I’m prepping for a test and simultaneous equations keep scrambling my brain. Here’s one from my practice set:
3x + 2y = 14
2x − y = 1
I tried substitution first. From the second equation I wrote y = 2x − 1, then plugged into the first: 3x + 2(2x − 1) = 14 → 3x + 4x − 2 = 14. Then I somehow turned that into 7x = 12 (which already feels suspicious), so x = 12/7. Plugging back, I got y = 2(12/7) − 1 = 24/7 − 7/7 = 17/7. But when I check in the first equation, 3*(12/7) + 2*(17/7) = 10, not 14. So… clearly I messed up. Where exactly did I go wrong in that substitution step?
I also tried elimination. I multiplied the second equation by 2 to get 4x − 2y = 2. Then I got confused about whether to add or subtract. I subtracted like this: (3x + 2y) − (4x − 2y) = 14 − 2, which gave me −x + 4y = 12… and that didn’t eliminate anything. If I add them instead, the y’s look like they cancel, but I keep second-guessing the signs. What am I supposed to add or subtract here to eliminate cleanly?
Follow-up: is there a quick sanity check to tell if an intermediate result like x = 12/7 even makes sense before I finish? And how do you decide between substitution and elimination so the numbers don’t get messier than they need to be?
Sorry if I’m overthinking this-I just want to fix whatever habit is causing these sign mistakes before the test.
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3 Responses
Your slip is just arithmetic: 3x + 4x − 2 = 14 gives 7x = 16 (not 12), so x = 16/7 and y = 2(16/7) − 1 = 25/7, and indeed 3·(16/7) + 2·(25/7) = 48/7 + 50/7 = 14.
For elimination, multiply the second by 2 to get 4x − 2y = 2 and add to 3x + 2y = 14 so the y’s cancel to 7x = 16 (a quick sanity check-7x should be 16); subtracting there would cancel x instead.
I think the only real slip was arithmetic right after your substitution: from 2x − y = 1 you had y = 2x − 1, and plugging into 3x + 2y = 14 gives 3x + 2(2x − 1) = 14, so 3x + 4x − 2 = 14, i.e., 7x − 2 = 14, hence 7x = 16 and x = 16/7 (not 12/7). Then y = 2(16/7) − 1 = 32/7 − 7/7 = 25/7, which checks: 3·(16/7) + 2·(25/7) = 14 and 2x − y = 1 as well. For elimination, doubling the second equation to 4x − 2y = 2 and then adding to 3x + 2y = 14 is the clean move, because +2y and −2y cancel: (3x + 2y) + (4x − 2y) = 14 + 2 ⇒ 7x = 16 ⇒ x = 16/7; subtracting instead doubles the y’s and creates −x + 4y, so nothing cancels. A quick sanity check I like: after substituting y = 2x − 1, simplify symbolically first to 7x − 2 = 14; that makes it obvious that 7x must be 16, so if you ever see 12 you know exactly where to look. Another fast check is to test your provisional x in that simplified form: with x = 12/7 you get 7x − 2 = 12 − 2 = 10, so it can’t satisfy the first equation. As for method choice, if a variable can be isolated with coefficient 1 or −1, substitution is usually tidy; if you can align coefficients to cancel in one step without fractions (like matching ±2y here), elimination is just as tidy. I might be overlooking a tiny nuance in how you’re organizing the steps, but keeping that “simplify-to-a-single-variable-expression first” habit tends to catch most sign slips before they snowball.
You were right there-after substituting y = 2x − 1, 3x + 2(2x − 1) = 14 becomes 3x + 4x − 2 = 14, and adding 2 to both sides gives 7x = 16 (not 12), so x = 16/7 and y = 2x − 1 = 25/7; quick check: 3·16/7 + 2·25/7 = 14 and 2·16/7 − 25/7 = 1.
For elimination, double the second to 4x − 2y = 2 and add it to the first to cancel y: (3x + 2y) + (4x − 2y) = 14 + 2 ⇒ 7x = 16, and a fast sanity check is that 2x − y = 1 says y is just under 2x, so both should be a few units positive-making x = 12/7 feel too small.