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

2025-07-20 17:18:28 319

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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