Solving Systems Of Equations Find The Solution For 12x + 15y = 34 And -6x + 5y = 3

Introduction

In mathematics, systems of equations are a fundamental concept, appearing in various fields, from algebra and calculus to linear programming and beyond. Solving a system of equations means finding the values for the unknown variables that satisfy all equations simultaneously. This article delves into the process of finding the solution for a specific system of linear equations. Our focus will be on employing the elimination method, a powerful technique for solving systems of equations. Understanding these methods is crucial for anyone studying mathematics, engineering, or related fields. This article will provide a comprehensive guide on how to tackle such problems, ensuring that you grasp the underlying principles and can apply them confidently. We will explore each step in detail, making it easy to follow even if you're new to the topic. The core of solving systems of equations lies in manipulating the equations in such a way that we can eliminate one variable, leaving us with a simpler equation in one variable that can be easily solved. Once we find the value of one variable, we can substitute it back into the original equations to find the value of the other variable. This process is both elegant and effective, and it forms the backbone of many mathematical and scientific calculations. So, let’s dive in and solve the given system of equations step-by-step.

Problem Statement

We are given the following system of linear equations:

egin{aligned} 12x + 15y &= 34 \\ -6x + 5y &= 3 \end{aligned}

Our goal is to find the values of x and y that satisfy both equations. To achieve this, we will use the elimination method. This method involves manipulating the equations so that when they are added together, one of the variables cancels out. This leaves us with a single equation in one variable, which we can then solve. Once we have the value of one variable, we can substitute it back into either of the original equations to solve for the other variable. This systematic approach allows us to break down a complex problem into simpler steps, making it easier to find the solution. The elimination method is particularly useful when dealing with systems of equations where the coefficients of one of the variables are multiples of each other, or can be easily made multiples of each other. In this case, we can see that the coefficients of x in the two equations are 12 and -6, which are multiples of each other. This makes the elimination method a very efficient choice for solving this system. By carefully planning our steps and paying attention to the arithmetic, we can arrive at the correct solution and gain a deeper understanding of how systems of equations work.

Solution using the Elimination Method

Step 1: Multiply the second equation by 2

To eliminate the variable x, we can multiply the second equation by 2. This will make the coefficient of x in the second equation -12, which is the opposite of the coefficient of x in the first equation. When we add the two equations together, the x terms will cancel out, leaving us with an equation in just y. This is the key idea behind the elimination method: to manipulate the equations so that adding them together simplifies the system. The new second equation becomes:

$2(-6x + 5y) = 2(3)$

$-12x + 10y = 6$

Multiplying both sides of an equation by a constant does not change its solution, so this manipulation is perfectly valid. It's a crucial step in setting up the elimination process. By performing this multiplication, we've created a situation where the x terms in the two equations are additive inverses of each other, making the elimination straightforward. This careful manipulation is what makes the elimination method such a powerful tool for solving systems of equations. It allows us to systematically reduce the complexity of the problem until we arrive at a solution. Understanding this principle is essential for mastering the technique and applying it to a wide range of problems.

Step 2: Add the modified second equation to the first equation

Now, we add the modified second equation to the first equation:

egin{aligned} (12x + 15y) + (-12x + 10y) &= 34 + 6 \\ 12x + 15y - 12x + 10y &= 40 \\ 25y &= 40 \end{aligned}

Notice how the x terms have canceled out, as planned. This is the core of the elimination method: strategically manipulating the equations to eliminate one variable. We're now left with a simple equation in y, which we can easily solve. This step demonstrates the power of algebraic manipulation in simplifying complex problems. By carefully choosing our operations, we can reduce a system of two equations in two unknowns to a single equation in one unknown. This makes the problem much more manageable and allows us to find the solution efficiently. The elimination method is particularly effective when dealing with linear equations, where the variables appear only to the first power. In these cases, the coefficients can be easily manipulated to achieve the desired cancellation.

Step 3: Solve for y

To solve for y, we divide both sides of the equation by 25:

y = rac{40}{25} = rac{8}{5}

We have now found the value of y. This is a significant step forward in solving the system of equations. Knowing the value of one variable allows us to substitute it back into one of the original equations to find the value of the other variable. This process of back-substitution is a common technique in algebra and is used to solve a variety of problems. The fraction $\frac{8}{5}$ represents the value of y that satisfies both equations in the system. This value is a key component of the overall solution. By carefully performing the algebraic steps, we have isolated y and determined its value. This demonstrates the power of algebraic techniques in solving mathematical problems and finding precise solutions.

Step 4: Substitute the value of y into one of the original equations to solve for x

Let's substitute $y = \frac{8}{5}$ into the second original equation:

$-6x + 5(\frac{8}{5}) = 3$

$-6x + 8 = 3$

Now, we solve for x:

egin{aligned} -6x &= 3 - 8 \\ -6x &= -5 \\ x &= rac{-5}{-6} \\ x &= rac{5}{6} \end{aligned}

We have now found the value of x. This completes the solution process for the system of equations. By substituting the value of y that we found earlier, we were able to isolate x and determine its value. This back-substitution technique is a crucial part of solving systems of equations and is used in many different mathematical contexts. The fraction $\frac{5}{6}$ represents the value of x that, along with the value of y, satisfies both equations in the system. This pair of values is the unique solution to the system, and it represents the point where the two lines represented by the equations intersect on a graph. Understanding this graphical interpretation can provide further insight into the nature of systems of equations and their solutions.

Final Answer

Therefore, the solution to the system of equations is:

$x = \frac{5}{6}$

$y = \frac{8}{5}$

So, the solution can be written as:

x = [5/6] y = [8/5]

We have successfully found the values of x and y that satisfy both equations in the system. This solution represents the point where the two lines represented by the equations intersect. To recap, we used the elimination method, which involved multiplying one of the equations by a constant to make the coefficients of one of the variables opposites, then adding the equations together to eliminate that variable. This allowed us to solve for one variable, and then we substituted that value back into one of the original equations to solve for the other variable. This systematic approach is a powerful tool for solving systems of equations and is widely used in mathematics and related fields. By understanding the steps involved and practicing the technique, you can confidently solve a wide range of systems of equations. The solution we have found is the unique solution to this particular system, meaning that there is only one pair of values for x and y that will satisfy both equations simultaneously. This is a common characteristic of systems of linear equations, and it reflects the geometric fact that two non-parallel lines intersect at exactly one point.

Conclusion

In this article, we have demonstrated how to solve a system of linear equations using the elimination method. We methodically worked through each step, from manipulating the equations to eliminate a variable to substituting the value of one variable to find the other. The solution we found, $x = \frac{5}{6}$ and $y = \frac{8}{5}$, represents the values that satisfy both equations simultaneously. Mastering the elimination method is a valuable skill in mathematics, as it provides a systematic approach to solving systems of equations. This method is not only applicable to linear equations but can also be adapted to solve more complex systems. The key to success lies in understanding the underlying principles and practicing the technique. By carefully manipulating the equations and paying attention to the algebraic steps, you can confidently solve a wide range of problems involving systems of equations. The ability to solve systems of equations is essential in many areas of mathematics, science, and engineering, making it a fundamental skill for anyone pursuing these fields. The elimination method, along with other techniques such as substitution and graphing, provides a powerful toolkit for tackling these problems and finding solutions. The solution we have found is the unique solution to this particular system, meaning that there is only one pair of values for x and y that will satisfy both equations simultaneously. This is a common characteristic of systems of linear equations, and it reflects the geometric fact that two non-parallel lines intersect at exactly one point.