Solving Linear Equations Find The Solution (a, B) For 3a + 6b = 45 And 2a - 2b = -12
Introduction: The Importance of Solving Linear Equation Systems
In the realm of mathematics, specifically within algebra, systems of linear equations hold a paramount position. These systems, composed of two or more linear equations, arise in various real-world applications, spanning from economics and engineering to computer science. The solution to a system of linear equations represents the point(s) where the lines or planes represented by the equations intersect. This intersection point provides critical information, such as equilibrium points in economic models or optimal solutions in engineering designs. Accurately determining these solutions is crucial for informed decision-making and problem-solving across diverse fields. Understanding the methods for solving these systems is not just an academic exercise; it's a fundamental skill with practical implications that underscore its significance in various disciplines. Mastering techniques like substitution, elimination, and graphical methods equips individuals with the tools to analyze and solve problems effectively in both theoretical and real-world scenarios. This article will delve into one such system, providing a step-by-step guide to finding its solution, while also highlighting the broader context and importance of these mathematical tools.
Understanding the System of Equations: 3a + 6b = 45 and 2a - 2b = -12
At the heart of our discussion lies the following system of linear equations:
3a + 6b = 45
2a - 2b = -12
This system comprises two linear equations, each containing two variables, a and b. Our objective is to find the values of a and b that simultaneously satisfy both equations. In simpler terms, we seek a pair of numbers (a, b) that, when substituted into both equations, make the equations true. This pair (a, b) represents the point of intersection between the two lines represented by the equations when graphed on a coordinate plane. The significance of finding this solution extends beyond the realm of pure mathematics. In practical applications, a and b could represent various quantities, such as the prices of two goods, the speeds of two objects, or the amounts of two ingredients in a mixture. The equations themselves might model constraints or relationships between these quantities. Therefore, solving the system allows us to determine the specific values of these quantities that meet all the given conditions. For instance, in economics, the equations could represent supply and demand curves, and the solution would represent the market equilibrium point. In engineering, the equations could describe the forces acting on a structure, and the solution would indicate the balance point. Understanding the context and the meaning of the variables adds a layer of practical relevance to the mathematical process of solving the system.
Method 1: Solving by Elimination: A Step-by-Step Guide
The elimination method, also known as the addition method, is a powerful technique for solving systems of linear equations. The core idea behind this method is to manipulate the equations in such a way that when they are added together, one of the variables is eliminated, leaving us with a single equation in one variable. This simplified equation can then be easily solved, and the solution can be substituted back into one of the original equations to find the value of the other variable. Let's apply this method to our system:
3a + 6b = 45
2a - 2b = -12
Step 1: Multiplying Equations to Match Coefficients
Our first goal is to make the coefficients of either a or b the same (or additive inverses) in both equations. To eliminate b, we can multiply the second equation by 3:
3 * (2a - 2b) = 3 * (-12)
6a - 6b = -36
Now our system looks like this:
3a + 6b = 45
6a - 6b = -36
Notice that the coefficients of b are now 6 and -6, which are additive inverses.
Step 2: Adding the Equations
Now we add the two equations together:
(3a + 6b) + (6a - 6b) = 45 + (-36)
9a = 9
As you can see, the b terms have canceled out, leaving us with a single equation in a.
Step 3: Solving for 'a'
Divide both sides of the equation by 9 to solve for a:
9a / 9 = 9 / 9
a = 1
We have found that a = 1.
Step 4: Substituting 'a' to Find 'b'
Substitute the value of a (which is 1) into either of the original equations to solve for b. Let's use the first equation:
3(1) + 6b = 45
3 + 6b = 45
Subtract 3 from both sides:
6b = 42
Divide both sides by 6:
b = 7
Therefore, b = 7.
Step 5: The Solution
The solution to the system of equations is (a, b) = (1, 7). This means that when a is 1 and b is 7, both equations in the system are satisfied.
Method 2: Solving by Substitution: Another Approach
The substitution method offers an alternative route to solving systems of linear equations. This technique involves solving one equation for one variable and then substituting that expression into the other equation. This process transforms the system into a single equation with one variable, which can be easily solved. Let's apply this method to the same system:
3a + 6b = 45
2a - 2b = -12
Step 1: Solve one equation for one variable
Let's solve the second equation for a:
2a - 2b = -12
2a = 2b - 12
a = b - 6
Now we have a expressed in terms of b.
Step 2: Substitute into the other equation
Substitute the expression for a (which is b - 6) into the first equation:
3(b - 6) + 6b = 45
Step 3: Solve for 'b'
Simplify and solve the equation for b:
3b - 18 + 6b = 45
9b - 18 = 45
9b = 63
b = 7
We have found that b = 7.
Step 4: Substitute 'b' to Find 'a'
Substitute the value of b (which is 7) back into the expression we found for a:
a = 7 - 6
a = 1
Therefore, a = 1.
Step 5: The Solution
The solution to the system of equations is (a, b) = (1, 7), which confirms the result we obtained using the elimination method.
Verifying the Solution: Ensuring Accuracy
It is always a good practice to verify the solution obtained by substituting the values of a and b back into the original equations. This step ensures that the solution is accurate and that no errors were made during the solving process. Let's substitute a = 1 and b = 7 into our original equations:
Equation 1:
3a + 6b = 45
3(1) + 6(7) = 45
3 + 42 = 45
45 = 45 (True)
Equation 2:
2a - 2b = -12
2(1) - 2(7) = -12
2 - 14 = -12
-12 = -12 (True)
Since the solution (a, b) = (1, 7) satisfies both equations, we can confidently conclude that it is the correct solution to the system.
Conclusion: The Power of Linear Equation Solving
In conclusion, we have successfully solved the system of linear equations:
3a + 6b = 45
2a - 2b = -12
using both the elimination and substitution methods. Both methods yielded the same solution, (a, b) = (1, 7), which we verified to be correct. This exercise highlights the power and versatility of these algebraic techniques. The ability to solve systems of linear equations is a fundamental skill in mathematics and has far-reaching applications in various fields. From determining equilibrium points in economic models to optimizing engineering designs, the principles and methods discussed in this article provide a foundation for solving complex problems and making informed decisions. Mastering these techniques empowers individuals to tackle challenges across diverse domains, underscoring the importance of linear equation solving in both academic and practical contexts. The journey through this mathematical process not only enhances problem-solving skills but also fosters a deeper appreciation for the elegance and utility of mathematics in the world around us.
