1.11-bo'lim: Tenglamalarni yechish, II qism
\(y\) uchun quyidagi tenglamalarning har birini yeching.
- \(\displaystyle x = \frac{{2y - 5}}{{6 - 7y}}\)
Yechim
Bu erda biz qilishimiz kerak bo'lgan yagona narsa barcha \(y\)-larni bir tomondan olish, \(y\)-ni hisobga olish va keyin \(y\) koeffitsientiga bo'lishdir.
\[\begin{align*}x & = \frac{{2y - 5}}{{6 - 7y}}\\ x\left( {6 - 7y} \right) & = 2y - 5\\ 6x - 7xy & = 2y - 5\\ 6x + 5 & = \left( {7x + 2} \right)y\\ y & = \frac{{6x + 5}}{{7x + 2}}\end{align*}\]Undagi o'zgaruvchilardan biri uchun tenglamalarni echish - bu siz Calculus kursida vaqti-vaqti bilan qiladigan narsadir, shuning uchun buni qila olishingizga ishonch hosil qiling.
- \(3{x^2}\left( {3 - 5y} \right) + \sin x = 3xy + 8\)
Yechim
Bu avvalgi muammo bilan bir xil tarzda hal qilinadi.
\[\begin{align*}3{x^2}\left( {3 - 5y} \right) + \sin x & = 3xy + 8\\ 9{x^2} - 15{x^2}y + \sin x & = 3xy + 8\\ 9{x^2} + \sin x - 8 & = \left( {3x + 15{x^2}} \right)y\\ y & = \frac{{9{x^2} + \sin x - 8}}{{3x + 15{x^2}}}\end{align*}\] - \(2{x^2} + 2{y^2} = 5\)
Yechim
Xuddi shu narsa, faqat oxirgi bosqichda ehtiyot bo'ling.
\[\begin{align*}2{x^2} + 2{y^2} & = 5\\ 2{y^2} & = 5 - 2{x^2}\\ {y^2} & = \frac{1}{2}\left( {5 - 2{x^2}} \right)\\ y & = \pm \sqrt {\frac{5}{2} - {x^2}} \end{align*}\]Yechimdagi "\( \pm \)" ni unutmang!