Solving Linear Equations With Elimination Method A Step-by-Step Guide
Hey guys! Let's dive into the fascinating world of linear equations and learn how to solve them using the elimination method. This method is super handy for finding the values of unknown variables when you have a pair of linear equations. We're going to break it down step by step, so you can ace your math problems in no time. So, let’s explore how we can solve linear equations effectively using the elimination method. This method is crucial for anyone looking to strengthen their algebra skills. Understanding the elimination method not only helps in solving academic problems but also builds a solid foundation for more advanced mathematical concepts. Linear equations are the backbone of numerous real-world applications, making this a must-know technique. Remember, the elimination method involves manipulating the equations so that either the x or y coefficients are the same (or opposites). This way, when you add or subtract the equations, one variable gets eliminated, leaving you with a single variable equation that’s easy to solve. Solving linear equations can sometimes feel like navigating a maze, but with the elimination method, you gain a powerful tool to simplify the process. The beauty of this method lies in its systematic approach, ensuring you don't miss any steps and arrive at the correct solution. It’s like having a roadmap that guides you through the problem, making the journey much smoother and more efficient. Mastering the elimination method involves a bit of practice, but the rewards are well worth the effort. You'll not only improve your problem-solving speed but also develop a deeper understanding of algebraic manipulations. This skill is invaluable for anyone pursuing studies in mathematics, physics, engineering, or any field that relies heavily on quantitative analysis.
Problem 1: Solving 3x - 2y + 4 = 0 and 2x - 3y = 11
Step 1: Rewrite the Equations
First, let’s make sure our equations are in the standard form, which is Ax + By = C. We have:
- Equation 1: 3x - 2y = -4
- Equation 2: 2x - 3y = 11
Step 2: Make the Coefficients Match
Our goal here is to make either the x or y coefficients the same (or opposites) in both equations. Let’s target the x coefficients. We can multiply Equation 1 by 2 and Equation 2 by 3:
- New Equation 1: 2 * (3x - 2y) = 2 * (-4) => 6x - 4y = -8
- New Equation 2: 3 * (2x - 3y) = 3 * (11) => 6x - 9y = 33
Step 3: Eliminate a Variable
Now that the x coefficients are the same (both are 6), we can subtract New Equation 2 from New Equation 1:
(6x - 4y) - (6x - 9y) = -8 - 33 6x - 4y - 6x + 9y = -41 5y = -41
Step 4: Solve for y
Divide both sides by 5 to solve for y:
y = -41 / 5
Step 5: Substitute and Solve for x
Now that we have the value of y, we can substitute it back into either of the original equations to find x. Let’s use Equation 1:
3x - 2(-41/5) = -4 3x + 82/5 = -4 3x = -4 - 82/5 3x = (-20 - 82) / 5 3x = -102 / 5 x = (-102 / 5) / 3 x = -34 / 5
So, the solution is x = -34/5 and y = -41/5. Understanding the step-by-step process of solving linear equations using the elimination method is crucial for mastering algebra. This method allows us to systematically reduce a system of two equations with two variables into a single equation with one variable, making it much easier to solve. By targeting the coefficients of either x or y, we can manipulate the equations through multiplication to create matching coefficients. This crucial step sets the stage for the elimination process. The act of eliminating a variable, either by addition or subtraction, is where the elimination method gets its name and its power. This simplification allows us to isolate and solve for the remaining variable. Once we've solved for one variable, substituting it back into one of the original equations allows us to find the value of the other variable. This substitution step highlights the interconnectedness of the two equations and ensures that we find a solution that satisfies both. The solution to the system of equations is the ordered pair (x, y) that satisfies both equations simultaneously. It represents the point where the two lines intersect on a graph. Understanding the geometric interpretation of solving systems of equations can provide a deeper insight into the algebraic process. The elimination method is not just a mathematical procedure; it's a systematic approach to problem-solving that can be applied in various contexts. Whether you're solving for unknowns in a scientific experiment or optimizing resources in a business scenario, the skills you gain from mastering the elimination method are invaluable.
