Solving Systems Of Equations Step-by-Step Guide

This article will delve into the methods for solving systems of equations, specifically focusing on three examples. We'll explore how to solve for the variables x and y in each system, providing a step-by-step explanation to enhance understanding. Mastering these techniques is crucial for various mathematical and scientific applications. This guide aims to equip you with the knowledge and skills to tackle similar problems effectively.

System 1: A Simple Start

Original System

( \begin{cases} 3y - 2 = 13 \ xy = 11 \end{cases} )

Solving for y

Solving for y is the first step. In this system of equations, our initial focus is to isolate y in the first equation, which is a straightforward linear equation. By isolating y first, we simplify the process of finding x later on. We begin with the equation 3y - 2 = 13. To isolate y, we must first eliminate the constant term, -2, from the left side of the equation. This is achieved by adding 2 to both sides of the equation, maintaining the balance and equality of the expression. This gives us 3y = 15. The next step involves removing the coefficient, 3, from the y term. This is done by dividing both sides of the equation by 3. This isolates y on one side of the equation, giving us y = 5. This value of y is crucial as it will be used in the subsequent steps to find the value of x. This method of isolating a variable is a fundamental technique in solving systems of equations and is applicable to various mathematical problems. By understanding this process, you are laying a strong foundation for tackling more complex algebraic problems. The value y = 5 is not just an intermediate step; it's a building block that allows us to unravel the solution to the entire system of equations. This approach of solving for one variable first is a common strategy that simplifies the overall process and makes the problem more manageable. This careful and methodical approach is key to success in algebra and beyond.

Substituting y and Solving for x

Now that we have determined the value of y to be 5, we proceed to substitute this value into the second equation, xy = 11. This substitution is a crucial step as it transforms the second equation from one with two unknowns (x and y) into an equation with only one unknown (x). This simplification is essential for solving the equation. By replacing y with 5, we get the equation 5x = 11. To solve for x, we need to isolate x on one side of the equation. This can be achieved by dividing both sides of the equation by 5. This operation maintains the equality of the equation while bringing us closer to the solution. Performing this division, we find that x = 11/5. This value represents the solution for x in the system of equations. It's important to note that this value is a fraction, which is a common occurrence in solving algebraic equations. The solution x = 11/5, when paired with the previously found value of y = 5, forms the complete solution to the system of equations. This process of substitution is a powerful technique in algebra, allowing us to solve systems of equations by reducing the number of unknowns in each equation. The ability to substitute known values into equations is a fundamental skill that is used extensively in mathematics and its applications. Understanding and mastering this technique is crucial for success in solving a wide range of mathematical problems.

Solution

The solution to the first system of equations is x = 11/5 and y = 5. This means that the pair of values (11/5, 5) is the unique solution that satisfies both equations in the system simultaneously. To verify this solution, we can substitute these values back into the original equations and check if the equations hold true. For the first equation, 3y - 2 = 13, substituting y = 5 gives us 3(5) - 2 = 15 - 2 = 13, which is true. For the second equation, xy = 11, substituting x = 11/5 and y = 5 gives us (11/5)(5) = 11, which is also true. Since both equations are satisfied, we can confidently say that the solution (11/5, 5) is correct. This process of verification is an important step in solving systems of equations, as it helps to ensure the accuracy of the solution. It is a good practice to always check your solutions, especially in complex problems where errors can easily occur. The solution represents the point of intersection of the two lines represented by the equations, providing a geometric interpretation of the algebraic solution. This understanding of the solution as a point of intersection is a valuable concept in mathematics and has applications in various fields.

System 2: Elimination Method

Original System

( \begin{cases} 3x + 2y = 25 \ x + 2y = 19 \end{cases} )

Eliminating y

Eliminating y is a strategic move in this system of equations. The key to solving this system efficiently lies in the coefficients of the y terms. Notice that both equations have the same coefficient for y, which is 2. This presents an opportunity to eliminate y by using the elimination method. The elimination method is a powerful technique in solving systems of equations, and it is particularly useful when the coefficients of one of the variables are the same or easily made the same. To eliminate y, we subtract the second equation from the first equation. This is a valid operation because subtracting equal quantities from equal quantities maintains the equality. When we subtract the second equation (x + 2y = 19) from the first equation (3x + 2y = 25), the y terms cancel out, leaving us with an equation in terms of x only. This results in the equation (3x - x) + (2y - 2y) = 25 - 19, which simplifies to 2x = 6. By eliminating y, we have reduced the system of two equations in two variables to a single equation in one variable, making it much easier to solve. This strategic elimination is a hallmark of the elimination method and demonstrates the power of choosing the right approach to solve a problem. The resulting equation, 2x = 6, is a simple linear equation that can be easily solved for x, paving the way for finding the value of y. The ability to identify and execute such strategic steps is crucial for success in solving systems of equations and other mathematical problems.

