Solving A System Of Equations: 6x - Y = 4 And 4x - 8 = 9y - 5

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Solving a System of Equations: 6x - y = 4 and 4x - 8 = 9y - 5

Hey guys! Today, we're diving into a classic math problem: solving a system of equations. Specifically, we're going to tackle the following system:

  • 6x - y = 4
  • 4x - 8 = 9y - 5

Systems of equations might seem intimidating at first, but don't worry! We'll break it down step by step, making it super easy to understand. We'll explore different methods to solve this, ensuring you're equipped to handle similar problems in the future. So, grab your pencils, 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. Simply put, it's a set of two or more equations that involve the same variables. Our goal is to find the values of these variables that satisfy all the equations in the system simultaneously. Think of it like finding the sweet spot where all the equations agree.

In our case, we have two equations with two variables, x and y. This is a common scenario, and we have several techniques at our disposal to find the solution. The solution will be a pair of values (x, y) that make both equations true.

Why are these important?

You might be wondering, "Why bother learning this stuff?" Well, systems of equations pop up in all sorts of real-world situations! From calculating the optimal mix of ingredients in a recipe to modeling complex physical systems, they're incredibly versatile tools. Mastering them opens doors to solving a wide range of problems.

For example, imagine you're running a business and need to figure out the break-even point – the point where your revenue equals your costs. This often involves setting up and solving a system of equations. Or, consider a scenario in physics where you're analyzing the motion of two objects. You might use a system of equations to describe their positions and velocities over time.

Key takeaway: Systems of equations are more than just abstract math problems; they're powerful tools for understanding and solving real-world challenges.

Method 1: Substitution

One of the most common methods for solving systems of equations is substitution. The basic idea is to solve one equation for one variable and then substitute that expression into the other equation. This eliminates one variable, leaving us with a single equation in one variable, which we can easily solve. Let's see how this works with our system:

  1. 6x - y = 4
  2. 4x - 8 = 9y - 5

Step-by-Step with Substitution

Step 1: Solve one equation for one variable.

Let's take the first equation (6x - y = 4) and solve it for y. This seems like a straightforward choice because the coefficient of y is -1, which will make the algebra a bit cleaner.

Add y to both sides and subtract 4 from both sides:

6x - 4 = y

So, we have y = 6x - 4.

Step 2: Substitute the expression into the other equation.

Now, we'll substitute this expression for y (6x - 4) into the second equation (4x - 8 = 9y - 5):

4x - 8 = 9(6x - 4) - 5

Notice that we've replaced y with the expression we found in step 1. This new equation only involves the variable x.

Step 3: Solve the resulting equation.

Now we need to simplify and solve for x:

4x - 8 = 54x - 36 - 5

Combine like terms:

4x - 8 = 54x - 41

Subtract 4x from both sides:

-8 = 50x - 41

Add 41 to both sides:

33 = 50x

Divide both sides by 50:

x = 33/50

So, we've found the value of x!

Step 4: Substitute the value back to find the other variable.

Now that we know x = 33/50, we can plug this value back into either of the original equations (or the expression we found for y in step 1) to find y. Let's use the expression y = 6x - 4:

y = 6(33/50) - 4

Simplify:

y = 198/50 - 4

y = 198/50 - 200/50

y = -2/50

y = -1/25

So, we've found y = -1/25.

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 both original equations to make sure they hold true. Let's do that:

  • Equation 1: 6x - y = 4
    • 6(33/50) - (-1/25) = 198/50 + 1/25 = 198/50 + 2/50 = 200/50 = 4 (Correct!)
  • Equation 2: 4x - 8 = 9y - 5
    • 4(33/50) - 8 = 132/50 - 400/50 = -268/50
    • 9(-1/25) - 5 = -9/25 - 125/25 = -134/25 = -268/50 (Correct!)

Since our solution satisfies both equations, we know we've found the correct answer.

Solution:

Therefore, the solution to the system of equations is x = 33/50 and y = -1/25, or as an ordered pair, (33/50, -1/25).

Guys, that was a bit of work, but we did it! We successfully solved the system of equations using the substitution method. Remember, the key is to carefully follow each step and double-check your work to avoid errors. You got this!

Method 2: Elimination

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

  1. 6x - y = 4
  2. 4x - 8 = 9y - 5

Step-by-Step with Elimination

Step 1: Rearrange the equations (if necessary).

To make the elimination method work smoothly, it's helpful to have the x and y terms aligned on the same side of the equation, and the constant terms on the other side. Let's rewrite the second equation to match the form of the first equation:

4x - 8 = 9y - 5

Add 8 to both sides:

4x = 9y + 3

Subtract 9y from both sides:

4x - 9y = 3

Now our system looks like this:

  1. 6x - y = 4
  2. 4x - 9y = 3

Step 2: Multiply one or both equations to make the coefficients of one variable opposites.

