Réitigh Q_{1}=600-4P_{A}-0.03M-12P_{A}+15G+6P_{B}+1.5N

$$Q _ { 1 } = 600 - 4 P _ { A } - 0.03 M - 12 P _ { A } + 15 G + 6 P _ { B } + 1.5 N$$

Réitigh do G.

$$G=\frac{M}{500}+\frac{Q_{1}}{15}+\frac{16P_{A}}{15}-\frac{N}{10}-\frac{2P_{B}}{5}-40$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Comhcheangail -4P_{A} agus -12P_{A} chun -16P_{A} a fháil.

$$Q_{1}=600-16P_{A}-0.03M+15G+6P_{B}+1.5N$$

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

$$600-16P_{A}-0.03M+15G+6P_{B}+1.5N=Q_{1}$$

Bain 600 ón dá thaobh.

$$-16P_{A}-0.03M+15G+6P_{B}+1.5N=Q_{1}-600$$

Cuir 16P_{A} leis an dá thaobh.

$$-0.03M+15G+6P_{B}+1.5N=Q_{1}-600+16P_{A}$$

Cuir 0.03M leis an dá thaobh.

$$15G+6P_{B}+1.5N=Q_{1}-600+16P_{A}+0.03M$$

Bain 6P_{B} ón dá thaobh.

$$15G+1.5N=Q_{1}-600+16P_{A}+0.03M-6P_{B}$$

Bain 1.5N ón dá thaobh.

$$15G=Q_{1}-600+16P_{A}+0.03M-6P_{B}-1.5N$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$15G=\frac{3M}{100}-\frac{3N}{2}+Q_{1}+16P_{A}-6P_{B}-600$$

Roinn an dá thaobh faoi 15.

$$\frac{15G}{15}=\frac{\frac{3M}{100}-\frac{3N}{2}+Q_{1}+16P_{A}-6P_{B}-600}{15}$$

Má roinntear é faoi 15 cuirtear an iolrúchán faoi 15 ar ceal.

$$G=\frac{\frac{3M}{100}-\frac{3N}{2}+Q_{1}+16P_{A}-6P_{B}-600}{15}$$

Roinn Q_{1}-600+16P_{A}+\frac{3M}{100}-6P_{B}-\frac{3N}{2} faoi 15.

$$G=\frac{M}{500}+\frac{Q_{1}}{15}+\frac{16P_{A}}{15}-\frac{N}{10}-\frac{2P_{B}}{5}-40$$

Réitigh do M.

$$M=-\frac{100Q_{1}}{3}-\frac{1600P_{A}}{3}+50N+200P_{B}+500G+20000$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Comhcheangail -4P_{A} agus -12P_{A} chun -16P_{A} a fháil.

$$Q_{1}=600-16P_{A}-0.03M+15G+6P_{B}+1.5N$$

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

$$600-16P_{A}-0.03M+15G+6P_{B}+1.5N=Q_{1}$$

Bain 600 ón dá thaobh.

$$-16P_{A}-0.03M+15G+6P_{B}+1.5N=Q_{1}-600$$

Cuir 16P_{A} leis an dá thaobh.

$$-0.03M+15G+6P_{B}+1.5N=Q_{1}-600+16P_{A}$$

Bain 15G ón dá thaobh.

$$-0.03M+6P_{B}+1.5N=Q_{1}-600+16P_{A}-15G$$

Bain 6P_{B} ón dá thaobh.

$$-0.03M+1.5N=Q_{1}-600+16P_{A}-15G-6P_{B}$$

Bain 1.5N ón dá thaobh.

$$-0.03M=Q_{1}-600+16P_{A}-15G-6P_{B}-1.5N$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$-0.03M=-\frac{3N}{2}+Q_{1}+16P_{A}-6P_{B}-15G-600$$

Roinn an dá thaobh den chothromóid faoi -0.03, arb ionann é sin agus an dá thaobh a mhéadú faoi dheilín an chodáin.

$$\frac{-0.03M}{-0.03}=\frac{-\frac{3N}{2}+Q_{1}+16P_{A}-6P_{B}-15G-600}{-0.03}$$

Má roinntear é faoi -0.03 cuirtear an iolrúchán faoi -0.03 ar ceal.

$$M=\frac{-\frac{3N}{2}+Q_{1}+16P_{A}-6P_{B}-15G-600}{-0.03}$$

Roinn Q_{1}-600+16P_{A}-15G-6P_{B}-\frac{3N}{2} faoi -0.03 trí Q_{1}-600+16P_{A}-15G-6P_{B}-\frac{3N}{2} a mhéadú faoi dheilín -0.03.

$$M=-\frac{100Q_{1}}{3}-\frac{1600P_{A}}{3}+50N+200P_{B}+500G+20000$$