Réitigh (T_{1}-T_{2})*0.8=T_2^4*5.620-8*0.05

$$( T _ { 1 } - T _ { 2 } ) \times 0.8 = T _ { 2 } ^ { 4 } \times 5.620 - 8 \times 0.05$$

Réitigh do T_1.

$$T_{1}=\frac{281T_{2}^{4}}{40}+T_{2}-0.5$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun T_{1}-T_{2} a mhéadú faoi 0.8.

$$0.8T_{1}-0.8T_{2}=T_{2}^{4}\times 5.62-8\times 0.05$$

Méadaigh 8 agus 0.05 chun 0.4 a fháil.

$$0.8T_{1}-0.8T_{2}=T_{2}^{4}\times 5.62-0.4$$

Cuir 0.8T_{2} leis an dá thaobh.

$$0.8T_{1}=T_{2}^{4}\times 5.62-0.4+0.8T_{2}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$0.8T_{1}=\frac{281T_{2}^{4}}{50}+\frac{4T_{2}}{5}-0.4$$

Roinn an dá thaobh den chothromóid faoi 0.8, arb ionann é sin agus an dá thaobh a mhéadú faoi dheilín an chodáin.

$$\frac{0.8T_{1}}{0.8}=\frac{\frac{281T_{2}^{4}}{50}+\frac{4T_{2}}{5}-0.4}{0.8}$$

Má roinntear é faoi 0.8 cuirtear an iolrúchán faoi 0.8 ar ceal.

$$T_{1}=\frac{\frac{281T_{2}^{4}}{50}+\frac{4T_{2}}{5}-0.4}{0.8}$$

Roinn \frac{281T_{2}^{4}}{50}-0.4+\frac{4T_{2}}{5} faoi 0.8 trí \frac{281T_{2}^{4}}{50}-0.4+\frac{4T_{2}}{5} a mhéadú faoi dheilín 0.8.

$$T_{1}=\frac{281T_{2}^{4}}{40}+T_{2}-\frac{1}{2}$$