Solving Systems Of Equations A Step By Step Guide
In the realm of mathematics, solving systems of equations is a fundamental skill with applications across various disciplines. Systems of equations arise when we have multiple equations with multiple unknowns, and our goal is to find the values of these unknowns that satisfy all equations simultaneously. This article delves into a step-by-step approach to solving the given system of equations:
0. 14x - 0.03y = -0.66
1. 05x + 0.12y = 0.81
We will explore the methods of elimination and substitution, providing a comprehensive understanding of the solution process. Let's embark on this mathematical journey together.
Understanding the Problem
Before diving into the solution, it's crucial to understand the problem at hand. We are presented with two linear equations, each containing two variables, x and y. The objective is to find the values of x and y that make both equations true. This point (x, y) represents the intersection of the two lines represented by the equations on a coordinate plane.
Linear equations, in their essence, describe straight lines when graphed. The beauty of a system of two linear equations lies in the fact that their solution corresponds to the point where the two lines intersect. Geometrically, if the lines intersect at a single point, the system has a unique solution. If the lines are parallel, there is no solution, and if they are the same line, there are infinitely many solutions. Before we solve, let's explore the methods available to us. The two primary algebraic techniques are substitution and elimination. We'll delve into each method with the aim of demystifying the process and turning you into a proficient equation solver. Understanding the underlying principles and the strategic application of these methods is key to unlocking the solutions to a wide array of mathematical problems.
Method 1 Elimination Method
The elimination method is a powerful technique for solving systems of equations. This method involves manipulating the equations to eliminate one variable, allowing us to solve for the other. The core idea is to multiply one or both equations by constants so that the coefficients of one variable are opposites. When we add the equations, that variable is eliminated, simplifying the system. Let's apply this method to our system:
0. 14x - 0.03y = -0.66
1. 05x + 0.12y = 0.81
Step 1: Clear the Decimals
To simplify the equations, let's first eliminate the decimals by multiplying both equations by 100:
1. 14x - 3y = -66 (Multiply the first equation by 100)
2. 5x + 12y = 81 (Multiply the second equation by 100)
Clearing the decimals immediately makes our equations friendlier to handle. By getting rid of these fractional coefficients, we set the stage for easier multiplication and addition steps later on. This seemingly small tweak can greatly reduce the chances of arithmetic errors and make the overall solution process smoother. The importance of this preparatory step cannot be overstated, especially in more complex systems of equations where the numbers involved can become unwieldy. By clearing decimals, we transform the problem into a more manageable form, making the subsequent steps of the elimination method far less cumbersome. This tactical approach underscores the value of looking for simplification opportunities at the outset of a problem.
Step 2: Eliminate y
To eliminate y, we need the coefficients of y to be opposites. Multiply the first equation by 4:
3. 4(14x - 3y) = 4(-66)
4. 56x - 12y = -264
Now we have:
5. 56x - 12y = -264
6. 5x + 12y = 81
The decision to eliminate y in this step is strategic. By observing the equations, we notice that the coefficients of y (-3 and 12) are easily manipulated to become additive inverses. Multiplying the first equation by 4 gives us -12y, which is the exact opposite of the 12y in the second equation. This clever manipulation sets the stage for a clean elimination when the equations are added together. Focusing on a variable with easily adjustable coefficients is a key problem-solving heuristic in the elimination method. It minimizes the complexity of the multipliers needed and streamlines the process. By targeting y for elimination, we've paved the way for a straightforward algebraic maneuver that will simplify the system and bring us closer to the solution.
Step 3: Add the Equations
Add the two equations together:
(56x - 12y) + (5x + 12y) = -264 + 81
This simplifies to:
6. 1x = -183
The addition of the equations is where the magic of the elimination method truly shines. By carefully manipulating the coefficients, we've created a situation where the y terms cancel each other out perfectly. This leaves us with a single equation in a single variable, x, which is easily solvable. The beauty of this step lies in its simplicity and directness. We've effectively reduced a two-variable problem to a one-variable problem, making the solution readily accessible. The act of adding equations together might seem like a basic algebraic operation, but in the context of systems of equations, it's a powerful tool for unraveling complexity. It's a testament to the elegance of mathematical methods that a strategic addition can lead to such a profound simplification, bringing us one giant leap closer to the values of x and y that satisfy both equations.
Step 4: Solve for x
Divide both sides by 61:
x = -183 / 61
x = -3
Solving for x after the elimination step is a moment of triumph in the problem-solving process. Having successfully eliminated y, we are left with a straightforward equation that directly reveals the value of x. The division operation isolates x, giving us a concrete numerical answer. This clarity and precision are hallmarks of the elimination method when it's applied strategically. The value of x = -3 is not just a number; it's a crucial piece of the puzzle that we're assembling. It represents the x-coordinate of the point where the two lines described by our original equations intersect. This tangible result reinforces the power of algebraic manipulation to uncover hidden relationships and solutions within mathematical systems. With x in hand, we're now poised to find y and complete our solution.
Step 5: Substitute x into one of the original equations to solve for y
Let's use the second original equation:
7. 05(-3) + 0.12y = 0.81
8. -0.15 + 0.12y = 0.81
Adding 0.15 to both sides:
9. 12y = 0.96
Dividing by 0.12:
y = 0.96 / 0.12
y = 8
Substituting the value of x back into one of the original equations is a pivotal step in completing the solution. It's the moment where we leverage the information gained from solving for one variable to unlock the value of the other. By choosing the second original equation, we embark on a journey of substitution and simplification. The beauty of this process lies in its ability to transform a seemingly complex equation into a solvable one. We replace x with its numerical value, and suddenly the equation contains only one unknown, y. The subsequent steps of adding and dividing are mechanical, but each one brings us closer to the final answer. The value y = 8 is not just a number; it's the y-coordinate that, together with x = -3, forms the complete solution to our system of equations. This act of substitution is a testament to the interconnectedness of mathematical concepts, where the solution of one part illuminates the path to the whole.