Problem 2 (i): Solving 3x - 5y = 4 and 2y + 7 = 9x
Step 1: Rewrite the Equations
Again, let’s rewrite the equations in the standard form Ax + By = C:
- Equation 1: 3x - 5y = 4
- Equation 2: 9x - 2y = 7
Step 2: Make the Coefficients Match
Let’s make the x coefficients match. We can multiply Equation 1 by 3:
- New Equation 1: 3 * (3x - 5y) = 3 * (4) => 9x - 15y = 12
- Equation 2: 9x - 2y = 7
Step 3: Eliminate a Variable
Now, subtract Equation 2 from New Equation 1:
(9x - 15y) - (9x - 2y) = 12 - 7 9x - 15y - 9x + 2y = 5 -13y = 5
Step 4: Solve for y
Divide both sides by -13 to solve for y:
y = -5 / 13
Step 5: Substitute and Solve for x
Substitute the value of y back into Equation 1:
3x - 5(-5/13) = 4 3x + 25/13 = 4 3x = 4 - 25/13 3x = (52 - 25) / 13 3x = 27 / 13 x = (27 / 13) / 3 x = 9 / 13
So, the solution is x = 9/13 and y = -5/13. In this specific problem, the strategic manipulation of equations to match coefficients is a key illustration of the elimination method's effectiveness. By focusing on the x coefficients, we were able to streamline the elimination process and arrive at a solution more efficiently. This highlights the flexibility of the elimination method – sometimes targeting one variable over another can simplify the algebra involved. The decision to eliminate x in this case simplified the subsequent steps, as it avoided dealing with larger numbers that could have arisen from manipulating the y coefficients instead. This thoughtful approach to problem-solving is an important aspect of mathematical proficiency. The substitution step, where the value of y is plugged back into the equation, is a crucial validation point. It ensures that the solutions we find for x and y are consistent across both original equations. This consistency check is a hallmark of careful algebraic work and helps prevent errors. The values x = 9/13 and y = -5/13 represent the unique point of intersection for the two lines defined by the original equations. This geometric interpretation provides a visual way to understand the solution to the system. It's also worth noting that, while this solution may not be immediately intuitive due to the fractions involved, it satisfies both equations and represents the precise answer. Problems like these reinforce the importance of being comfortable working with fractions in algebraic manipulations. The elimination method, as demonstrated here, is a powerful tool not only for solving specific equations but also for developing mathematical intuition and problem-solving skills that are applicable in many other areas.
Problem 2 (ii): Solving 4x - 3y - 8 = 0 and 6x - y - 29 = 0
Step 1: Rewrite the Equations
Let's rewrite the equations in the standard form Ax + By = C:
- Equation 1: 4x - 3y = 8
- Equation 2: 6x - y = 29
Step 2: Make the Coefficients Match
This time, let’s make the y coefficients match. To do this, we can multiply Equation 2 by -3:
- Equation 1: 4x - 3y = 8
- New Equation 2: -3 * (6x - y) = -3 * (29) => -18x + 3y = -87
Step 3: Eliminate a Variable
Now, add Equation 1 and New Equation 2:
(4x - 3y) + (-18x + 3y) = 8 + (-87) 4x - 3y - 18x + 3y = -79 -14x = -79
Step 4: Solve for x
Divide both sides by -14 to solve for x:
x = 79 / 14
Step 5: Substitute and Solve for y
Substitute the value of x back into Equation 2:
6(79/14) - y = 29 474/14 - y = 29 237/7 - y = 29 y = 237/7 - 29 y = (237 - 203) / 7 y = 34 / 7
Thus, the solution is x = 79/14 and y = 34/7. This final problem underscores the importance of strategic decision-making within the elimination method. Choosing to match the y coefficients in this instance led to a different, yet equally effective, path to the solution. This highlights the flexibility and adaptability of the method. The process of multiplying Equation 2 by -3 is a clever maneuver, as it not only matches the y coefficients but also sets them up to cancel out through addition, simplifying the equation. This is a prime example of how algebraic manipulations can be used to streamline problem-solving. Adding the equations together to eliminate y demonstrates another facet of the elimination method's versatility. It's not always about subtraction; sometimes addition is the more efficient route. The subsequent steps, where we solve for x and then substitute back to find y, reinforce the systematic nature of the method. Each step builds upon the previous one, leading to a well-defined solution. The solutions x = 79/14 and y = 34/7, while again involving fractions, are the precise values that satisfy both equations. This demonstrates the power of the elimination method to handle even non-integer solutions. Throughout this problem, we've seen how the elimination method is not just a rote procedure but a thoughtful approach to algebraic problem-solving. It requires careful planning, strategic manipulation, and a systematic execution to arrive at the correct solution.
And there you have it! We’ve successfully solved pairs of linear equations using the elimination method. Remember, practice makes perfect, so keep working on these problems. You’ve got this! By mastering the elimination method, you gain a versatile tool for solving linear equations, applicable in various fields beyond the classroom. Keep practicing, and you'll become a pro at solving these types of problems!
