Solving Systems: Isolating X Or Y

by Alex Johnson 34 views

Let's dive into the world of system of equations, a fundamental concept in algebra! We're going to take a look at a specific problem: setting up equations with either the x or y variable isolated. This is a crucial step in solving these systems. We'll explore how to manipulate the given equations to our advantage. The core idea is to eliminate one variable, allowing us to solve for the other. We then use the result to solve for the remaining variable. It is a methodical process. This method, often called the elimination method, simplifies the problem, making it easier to find the solution. Remember that solving a system of equations means finding the values of x and y that satisfy all the equations simultaneously. Let's make this clear and straightforward so that you can tackle these problems confidently. With a little practice, you'll be solving these equations like a pro! So, buckle up, and let's get started. We'll break down each step so that you understand the underlying concepts.

Setting Up the Equations

Our system of equations is:

  1. 4x + 2y = -34
  2. -6x + 2y = 6

Our goal is to eliminate either x or y. Looking at the equations, we notice that the y terms have the same coefficient (2). This makes it easy to eliminate y. We can subtract one equation from the other. Alternatively, you could multiply one or both equations by a constant and then either add or subtract them. It is important to know the flexibility. This is based on the specific coefficients in each equation. Here's how to eliminate y: Subtract equation 2 from equation 1. This means, (4x + 2y) - (-6x + 2y) = -34 - 6. Now, let's simplify this. This gives us 4x + 6x = -40, which simplifies to 10x = -40. In this step, we've successfully eliminated y, leaving us with a simple equation in terms of x. Solving for x is now a breeze. Remember that there are many different methods that you can use. You can also solve for y first. It just depends on the specific equations given. The process is very flexible.

Now, let's solve for x.

Dividing both sides of 10x = -40 by 10, we get x = -4. The value of x that satisfies both equations. We are halfway there! We've found the value of one variable. Now, we'll substitute this value back into either of the original equations. This will allow us to find the value of y. This step is crucial. This helps us ensure that our solution is correct. We could have done the same process and eliminated x first. You can always select what is best for the situation.

Solving for y

Let's substitute x = -4 into the first equation: 4x + 2y = -34. This becomes 4(-4) + 2y = -34, which simplifies to -16 + 2y = -34. Now, add 16 to both sides, which gives us 2y = -18. Finally, divide both sides by 2 to solve for y. We then get y = -9. Therefore, the solution to the system of equations is x = -4 and y = -9. This means that the point (-4, -9) is where the two lines represented by the equations intersect on a graph. This is the only point that satisfies both equations simultaneously. So, always remember to verify your answer by substituting the values of x and y back into the original equations. This is a great way to catch any errors. Let's take a moment and test our solution. For equation 1: 4(-4) + 2(-9) = -16 - 18 = -34. This matches the right-hand side of the equation. Similarly, for equation 2: -6(-4) + 2(-9) = 24 - 18 = 6. This also matches. It is a great method to confirm that your solution is correct. This is how you confidently solve a system of equations by isolating x or y.

Step-by-Step Guide

Let's break down the process into easy-to-follow steps:

  1. Identify the Equations: We have two equations: 4x + 2y = -34 and -6x + 2y = 6.
  2. Choose a Variable to Eliminate: In this case, y is the easiest to eliminate because the coefficients are the same. In some cases, you may need to manipulate the equations. It is important to know that you must multiply one or both equations by a constant to make the coefficients of either x or y opposites.
  3. Eliminate the Chosen Variable: Subtract the second equation from the first: (4x + 2y) - (-6x + 2y) = -34 - 6. This simplifies to 10x = -40.
  4. Solve for the Remaining Variable: Divide both sides of 10x = -40 by 10 to get x = -4.
  5. Substitute to Find the Other Variable: Substitute x = -4 into one of the original equations. We used the first equation: 4(-4) + 2y = -34. Simplify to find y = -9.
  6. Verify the Solution: Plug both x and y back into both original equations to make sure they work. Always be careful to do this step. It avoids many mistakes.

This method is effective because it reduces a two-variable problem to a one-variable problem. It's an efficient way to find the values that satisfy all given equations. The process is very flexible. Sometimes, you may not want to eliminate the y variable. All of this depends on the specific problems you are working on. You will get more comfortable as you practice. This method is fundamental to algebra.

Practice Problems

To solidify your understanding, let's look at some practice problems. Try solving these on your own, and then compare your solutions with the solutions given.

  1. 2x + y = 7 and x - y = 2
  2. 3x - 2y = 1 and x + 2y = 7
  3. 5x + 3y = 8 and 2x - 3y = -1

Remember to first eliminate a variable by either adding or subtracting the equations. Then, solve for the remaining variable. Substitute the value back into one of the original equations to find the other variable. Always check your answers by substituting the values into both original equations. Practicing these problems will improve your ability to solve systems of equations efficiently.

Solutions

  1. Solution: x = 3, y = 1. To solve, add the two equations together. This eliminates y and gives you 3x = 9, which leads to x = 3. Substituting x back into one of the equations gives you y = 1.
  2. Solution: x = 2, y = 5. Add the two equations. This eliminates y and gives you 4x = 8, which means x = 2. Substituting x back into one of the equations gives you y = 5.
  3. Solution: x = 1, y = 1. Add the two equations together, eliminating y, which gives you 7x = 7, so x = 1. Then substitute x back into one of the equations to find y = 1.

These practice problems demonstrate how to solve systems of equations using the elimination method. Always be careful in solving the equations. Always double-check your answers. The process becomes easier with practice. It is fundamental in algebra.

Other Methods

While we focused on the elimination method, other methods can be used to solve systems of equations. The substitution method involves solving one equation for one variable and substituting that expression into the other equation. The graphing method involves graphing both equations and finding the point of intersection. Each method has its advantages. The most appropriate method depends on the specific equations. In the substitution method, one of the equations is solved for one variable and substituted into the other equation. This process eliminates one of the variables. The result is a single-variable equation that can be easily solved. For example, if you have y = 2x + 1 and x + y = 4, you substitute the first equation for y in the second equation. This gives you x + (2x + 1) = 4. Solving this equation gives you x = 1. Then, you can substitute back to find y. The graphing method involves graphing both equations on the same coordinate plane. The solution to the system is the point where the lines intersect. This method is visual. It helps to understand the graphical representation. While the elimination method is often the most efficient for systems with simple coefficients, the substitution method is great if one of the equations is easily solved for one of the variables. The graphing method is very visual. It is great for understanding the concepts.

Conclusion

Mastering how to set up and solve systems of equations. It is a fundamental skill in algebra. The elimination method is a powerful and versatile tool. It allows you to systematically solve for unknown variables. This method is a crucial step in algebra. We have covered the step-by-step process. We have provided practice problems to help you solidify your understanding. Remember to practice regularly and explore different methods. Keep working on these problems. As you continue to work on these problems, you will become more confident and accurate. Always remember to check your solutions. You are well-equipped to tackle more complex algebraic challenges. Keep practicing! You've got this!

For more in-depth explanations and additional practice problems, check out this trusted resource: Khan Academy - Systems of Equations