Solve The Following Systems Of Equations: 1. X + 2y + 1 = 0 , 2x - 3y - 12 = 0 2. 3x - 2y + 3 = 0 , 4x + 3y - 47 = 0 3. 3x + 2y + 25 = 0 , 2x + Y + 10 = 0

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In mathematics, solving systems of equations is a fundamental skill. It involves finding the values of the variables that satisfy all equations in the system simultaneously. This article will delve into solving systems of linear equations, providing step-by-step solutions and explanations for a better understanding. We'll explore different methods, including substitution and elimination, to tackle various systems effectively. Understanding these methods and applying them correctly is crucial for success in algebra and beyond.

1. Solving the System: x + 2y + 1 = 0 and 2x - 3y - 12 = 0

Let's start with the first system of equations:

 x + 2y + 1 = 0
 2x - 3y - 12 = 0

To solve this system, we can use either the substitution or elimination method. Let's use the substitution method for this example. First, we isolate x in the first equation:

 x = -2y - 1

Now, we substitute this expression for x into the second equation:

 2(-2y - 1) - 3y - 12 = 0

Expanding and simplifying the equation, we get:

 -4y - 2 - 3y - 12 = 0
 -7y - 14 = 0
 -7y = 14
 y = -2

Now that we have the value of y, we can substitute it back into the equation x = -2y - 1 to find the value of x:

 x = -2(-2) - 1
 x = 4 - 1
 x = 3

Therefore, the solution to the first system of equations is x = 3 and y = -2. We can verify this solution by plugging these values back into the original equations:

  • Equation 1: 3 + 2(-2) + 1 = 3 - 4 + 1 = 0 (Correct)
  • Equation 2: 2(3) - 3(-2) - 12 = 6 + 6 - 12 = 0 (Correct)

The solution (x, y) = (3, -2) satisfies both equations, confirming our answer. This step-by-step approach highlights the importance of algebraic manipulation and substitution in solving linear systems. Mastering this technique provides a solid foundation for more complex mathematical problems. The key is to break down the problem into manageable steps, ensuring each step is performed accurately to reach the correct solution. Understanding the underlying principles allows for flexibility in choosing the most efficient method for solving different types of systems.

2. Solving the System: 3x - 2y + 3 = 0 and 4x + 3y - 47 = 0

Next, let's tackle the second system of equations:

 3x - 2y + 3 = 0
 4x + 3y - 47 = 0

For this system, the elimination method might be more convenient. The elimination method involves manipulating the equations to eliminate one of the variables. To eliminate y, we can multiply the first equation by 3 and the second equation by 2:

 3(3x - 2y + 3) = 9x - 6y + 9 = 0
 2(4x + 3y - 47) = 8x + 6y - 94 = 0

Now, add the two equations together:

 (9x - 6y + 9) + (8x + 6y - 94) = 0
 17x - 85 = 0
 17x = 85
 x = 5

Now that we have the value of x, we can substitute it back into either of the original equations to find the value of y. Let's use the first equation:

 3(5) - 2y + 3 = 0
 15 - 2y + 3 = 0
 18 - 2y = 0
 -2y = -18
 y = 9

Therefore, the solution to the second system of equations is x = 5 and y = 9. Let's verify this solution:

  • Equation 1: 3(5) - 2(9) + 3 = 15 - 18 + 3 = 0 (Correct)
  • Equation 2: 4(5) + 3(9) - 47 = 20 + 27 - 47 = 0 (Correct)

Our solution (x, y) = (5, 9) satisfies both equations. This exemplifies how the elimination method simplifies the process by strategically removing variables, making it easier to solve for the remaining ones. This method shines when the coefficients of one variable are multiples of each other, or when a simple multiplication can achieve this. The strategic choice of method is crucial for efficient problem-solving in algebra.

3. Solving the System: 3x + 2y + 25 = 0 and 2x + y + 10 = 0

Finally, let's solve the third system of equations:

 3x + 2y + 25 = 0
 2x + y + 10 = 0

For this system, we can again use the elimination method. To eliminate y, we can multiply the second equation by -2:

 -2(2x + y + 10) = -4x - 2y - 20 = 0

Now, add this modified equation to the first equation:

 (3x + 2y + 25) + (-4x - 2y - 20) = 0
 -x + 5 = 0
 -x = -5
 x = 5

Now that we have the value of x, substitute it back into the second equation to find y:

 2(5) + y + 10 = 0
 10 + y + 10 = 0
 y + 20 = 0
 y = -20

Thus, the solution to the third system of equations is x = 5 and y = -20. Let's verify this solution:

  • Equation 1: 3(5) + 2(-20) + 25 = 15 - 40 + 25 = 0 (Correct)
  • Equation 2: 2(5) + (-20) + 10 = 10 - 20 + 10 = 0 (Correct)

Therefore, the solution (x, y) = (5, -20) is confirmed. This system further demonstrates the effectiveness of the elimination method, particularly when dealing with equations where variables can be easily eliminated through multiplication and addition. The consistent application of algebraic principles, coupled with careful execution, ensures accurate solutions even in systems with larger numbers or negative values. This reinforces the importance of a methodical approach in problem-solving.

Conclusion

In this article, we've explored how to solve systems of linear equations using both substitution and elimination methods. Each system presented different challenges, highlighting the importance of understanding the nuances of each method and choosing the most efficient approach. Mastery of these techniques is essential for further studies in mathematics and related fields. Solving systems of equations is a cornerstone of algebra, and with practice, these methods become second nature, enabling you to tackle complex problems with confidence. The ability to solve such systems is not just a mathematical skill but also a valuable tool for analytical thinking and problem-solving in various real-world scenarios.