Understanding Systems of Equations

Before we jump into the solution, let's quickly discuss 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. The goal is to find the values for those variables that satisfy all equations in the system simultaneously. Think of it as a quest to find the one set of values that makes all the equations happy.

In our case, we have two equations with two variables, x and y. This is a very common type of system, and there are several methods we can use to solve it. We'll focus on the substitution method and the elimination method, as these are two of the most powerful and widely used techniques. Mastering these methods will give you a solid foundation for tackling more complex systems in the future.

Why are systems of equations important, you ask? Well, they pop up everywhere in real life! From calculating the optimal mix of ingredients in a recipe to modeling complex physical systems, systems of equations are essential tools in science, engineering, economics, and many other fields. So, by learning how to solve them, you're not just mastering a mathematical skill; you're equipping yourself with a powerful problem-solving tool that can be applied in countless situations. Isn't that cool?

Method 1: The Substitution Method

The substitution method is a clever technique that involves solving one equation for one variable and then substituting that expression into the other equation. This effectively reduces the system to a single equation with a single variable, which we can then easily solve. Let's see how it works with our system:

{3x+2y=125x−y=7{ \begin{cases} 3x+2y = 12\\ 5x-y=7 \end{cases} }

Step 1: Solve one equation for one variable. Look for an equation where it's easy to isolate a variable. In our case, the second equation, 5x - y = 7, looks promising because we can easily solve for y by adding y to both sides and subtracting 7 from both sides:

y = 5x - 7

Great! We've now expressed y in terms of x. This is our key to unlocking the system.

Step 2: Substitute the expression into the other equation. Now, we'll substitute this expression for y into the first equation, 3x + 2y = 12. This is the crucial step where we eliminate one variable:

3x + 2(5x - 7) = 12

Notice how we've replaced y with (5x - 7). We now have an equation with only one variable, x. Time to solve it!

Step 3: Solve the resulting equation. Let's simplify and solve for x:

3x + 10x - 14 = 12 13x - 14 = 12 13x = 26 x = 2

Fantastic! We've found the value of x: x = 2. We're halfway there!

Step 4: Substitute the value back to find the other variable. Now that we know x = 2, we can substitute this value back into either of the original equations (or the expression we found in Step 1) to find y. Let's use the expression y = 5x - 7, as it's already set up to solve for y:

y = 5(2) - 7 y = 10 - 7 y = 3

Awesome! We've found the value of y: y = 3. We now have both x and y.

Step 5: Check your solution. It's always a good idea to check your solution by plugging the values of x and y back into the original equations to make sure they hold true. Let's do that:

Equation 1: 3x + 2y = 12 3(2) + 2(3) = 6 + 6 = 12 (Correct!)

Equation 2: 5x - y = 7 5(2) - 3 = 10 - 3 = 7 (Correct!)

Our solution x = 2 and y = 3 satisfies both equations. We've successfully solved the system using the substitution method!

Method 2: The Elimination Method

The elimination method, also known as the addition method, is another powerful technique for solving systems of equations. The idea here is to manipulate the equations so that when you add them together, one of the variables cancels out (is eliminated). This leaves you with a single equation with a single variable, which you can then solve. Let's apply this method to our system:

{3x+2y=125x−y=7{ \begin{cases} 3x+2y = 12\\ 5x-y=7 \end{cases} }

Step 1: Multiply one or both equations to make the coefficients of one variable opposites. We need to find a variable whose coefficients we can easily make opposites. Notice that the coefficient of y in the first equation is 2, and in the second equation, it's -1. If we multiply the second equation by 2, the coefficient of y will become -2, which is the opposite of 2. So, let's multiply the entire second equation by 2:

2(5x - y) = 2(7) 10x - 2y = 14

Now our system looks like this:

{3x+2y=1210x−2y=14{ \begin{cases} 3x+2y = 12\\ 10x-2y=14 \end{cases} }

See how the coefficients of y are now opposites? This is exactly what we wanted!

Step 2: Add the equations together. Now, we'll add the two equations together, term by term:

(3x + 2y) + (10x - 2y) = 12 + 14 13x = 26

Notice that the y terms have canceled out, leaving us with a single equation in x. This is the magic of the elimination method!

Step 3: Solve the resulting equation. Let's solve for x:

13x = 26 x = 2

Excellent! We've found x = 2. Just like with the substitution method, we're halfway to our solution.

Step 4: Substitute the value back to find the other variable. Now we substitute x = 2 back into either of the original equations to solve for y. Let's use the first equation, 3x + 2y = 12:

3(2) + 2y = 12 6 + 2y = 12 2y = 6 y = 3

Wonderful! We've found y = 3. We have both x and y.

Step 5: Check your solution. As always, let's check our solution by plugging x = 2 and y = 3 back into the original equations:

Equation 1: 3x + 2y = 12 3(2) + 2(3) = 6 + 6 = 12 (Correct!)

Equation 2: 5x - y = 7 5(2) - 3 = 10 - 3 = 7 (Correct!)

Our solution x = 2 and y = 3 satisfies both equations. We've successfully solved the system using the elimination method as well!

The Solution

Using both the substitution and elimination methods, we've arrived at the same solution: x = 2 and y = 3. This means that the point (2, 3) is the intersection of the two lines represented by our equations. Graphically, this is where the two lines cross on a coordinate plane.

We can write our solution as an ordered pair: (2, 3). This clearly shows the values of x and y that satisfy the system of equations.

When to Use Each Method

Now that we've explored both the substitution and elimination methods, you might be wondering when to use each one. Here's a little guidance:

  • Substitution Method: This method is particularly useful when one of the equations is already solved for one variable, or when it's easy to isolate a variable in one of the equations. In our example, solving for y in the second equation (5x - y = 7) was straightforward, making substitution a good choice.
  • Elimination Method: This method shines when the coefficients of one of the variables are either the same or easily made the same (or opposites) by multiplying one or both equations. In our example, multiplying the second equation by 2 made the y coefficients opposites, setting up the elimination method perfectly.

Ultimately, the best method depends on the specific system of equations you're dealing with. With practice, you'll develop an intuition for which method will be the most efficient in each situation. And remember, sometimes you can even use a combination of both methods!

Conclusion

Solving systems of equations might seem daunting at first, but with a little practice and the right techniques, it becomes a manageable and even enjoyable task. We've explored two powerful methods – substitution and elimination – and seen how they can be used to find the solution to a system of two equations with two variables.

Remember the key steps:

  • Substitution: Solve for one variable, substitute into the other equation, solve for the remaining variable, and substitute back to find the first variable.
  • Elimination: Manipulate the equations to make the coefficients of one variable opposites, add the equations, solve for the remaining variable, and substitute back to find the first variable.

Don't be afraid to try both methods and see which one works best for you in different situations. And most importantly, practice, practice, practice! The more you work with systems of equations, the more confident and skilled you'll become.

So, there you have it! You're now equipped with the knowledge and skills to solve systems of equations like a pro. Go forth and conquer those mathematical challenges! Keep practicing, keep exploring, and have fun with math! You got this!