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In order to solve the following system of equations by addition, which of the following could you do before adding the equations so that one variable will be eliminated when you add them? \( 2 x-4 y=5 \) \( 6 x-3 y=10 \) A. Multiply the top equation by -3 . B. Multiply the top equation by -3 and the bottom equation by 2 . C. Multiply the top equation by 3 and the bottom equation by 4 . D. Multiply the top equation by -2 .

Ask by Cook Parsons. in the United States
Feb 26,2025

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Option A: Multiply the top equation by –3.

Solución

We start with the system of equations:   (1) 2x – 4y = 5   (2) 6x – 3y = 10 Our goal is to manipulate one or both equations so that when we add them, one variable is eliminated. Looking at option A: Multiply the top equation by –3. Multiplying equation (1) by –3 gives:   –3(2x – 4y) = –3(5)   –6x + 12y = –15 Now we have:   Modified (1): –6x + 12y = –15   (2):   6x – 3y = 10 When we add these equations:   (–6x + 12y) + (6x – 3y) = (–15) + 10   (–6x + 6x) + (12y – 3y) = –5   0x + 9y = –5   9y = –5 The x terms cancel out, eliminating x. We can now solve for y. Thus, option A is a valid step to eliminate one variable when adding the equations.

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The Deep Dive

Let's consider option B – if you multiply the top equation \(2x - 4y = 5\) by -3 and the bottom equation \(6x - 3y = 10\) by 2, you'll get: - Top equation: \( -6x + 12y = -15 \) - Bottom equation: \( 12x - 6y = 20 \) Now when you add them together, \( -6x + 12y + 12x - 6y \) simplifies perfectly, letting you eliminate \(x\)! So, make sure to use the right coefficients for a smooth elimination process. If you consider the first equation multiplied by -3, while trying other options, be cautious of the coefficients. It's easy to accidentally create different terms that don't cancel out. Always keep track of how the changes affect both equations, ensuring one variable vanishes upon addition!

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