Can You Solve System Of Linear Equations By Elimination With Fractions?

2025-07-20 17:18:28 233

3 Answers

Yolanda
Yolanda
2025-07-21 09:56:57
Elimination with fractions can seem intimidating at first, but once you break it down, it’s just an extension of regular elimination. I remember tutoring a friend who was struggling with this, and the lightbulb moment came when they realized it’s all about balancing the equations. Take the system (2/5)x + (1/2)y = 3 and (3/10)x - y = 1. First, find the LCD for each equation—10 for the first and 10 for the second. Multiply the first by 10 to get 4x + 5y = 30 and the second by 10 to get 3x - 10y = 10. Now, you can manipulate these like any other system. To eliminate 'y,' multiply the first new equation by 2: 8x + 10y = 60. Add this to the second new equation: 11x = 70, so x ≈ 6.36. Plug this back to solve for y ≈ 0.91.

Another trick is to cross-multiply if the fractions are complex, like (x/3) + (y/4) = 5 becoming 4x + 3y = 60. the goal is always to reduce the system to integers before elimination. Practice with simpler systems first, like (1/2)x + y = 4 and x - (1/3)y = 2, to build confidence. Over time, handling fractions becomes second nature, and the process feels just as smooth as working with whole numbers.
Sawyer
Sawyer
2025-07-26 05:03:15
Solving systems of linear equations with fractions using elimination is totally doable, and I’ve done it plenty of times in my math adventures. The key is to eliminate the fractions early to simplify the equations. Multiply each term by the least common denominator to convert the fractions into whole numbers. For example, if you have (1/2)x + (1/3)y = 5 and (1/4)x - (1/6)y = 2, multiply the first equation by 6 and the second by 12 to clear the denominators. This gives 3x + 2y = 30 and 3x - 2y = 24. Then, add or subtract the equations to eliminate one variable. Here, adding them cancels 'y,' leaving 6x = 54, so x = 9. Substitute back to find y = 1.5. It’s a bit more work with fractions, but the method stays reliable.
Nina
Nina
2025-07-26 15:33:45
I love the elegance of elimination, even with fractions! Here’s how I approach it: start by tidying up the equations. For instance, (3/4)x - (1/2)y = 1 and (1/8)x + (3/4)y = 5 can be rewritten by multiplying the first by 4 (3x - 2y = 4) and the second by 8 (x + 6y = 40). Now, align the coefficients. Multiply the second new equation by 3 to match the 'x' coefficient: 3x + 18y = 120. Subtract the first new equation from this: 20y = 116, so y = 5.8. Substitute back to find x = 5.2.

Fractions add a layer, but they don’t change the core strategy. If denominators are messy, like in (2/7)x + (5/9)y = 3, multiply by 63 (LCD) to get 18x + 35y = 189. Pair this with another scaled equation, and proceed as usual. The extra step of clearing fractions upfront saves headaches later. It’s like prepping ingredients before cooking—annoying but worth it.
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I remember struggling with math until I discovered the elimination method for solving linear equations. It’s straightforward and doesn’t require complex formulas like substitution does. You just line up the equations, eliminate one variable by adding or subtracting, and solve for the other. It’s especially handy when dealing with equations that have coefficients that cancel out easily. For example, if you have 2x + 3y = 5 and 2x - y = 1, you can subtract the second equation from the first to eliminate x instantly. This method feels like tidying up a messy room—everything falls into place neatly. Plus, it’s less prone to arithmetic errors since you’re working with whole equations at once.

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I remember learning this method in class, and it's actually pretty straightforward once you get the hang of it. The elimination method is about getting rid of one variable so you can solve for the other. You start by writing both equations clearly. Then, you adjust them so one of the variables cancels out when you add or subtract the equations. For example, if you have 2x + 3y = 5 and 4x + 6y = 10, you can multiply the first equation by 2 to match the coefficients of x. Then subtract the first from the second, and the x terms cancel out, leaving you with an equation in y. Solve for y, then plug that back into one of the original equations to find x. It's like solving a puzzle where you remove pieces step by step until the picture becomes clear.

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3 Answers2025-07-20 21:42:37
I remember struggling with elimination in linear equations when I first learned it. One common mistake is not aligning variables properly before subtracting or adding equations. People often forget to multiply every term in an equation by the same number when trying to eliminate a variable, which throws off the entire solution. Another error is mixing up signs when combining equations, leading to incorrect results. Sometimes, students eliminate the wrong variable first, making the problem more complicated than it needs to be. It’s also easy to forget to check the solution by plugging it back into the original equations. These small oversights can turn a straightforward problem into a frustrating mess.

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I remember struggling with this exact question in my math class. Elimination just clicked better for me because it felt more straightforward when dealing with multiple variables. With substitution, I kept getting tangled up in rearranging equations, especially if they had fractions or complex terms. Elimination lets you add or subtract equations to cancel out a variable, which is cleaner when the coefficients line up nicely. For example, if you have 2x + 3y = 12 and 2x - y = 4, you can subtract the second equation from the first to eliminate x instantly. It’s like tidying up a messy room—sometimes it’s easier to remove the clutter all at once rather than piece by piece. Plus, elimination scales better for larger systems. If you’re dealing with three or more equations, substitution becomes a nightmare of nested substitutions, but elimination keeps things manageable by systematically zeroing out variables.

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3 Answers2025-07-20 08:16:48
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3 Answers2025-07-20 23:45:05
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