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4 Responses
Both orders are equivalent if you’re loyal to parentheses, but the tidiest route is to clear denominators ASAP: from y = (2x − 5)/3 write 3y = 2x − 5 so x = (3y + 5)/2, then 4y − (3y + 5)/2 = 7 gives y = 19/5 and x = 41/5 (same result as substituting first and then multiplying everything by 3). When I was learning, I kept donating minus signs to the fraction gremlins until I made a rule-always wrap the substituted chunk in big parentheses and clear the common denominator in one swoop; quick refresher: https://www.khanacademy.org/math/algebra/systems-of-equations-algebra/substitution-alg1/v/solving-systems-by-substitution-1.
Short answer: either order is fine; math doesn’t care. If you’re getting different answers, it’s a parentheses/sign mistake. The cleanest workflow is: substitute with full parentheses, then clear denominators in one shot. For your example: 4[(2x − 5)/3] − x = 7, multiply both sides by 3 right away to avoid fraction chaos: 4(2x − 5) − 3x = 21 → 8x − 20 − 3x = 21 → 5x = 41 → x = 41/5. Then y = (2x − 5)/3 = (82/5 − 25/5)/3 = 57/15 = 19/5. Same result if you first multiply the second equation by 3 (12y − 3x = 21) and then substitute, or if you rewrite 3y = 2x − 5 and instead solve for x = (3y + 5)/2 and substitute that.
Two practical tips to stop sign carnage: always drop substituted expressions in parentheses, and when you clear fractions, multiply the entire equation by the LCM in one move-don’t piecemeal it. If there’s a leading minus on a parenthesis, distribute it before you combine like terms. A quick walkthrough of the method is here: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/x2f8bb11595b61c86:solving-systems-by-substitution/v/solving-systems-with-substitution. Hope this helps!
Fractions are just outfits your numbers wear-cute but fiddly-so pick the path that keeps your sleeves clean. Both orders are valid and should give the same solution if parentheses get proper hugs. Option A (clear first): from y = (2x − 5)/3, multiply both sides by 3 to get 3y = 2x − 5, so x = (3y + 5)/2; plug into 4y − x = 7 to get 4y − (3y + 5)/2 = 7, multiply by 2: 8y − (3y + 5) = 14, so 5y − 5 = 14 → y = 19/5, then x = (3y + 5)/2 = 41/5. Option B (substitute first): 4((2x − 5)/3) − x = 7; multiply everything by 3 to clear the denominator: 4(2x − 5) − 3x = 21 → 8x − 20 − 3x = 21 → 5x = 41 → x = 41/5, and y = (2x − 5)/3 = (82/5 − 25/5)/3 = 57/15 = 18/5-oops, brain blink-57/15 reduces to 19/5. Tips to tame the signs: always wrap the substituted expression in parentheses, think of “−something” as + (−1)·something, and when clearing denominators, multiply every term (even the lonely constant on the right) by the same number. I slightly prefer clearing the fraction-y equation first because it keeps the substitution chunk shorter, but either route is equally legit. Hope this helps!
Either order works; if the algebra is tidy, you’ll get the same solution. In practice, clearing denominators early usually reduces small mistakes, and always wrapping a substituted expression in parentheses helps with signs. Also, write one operation per line and distribute carefully when a minus sign is outside parentheses.
For your system y = (2x − 5)/3 and 4y − x = 7:
– Substitute directly: 4·(2x − 5)/3 − x = 7. Multiply both sides by 3 to clear the fraction: 4(2x − 5) − 3x = 21 ⇒ 8x − 20 − 3x = 21 ⇒ 5x = 41 ⇒ x = 41/5. Then y = (2x − 5)/3 = (82/5 − 25/5)/3 = (57/5)/3 = 19/5. If you clear denominators first in the second equation, 3·(4y − x = 7) gives 12y − 3x = 21, and substituting y = (2x − 5)/3 yields 12·(2x − 5)/3 − 3x = 21 ⇒ 4(2x − 5) − 3x = 21, leading to the same x = 41/5, y = 19/5. The key is: when you substitute, use parentheses; when you clear denominators, multiply every term on both sides. A short refresher on substitution with fractions is here: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/systems-of-equations-with-substitution/a/solving-systems-with-substitution