\[x = \frac{{1 + 9y}}{{4 - y}}\]

\[\begin{align*}x & = \frac{{1 + 9y}}{{4 - y}}\\ x\left( {4 - y} \right) & = 1 + 9y\\ & 4x - xy = 1 + 9y\\ 4x - 1 & = 9y + xy\\ 4x - 1 & = \left( {9 + x} \right)y\\ y & = \frac{{4x - 1}}{{9 + x}}\hspace{0.25in}\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{h^{ - 1}}\left( x \right) = \frac{{4x - 1}}{{9 + x}}}}\end{align*}\]

E'tibor bering, bunday muammolardagi Algebra ko'pincha juda chalkash bo'lishi mumkin, ammo bu sizni bu muammolarni hal qila olmasligingizga qaror qilishiga yo'l qo'ymang. Tartibsiz algebra Kalkulyator sinfida juda keng tarqalgan hodisa bo'ladi, shuning uchun siz unga ko'nikishingiz kerak bo'ladi!

Va nihoyat, ishimizni tekshirish uchun \(\left( {h \circ {h^{ - 1}}} \right)\left( x \right)\) yoki \(\left( {{h^{ - 1}} \circ h} \right)\left( x \right)\) hisoblang.

Ikkala kompozitsiya ham bajarilishi mumkin, keling, bu holda \(\left( {{h^{ - 1}} \circ h} \right)\left( x \right)\) ni bajaraylik. Oldingi bosqichda bo'lgani kabi, bu erda ham Algebra tartibsiz bo'ladi va aslida chalkashroq bo'ladi.

\[\begin{align*}\left( {{h^{ - 1}} \circ h} \right)\left( x \right) & = {h^{ - 1}}\left[ {h\left( x \right)} \right]\\ & = \frac{{4\left[ {\frac{{1 + 9x}}{{4 - x}}} \right] - 1}}{{9 + \left[ {\frac{{1 + 9x}}{{4 - x}}} \right]}}\,\,\,\frac{{4 - x}}{{4 - x}}\\ & = \frac{{4\left( {1 + 9x} \right) - \left( {4 - x} \right)}}{{9\left( {4 - x} \right) + 1 + 9x}}\\ & = \frac{{4 + 36x - 4 + x}}{{36 - 9x + 1 + 9x}}\\ & = \frac{{37x}}{{37}}\\ & = x\end{align*}\]

Soddalashtirishni amalga oshirish uchun biz ba'zi maxrajlarni tozalash uchun boshlang'ich kasrning payini va maxrajini \(4 - x\) ga ko'paytirdik. Bu, o'z navbatida, adolatli darajada soddalashtirish imkonini berdi.

Shunday qilib, biz \(x\)-ni kompozitsiyadan chiqardik va shuning uchun biz o'z ishimizni to'g'ri bajarganimizni bilamiz.