Réitigh 4/18-2x_{1}[(1+x_{1})(9-x_{1})-8x_{2}]=0

$$\frac { 4 } { 18 - 2 x _ { 1 } } [ ( 1 + x _ { 1 } ) ( 9 - x _ { 1 } ) - 8 x _ { 2 } ] = 0$$

Réitigh do x_2.

$$x_{2}=-\frac{x_{1}^{2}}{8}+x_{1}+\frac{9}{8}$$

$$x_{1}\neq 9$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh an dá thaobh den chothromóid faoi 2\left(-x_{1}+9\right).

$$4\left(\left(1+x_{1}\right)\left(9-x_{1}\right)-8x_{2}\right)=0$$

Úsáid an t-airí dáileach chun 1+x_{1} a mhéadú faoi 9-x_{1} agus chun téarmaí comhchosúla a chumasc.

$$4\left(9+8x_{1}-x_{1}^{2}-8x_{2}\right)=0$$

Úsáid an t-airí dáileach chun 4 a mhéadú faoi 9+8x_{1}-x_{1}^{2}-8x_{2}.

$$36+32x_{1}-4x_{1}^{2}-32x_{2}=0$$

Bain 36 ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

$$32x_{1}-4x_{1}^{2}-32x_{2}=-36$$

Bain 32x_{1} ón dá thaobh.

$$-4x_{1}^{2}-32x_{2}=-36-32x_{1}$$

Cuir 4x_{1}^{2} leis an dá thaobh.

$$-32x_{2}=-36-32x_{1}+4x_{1}^{2}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$-32x_{2}=4x_{1}^{2}-32x_{1}-36$$

Roinn an dá thaobh faoi -32.

$$\frac{-32x_{2}}{-32}=\frac{4\left(x_{1}-9\right)\left(x_{1}+1\right)}{-32}$$

Má roinntear é faoi -32 cuirtear an iolrúchán faoi -32 ar ceal.

$$x_{2}=\frac{4\left(x_{1}-9\right)\left(x_{1}+1\right)}{-32}$$

Roinn 4\left(-9+x_{1}\right)\left(1+x_{1}\right) faoi -32.

$$x_{2}=-\frac{\left(x_{1}-9\right)\left(x_{1}+1\right)}{8}$$

Réitigh do x_1.

$$\left\{\begin{matrix}x_{1}=\sqrt{25-8x_{2}}+4\text{, }&x_{2}\neq 0\text{ and }x_{2}\leq \frac{25}{8}\\x_{1}=-\sqrt{25-8x_{2}}+4\text{, }&x_{2}\leq \frac{25}{8}\end{matrix}\right.$$