How To Solve System Of Linear Equations By Elimination Step By Step?

2025-07-20 14:21:31 383
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3 Answers

Sadie
Sadie
2025-07-21 02:37:49
Solving systems of linear equations by elimination is one of those math techniques that feels like magic once you get the hang of it. I remember struggling with it at first, but now it's my go-to method. Here's how I do it: Start by writing both equations clearly. For example, 2x + 3y = 8 and 4x - y = 6. the goal is to eliminate one variable by making the coefficients opposites. Multiply the second equation by 3 to get 12x - 3y = 18. Now, add it to the first equation: 2x + 3y + 12x - 3y = 8 + 18. The y terms cancel out, leaving 14x = 26. Solve for x by dividing both sides by 14, giving x ≈ 1.857. Plug this back into one of the original equations to find y. Using 4x - y = 6, substitute x: 4(1.857) - y = 6 → 7.428 - y = 6 → y ≈ 1.428. And there you have it, the solution is (1.857, 1.428). Practice with different systems to build confidence.
Ryder
Ryder
2025-07-23 14:36:33
Elimination is a powerful method for solving systems of linear equations, especially when substitution might be messy. I love how systematic it feels. Let's break it down with an example: 3x + 2y = 12 and 5x - 2y = 4. The key is to align the equations vertically and look for terms that can cancel each other out. Here, the y terms are already opposites (+2y and -2y), so adding the equations eliminates y: 3x + 5x = 12 + 4 → 8x = 16 → x = 2. Now, substitute x = 2 into the first equation: 3(2) + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3. The solution is (2, 3).

Sometimes, you need to adjust the equations first. Take 2x + 5y = 3 and 3x + 4y = 7. Neither variable cancels out directly. Multiply the first equation by 3 and the second by 2 to match the x coefficients: 6x + 15y = 9 and 6x + 8y = 14. Subtract the second from the first: (6x - 6x) + (15y - 8y) = 9 - 14 → 7y = -5 → y = -5/7. Now, plug y into one of the original equations to find x. It's a bit more work, but the method is reliable.

Remember, elimination works best when the equations are neatly arranged. If the numbers seem unwieldy, don't hesitate to multiply entire equations to create cancelable terms. This method shines in standardized tests and real-world problems where speed and accuracy matter.
Liam
Liam
2025-07-25 01:10:55
I've always found elimination to be the most satisfying way to solve linear systems. Here's my step-by-step approach with a practical example. Consider the equations x + y = 5 and 2x - y = 1. The y terms are set up perfectly for elimination—one positive, one negative. Add the two equations: x + 2x + y - y = 5 + 1 → 3x = 6 → x = 2. Substitute x = 2 into the first equation: 2 + y = 5 → y = 3. The solution is (2, 3). Easy, right?

For trickier systems, like 4x + 3y = 10 and 2x + 5y = 12, you'll need to manipulate one equation. Multiply the second equation by 2: 4x + 10y = 24. Now subtract the first equation: (4x - 4x) + (10y - 3y) = 24 - 10 → 7y = 14 → y = 2. Plug y = 2 into 2x + 5y = 12 → 2x + 10 = 12 → 2x = 2 → x = 1. The solution is (1, 2).

The beauty of elimination lies in its adaptability. Whether the coefficients are small or large, the process remains consistent. It's a cornerstone of algebra that paves the way for more advanced math.
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