Method 2 Substitution Method
The substitution method is another powerful technique for solving systems of equations. This method involves solving one equation for one variable and substituting that expression into the other equation. This results in a single equation with a single variable, which can be easily solved. Let's apply this method to our system:
0. 14x - 0.03y = -0.66
1. 05x + 0.12y = 0.81
Step 1: Clear the Decimals
As before, let's clear the decimals by multiplying both equations by 100:
10. 14x - 3y = -66
11. 5x + 12y = 81
Step 2: Solve one equation for one variable
Let's solve the first equation for x:
12. 14x = 3y - 66
13. x = (3y - 66) / 14
Choosing to solve for x in the first equation is a strategic decision that can streamline the substitution process. By isolating x, we create a direct expression for its value in terms of y. This expression will serve as the key to unlocking the solution in the next step. The act of solving for a variable is a fundamental algebraic skill, and in the context of substitution, it's particularly important to choose the variable and equation that will lead to the simplest expression. Sometimes, this might involve looking for coefficients that are easy to work with or equations where a variable is already close to being isolated. The expression x = (3y - 66) / 14 is more than just a formula; it's a bridge that connects the two equations in our system. It allows us to express one variable in terms of the other, paving the way for a substitution that will simplify the problem.
Step 3: Substitute the expression for x into the second equation
Substitute x = (3y - 66) / 14 into the second equation:
14. 5((3y - 66) / 14) + 12y = 81
The substitution of the expression for x into the second equation is the heart of the substitution method. It's the step where we transform a two-variable problem into a single-variable problem. By replacing x with its equivalent expression in terms of y, we create an equation that contains only y as an unknown. This act of substitution is a powerful technique for simplifying systems of equations, and it requires careful attention to detail to ensure that the expression is correctly placed and the equation is properly balanced. The resulting equation, 5((3y - 66) / 14) + 12y = 81, might look complex at first glance, but it's a pivotal step in the solution process. It's the gateway to finding the value of y, which will then allow us to determine the value of x. This substitution is not just a mechanical process; it's a strategic move that unlocks the solution by reducing the complexity of the system.
Step 4: Simplify and solve for y
Multiply both sides by 14 to eliminate the fraction:
15. 5(3y - 66) + 168y = 1134
16. 15y - 330 + 168y = 1134
17. 183y = 1464
18. y = 1464 / 183
19. y = 8
Simplifying and solving for y is the culmination of the substitution process. It's the stage where we unravel the equation we created in the previous step and isolate y. The initial act of multiplying both sides by 14 is a strategic move to eliminate the fraction, making the equation easier to handle. The subsequent steps of distributing, combining like terms, and isolating y are all standard algebraic techniques, but they are crucial for arriving at the correct solution. Each manipulation brings us closer to the value of y, and the final result, y = 8, is a significant milestone in our problem-solving journey. This value is not just a number; it's the y-coordinate of the point where the two lines described by our original equations intersect. With y in hand, we're now ready to complete the solution by finding the corresponding value of x.
Step 5: Substitute the value of y back into the expression for x
Substitute y = 8 into x = (3y - 66) / 14:
20. x = (3(8) - 66) / 14
21. x = (24 - 66) / 14
22. x = -42 / 14
23. x = -3
The final act of substituting the value of y back into the expression for x is the concluding step in the substitution method. It's the moment where we leverage the value we found for y to unlock the value of x. By plugging y = 8 into the equation x = (3y - 66) / 14, we set in motion a series of arithmetic operations that lead us to the value of x. The calculations are straightforward, but they are imbued with the satisfaction of bringing the solution to completion. The result, x = -3, is not just a number; it's the x-coordinate that, together with y = 8, forms the complete solution to our system of equations. This final substitution is a testament to the interconnectedness of the variables in the system, and it underscores the power of algebraic manipulation to reveal hidden relationships and solutions. With both x and y in hand, we can confidently declare that we have solved the system.
Solution
Both methods lead us to the same solution:
x = -3
y = 8
Therefore, the solution to the system of equations is x = -3 and y = 8. This means that the point (-3, 8) is the intersection of the two lines represented by the equations. This is the one and only pair of values for x and y that satisfies both equations simultaneously. The fact that both the elimination and substitution methods converge on the same solution underscores the consistency and reliability of these algebraic techniques. It also provides a level of confidence in our answer, knowing that we have arrived at the solution through two independent paths.
Checking the Solution: It's always a good practice to check our solution by substituting the values of x and y back into the original equations. If both equations hold true, then we can be confident that we have found the correct solution.
Conclusion
Solving systems of equations is a fundamental skill in mathematics with applications in various fields. We have explored two powerful methods, elimination and substitution, to solve the given system of equations. Both methods demonstrate the elegance and effectiveness of algebraic techniques in finding solutions to complex problems. Understanding these methods is crucial for tackling more advanced mathematical concepts and real-world applications.
In conclusion, mastering the art of solving systems of equations is a valuable asset in any mathematical toolkit. Whether you prefer the strategic elimination of variables or the direct substitution of expressions, the ability to solve these systems opens doors to a deeper understanding of mathematics and its applications. The solution x = -3 and y = 8 is not just a pair of numbers; it's a testament to the power of algebraic manipulation and the beauty of mathematical problem-solving.
