Solving 3.5a-4.25b=-14.25 And 4.25a-3.5b=-9 Find A-b
In this article, we will explore how to solve a system of linear equations and then determine the value of a specific expression involving the variables. We'll use the given system of equations as an example and walk through the steps to find the solution. Understanding these techniques is crucial for various mathematical and real-world applications.
Understanding Systems of Equations
Systems of equations are sets of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that make all the equations in the system true simultaneously. There are several methods to solve systems of equations, including substitution, elimination, and graphical methods. In this article, we will primarily focus on the elimination method, which is particularly useful when the coefficients of the variables allow for easy cancellation.
Linear equations, which form the basis of many systems, represent a relationship between variables that can be graphed as a straight line. When dealing with two variables, like a and b, the solution to a system of two linear equations represents the point where the two lines intersect on a coordinate plane. This intersection point provides the values of a and b that satisfy both equations.
To effectively solve systems of equations, it’s important to understand the properties of equality. We can add, subtract, multiply, or divide both sides of an equation by the same non-zero value without changing the solution set. This principle is fundamental to the elimination method, where we manipulate the equations to eliminate one variable and solve for the other. The solution process often involves careful algebraic manipulation and attention to detail to avoid errors.
The importance of understanding systems of equations extends beyond the classroom. These systems are used to model real-world scenarios in various fields, such as economics, engineering, and computer science. For example, in economics, systems of equations can be used to model supply and demand curves, while in engineering, they can be used to analyze circuits or structural systems. Therefore, mastering the techniques to solve systems of equations is a valuable skill.
The Given System of Equations
Let's consider the system of equations provided:
3. 5a - 4.25b = -14.25
4. 25a - 3.5b = -9
This is a system of two linear equations with two variables, a and b. Our goal is to find the values of a and b that satisfy both equations. We will use the elimination method to solve this system. The elimination method involves manipulating the equations so that when we add or subtract them, one of the variables is eliminated. This leaves us with a single equation in one variable, which we can easily solve.
To apply the elimination method, we first need to find a common multiple for the coefficients of either a or b in both equations. Looking at the coefficients, we can see that multiplying the first equation by 4.25 and the second equation by 3.5 will give us the same coefficient for a in both equations. This will allow us to eliminate a when we subtract the equations.
Before proceeding, it's crucial to ensure that we perform the same operation on every term in the equation to maintain equality. This includes multiplying both sides of the equation by the chosen constant. This step is vital to avoid errors and ensure the solution we obtain is correct. It's also important to keep track of the signs of the coefficients, as this will affect whether we add or subtract the equations to eliminate a variable.
Once we've eliminated one variable, we'll have an equation with only one variable, which we can solve using basic algebraic techniques. After finding the value of one variable, we can substitute it back into either of the original equations to solve for the other variable. This process of back-substitution is a key step in solving systems of equations.
Solving the System Using Elimination
To eliminate a, we'll multiply the first equation by 4.25 and the second equation by 3.5:
First equation multiplied by 4.25:
5. 25(3.5a - 4.25b) = 4.25(-14.25)
6. 875a - 18.0625b = -60.5625
Second equation multiplied by 3.5:
7. 5(4.25a - 3.5b) = 3.5(-9)
8. 875a - 12.25b = -31.5
Now we subtract the second modified equation from the first:
(14.875a - 18.0625b) - (14.875a - 12.25b) = -60.5625 - (-31.5)
Simplifying, we get:
-5.8125b = -29.0625
Now, divide both sides by -5.8125 to solve for b:
b = -29.0625 / -5.8125
b = 5
Now that we have the value of b, we can substitute it into one of the original equations to solve for a. Let's use the first equation:
9. 5a - 4.25(5) = -14.25
10. 5a - 21.25 = -14.25
Add 21.25 to both sides:
11. 5a = 7
Divide by 3.5:
a = 7 / 3.5
a = 2
So, the solution to the system of equations is a = 2 and b = 5.
The elimination process involved several key steps, each crucial to arriving at the correct solution. First, we identified the coefficients of a and b and determined the multipliers needed to make the coefficients of a the same in both equations. This required careful attention to decimal places and the overall structure of the equations. Then, we performed the multiplication, ensuring each term in the equation was multiplied correctly.
The subtraction step was also critical. We had to subtract the modified equations in the correct order, paying close attention to the signs. A small error in sign can lead to a completely different solution. After simplifying, we ended up with an equation in just one variable, b, which we solved by dividing both sides by the coefficient of b.
The substitution step was the final piece of the puzzle. By plugging the value of b back into one of the original equations, we created a new equation with only a as the unknown. This equation was then solved using basic algebraic steps, giving us the value of a. Together, the values of a and b form the solution to the system of equations.
Finding a - b
The problem asks for the value of a - b. Now that we have found a = 2 and b = 5, we can easily calculate this:
a - b = 2 - 5 = -3
Therefore, the value of a - b is -3.
Calculating a - b is a straightforward step once we have the values of a and b. This subtraction provides a specific value that answers the question posed. In this case, subtracting the value of b from a gives us a negative result, which indicates that b is greater than a. This kind of simple arithmetic operation is often the final step in solving more complex problems, emphasizing the importance of mastering basic mathematical skills.
The final answer, -3, represents a specific numerical value derived from the solution of the system of equations. It’s a single number that encapsulates the relationship between the values of a and b that satisfy the original system. Understanding how to arrive at this value through a combination of algebraic manipulation and arithmetic calculation is a fundamental aspect of problem-solving in mathematics.
Conclusion
We have successfully solved the system of equations and found that the value of a - b is -3. This process involved several key steps, including using the elimination method to solve for the variables and then performing a simple subtraction. Understanding these techniques is essential for solving various mathematical problems and real-world applications.
Solving systems of equations is a core skill in mathematics, with applications spanning various fields. The ability to manipulate equations, eliminate variables, and find solutions is invaluable. This article demonstrated the elimination method, a powerful technique that can be applied to many systems of equations. By mastering this method, students and professionals alike can tackle complex problems and gain a deeper understanding of mathematical relationships.
Furthermore, the ability to solve for expressions like a - b highlights the importance of understanding the question being asked. Often, the solution to a system of equations is not the final answer; rather, it's a stepping stone to finding another value or understanding a broader concept. In this case, finding a and b was necessary, but the ultimate goal was to find the difference between them.
In conclusion, the process of solving systems of equations and calculating related expressions reinforces fundamental mathematical principles. These skills are not only crucial for academic success but also for problem-solving in various real-world contexts. By practicing and mastering these techniques, individuals can enhance their analytical abilities and approach complex problems with confidence.