Solving for x and Substituting

With the equation 2x = 6, solving for x becomes a straightforward process. This equation is a simple linear equation, and the goal is to isolate x on one side of the equation. To do this, we divide both sides of the equation by 2. This operation maintains the balance of the equation while isolating x. Dividing both sides by 2, we get x = 6 / 2, which simplifies to x = 3. This value of x is a crucial piece of information, as it will be used in the next step to find the value of y. Once we have found x, the next step is to substitute this value into one of the original equations to solve for y. We can choose either of the original equations; however, it is often easier to choose the equation that appears simpler. In this case, the second equation, x + 2y = 19, seems simpler. Substituting x = 3 into this equation gives us 3 + 2y = 19. This is now an equation in terms of y only, which can be easily solved. To solve for y, we first subtract 3 from both sides of the equation, which gives us 2y = 16. Then, we divide both sides by 2 to isolate y, which results in y = 8. This value of y, along with the previously found value of x = 3, gives us the complete solution to the system of equations. The process of substituting a known value into an equation to solve for another unknown is a fundamental technique in algebra and is used extensively in solving systems of equations and other mathematical problems. The ability to perform this substitution accurately and efficiently is a key skill for success in algebra.

Solution

The solution to the second system of equations is x = 3 and y = 8. This means that the ordered pair (3, 8) is the unique solution that satisfies both equations in the system simultaneously. To verify this solution, we can substitute these values back into the original equations and check if the equations hold true. For the first equation, 3x + 2y = 25, substituting x = 3 and y = 8 gives us 3(3) + 2(8) = 9 + 16 = 25, which is true. For the second equation, x + 2y = 19, substituting x = 3 and y = 8 gives us 3 + 2(8) = 3 + 16 = 19, which is also true. Since both equations are satisfied, we can confidently say that the solution (3, 8) is correct. This process of verification is an essential step in solving systems of equations, as it helps to ensure the accuracy of the solution. It is a good practice to always check your solutions, especially in complex problems where errors can easily occur. The solution represents the point of intersection of the two lines represented by the equations, providing a geometric interpretation of the algebraic solution. This understanding of the solution as a point of intersection is a valuable concept in mathematics and has applications in various fields.

System 3: Another Elimination Example

Original System

( \begin{cases} 2x + 5y = 19 \ x + 2y = 8 \end{cases} )

Adjusting Equations for Elimination

Adjusting equations is crucial here to effectively use elimination. In this system, the coefficients of neither x nor y are the same or easily made the same by a simple addition or subtraction. This means we need to manipulate one or both equations before we can eliminate a variable. The goal is to make the coefficients of either x or y opposites or equal so that when we add or subtract the equations, one variable will be eliminated. A common strategy is to multiply one or both equations by a constant. This operation changes the coefficients of the variables but does not change the solution of the system, as long as we multiply both sides of the equation by the same constant. In this case, we can choose to eliminate x. To do this, we can multiply the second equation by -2. This will make the coefficient of x in the second equation -2, which is the opposite of the coefficient of x in the first equation. Multiplying the entire second equation (x + 2y = 8) by -2 gives us -2x - 4y = -16. Now, we have a modified system where the coefficients of x are opposites. This adjustment is a critical step in the elimination method, as it sets the stage for eliminating a variable and simplifying the system. The ability to strategically adjust equations in this way is a key skill in solving systems of equations and other algebraic problems. This step requires careful attention to detail to ensure that the multiplication is performed correctly and that the resulting equations are accurate.

Eliminating x and Solving for y

With the equations adjusted, eliminating x is the next strategic step. We now have the system:

( \begin{cases} 2x + 5y = 19 \ -2x - 4y = -16 \end{cases} )

Notice that the coefficients of x are now opposites (2 and -2). This is exactly what we wanted, as it allows us to eliminate x by adding the two equations together. When we add the equations, the x terms cancel out, leaving us with an equation in terms of y only. Adding the equations gives us (2x - 2x) + (5y - 4y) = 19 - 16, which simplifies to y = 3. This elimination process is a powerful technique in solving systems of equations, as it reduces the system from two variables to one, making it much easier to solve. Now that we have found y, the next step is to substitute this value into one of the original equations to solve for x. We can choose either of the original equations; however, it is often easier to choose the equation that appears simpler. In this case, the second equation, x + 2y = 8, seems simpler. Substituting y = 3 into this equation gives us x + 2(3) = 8, which simplifies to x + 6 = 8. To solve for x, we subtract 6 from both sides of the equation, which gives us x = 2. This value of x, along with the previously found value of y = 3, gives us the complete solution to the system of equations. The process of adding equations to eliminate a variable is a fundamental technique in algebra and is used extensively in solving systems of equations and other mathematical problems.

Solution

The solution to the third system of equations is x = 2 and y = 3. This means that the ordered pair (2, 3) is the unique solution that satisfies both equations in the system simultaneously. To verify this solution, we can substitute these values back into the original equations and check if the equations hold true. For the first equation, 2x + 5y = 19, substituting x = 2 and y = 3 gives us 2(2) + 5(3) = 4 + 15 = 19, which is true. For the second equation, x + 2y = 8, substituting x = 2 and y = 3 gives us 2 + 2(3) = 2 + 6 = 8, which is also true. Since both equations are satisfied, we can confidently say that the solution (2, 3) is correct. This process of verification is an essential step in solving systems of equations, as it helps to ensure the accuracy of the solution. It is a good practice to always check your solutions, especially in complex problems where errors can easily occur. The solution represents the point of intersection of the two lines represented by the equations, providing a geometric interpretation of the algebraic solution. This understanding of the solution as a point of intersection is a valuable concept in mathematics and has applications in various fields.

Conclusion

In conclusion, this article has demonstrated methods for solving systems of linear equations. We explored solving by substitution and elimination, illustrating each with detailed examples. Mastering these techniques is essential for a strong foundation in mathematics and its applications.