Réitigh 36/4x-2+8x=12y-3/5

$$\frac { 36 } { 4 x - 2 } + 8 x = \frac { 12 y - 3 } { 5 }$$

Réitigh do y.

$$y=-\frac{80x^{2}-34x+87}{12\left(1-2x\right)}$$

$$x\neq \frac{1}{2}$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Iolraigh an dá thaobh den chothromóid faoi 10\left(2x-1\right), an comhiolraí is lú de 4x-2,5.

$$5\times 36+8x\times 10\left(2x-1\right)=\left(4x-2\right)\left(12y-3\right)$$

Méadaigh 5 agus 36 chun 180 a fháil.

$$180+8x\times 10\left(2x-1\right)=\left(4x-2\right)\left(12y-3\right)$$

Méadaigh 8 agus 10 chun 80 a fháil.

$$180+80x\left(2x-1\right)=\left(4x-2\right)\left(12y-3\right)$$

Úsáid an t-airí dáileach chun 80x a mhéadú faoi 2x-1.

$$180+160x^{2}-80x=\left(4x-2\right)\left(12y-3\right)$$

Úsáid an t-airí dáileach chun 4x-2 a mhéadú faoi 12y-3.

$$180+160x^{2}-80x=48xy-12x-24y+6$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$48xy-12x-24y+6=180+160x^{2}-80x$$

Cuir 12x leis an dá thaobh.

$$48xy-24y+6=180+160x^{2}-80x+12x$$

Comhcheangail -80x agus 12x chun -68x a fháil.

$$48xy-24y+6=180+160x^{2}-68x$$

Bain 6 ón dá thaobh.

$$48xy-24y=180+160x^{2}-68x-6$$

Dealaigh 6 ó 180 chun 174 a fháil.

$$48xy-24y=174+160x^{2}-68x$$

Comhcheangail na téarmaí ar fad ina bhfuil y.

$$\left(48x-24\right)y=174+160x^{2}-68x$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(48x-24\right)y=160x^{2}-68x+174$$

Roinn an dá thaobh faoi 48x-24.

$$\frac{\left(48x-24\right)y}{48x-24}=\frac{160x^{2}-68x+174}{48x-24}$$

Má roinntear é faoi 48x-24 cuirtear an iolrúchán faoi 48x-24 ar ceal.

$$y=\frac{160x^{2}-68x+174}{48x-24}$$

Roinn 174+160x^{2}-68x faoi 48x-24.

$$y=\frac{80x^{2}-34x+87}{12\left(2x-1\right)}$$

Réitigh do x. (complex solution)

$$x=\frac{\sqrt{144y^{2}-552y-6671}}{80}+\frac{3y}{20}+\frac{17}{80}$$

$$x=-\frac{\sqrt{144y^{2}-552y-6671}}{80}+\frac{3y}{20}+\frac{17}{80}$$

Réitigh do x.

$$x=\frac{\sqrt{144y^{2}-552y-6671}}{80}+\frac{3y}{20}+\frac{17}{80}$$

$$x=-\frac{\sqrt{144y^{2}-552y-6671}}{80}+\frac{3y}{20}+\frac{17}{80}\text{, }y\geq 5\sqrt{2}+\frac{23}{12}\text{ or }y\leq \frac{23}{12}-5\sqrt{2}$$