Our goal is to make either the x coefficients or the y coefficients opposites (e.g., 6 and -6, or 2 and -2). Let's focus on the y coefficients. In the first equation, the coefficient of y is -1, and in the second equation, it's -9. To make them opposites, we can multiply the first equation by -9:

-9 * (6x - y) = -9 * 4

-54x + 9y = -36

Now our system is:

  1. -54x + 9y = -36
  2. 4x - 9y = 3

Notice that the y coefficients are now 9 and -9, which are opposites!

Step 3: Add the equations together.

Now we add the two equations vertically:

 -54x + 9y = -36
+  4x - 9y = 3
----------------
 -50x + 0y = -33

The y terms cancel out, leaving us with:

-50x = -33

Step 4: Solve the resulting equation.

Divide both sides by -50:

x = -33/-50

x = 33/50

We've found x again! Notice that we got the same value for x as we did using the substitution method, which is a good sign.

Step 5: Substitute the value back to find the other variable.

Just like with substitution, we now plug x = 33/50 back into either of the original equations to find y. Let's use the first original equation (6x - y = 4):

6(33/50) - y = 4

198/50 - y = 4

Subtract 198/50 from both sides:

-y = 4 - 198/50

-y = 200/50 - 198/50

-y = 2/50

y = -2/50

y = -1/25

We found y = -1/25, which matches our result from the substitution method!

Step 6: Check your solution.

As before, it's crucial to check our solution by plugging the values of x and y back into both original equations. We already did this in the substitution method section, and we know the solution checks out.

Solution:

Therefore, the solution to the system of equations is x = 33/50 and y = -1/25, or as an ordered pair, (33/50, -1/25).

Awesome! We've successfully solved the same system of equations using the elimination method. See, there's more than one way to crack a math problem!

Method 3: Graphing (Conceptual)

While not always the most precise method for finding exact solutions, graphing provides a valuable visual understanding of systems of equations. The idea is simple: each equation represents a line on a graph. The solution to the system is the point where the lines intersect. If the lines don't intersect (they're parallel), the system has no solution. If the lines are the same, there are infinitely many solutions.

Let's think about how this would apply to our system:

  1. 6x - y = 4
  2. 4x - 8 = 9y - 5

To graph these equations, we could rewrite them in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

  • Equation 1: 6x - y = 4 => y = 6x - 4
  • Equation 2: 4x - 8 = 9y - 5 => 9y = 4x - 3 => y = (4/9)x - 1/3

If we were to graph these two lines, we would see that they intersect at the point (33/50, -1/25), which confirms our solutions obtained through substitution and elimination.

Key takeaway: Graphing provides a visual representation of the solution, making it easier to understand the relationship between the equations. While it might not always give you the exact answer, it's a great tool for visualizing the problem and estimating the solution.

Choosing the Best Method

So, we've explored three different methods for solving systems of equations: substitution, elimination, and graphing. You might be wondering, "Which method is the best one?" The truth is, there's no single "best" method. The most efficient approach often depends on the specific system of equations you're dealing with.

  • Substitution: This method is particularly useful when one of the equations is already solved (or easily solved) for one variable in terms of the other. For example, if you have an equation like y = 3x + 2, substitution is a natural choice.
  • Elimination: Elimination shines when the coefficients of one of the variables are opposites or easy to make opposites by multiplication. It's also a good option when both equations are in standard form (Ax + By = C).
  • Graphing: Graphing is excellent for visualizing the system and understanding the nature of the solutions. It's particularly helpful when you want to estimate the solution or determine if the system has no solution or infinitely many solutions. However, it might not be the most accurate method for finding exact solutions, especially if the solutions are not integers.

In our example, both substitution and elimination worked well. Graphing would have provided a visual confirmation of our solution. The best advice I can give you is to practice using all three methods so you can develop a feel for which one is most appropriate in different situations.

Practice Makes Perfect

Alright, you guys, we've covered a lot of ground! We've explored what systems of equations are, why they're important, and three different methods for solving them. But the real key to mastering this topic is practice. The more you work through problems, the more comfortable and confident you'll become.

I encourage you to find more examples of systems of equations and try solving them using all three methods. Pay attention to which methods seem to work best for different types of problems. Don't be afraid to make mistakes – that's how we learn! And remember, if you get stuck, there are tons of resources available online and in textbooks to help you out.

Solving systems of equations is a fundamental skill in mathematics, and it's a skill that will serve you well in many areas of your life. So keep practicing, keep exploring, and keep having fun with math!

Conclusion

In this article, we've thoroughly explored how to solve the system of equations 6x - y = 4 and 4x - 8 = 9y - 5. We've demonstrated the substitution and elimination methods, providing step-by-step explanations and checks to ensure accuracy. We also touched upon the graphing method for a visual understanding. Remember, the best method depends on the specific problem, so practice is key to mastering these techniques. Keep up the great work, and happy problem-solving!