Réitigh Q_{1}=600-4P_{A}-003M-12P_{A}+15G+6P_{B}+15N

$$Q _ { 1 } = 600 - 4 P _ { A } - 003 M - 12 P _ { A } + 15 G + 6 P _ { B } + 15 N$$

Réitigh do G.

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

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh 0 agus 0 chun 0 a fháil.

$$Q_{1}=600-4P_{A}-0\times 3M-12P_{A}+15G+6P_{B}+15N$$

Méadaigh 0 agus 3 chun 0 a fháil.

$$Q_{1}=600-4P_{A}-0M-12P_{A}+15G+6P_{B}+15N$$

Is ionann rud ar bith méadaithe faoi nialas agus nialas.

$$Q_{1}=600-4P_{A}-0-12P_{A}+15G+6P_{B}+15N$$

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

$$600-4P_{A}-0-12P_{A}+15G+6P_{B}+15N=Q_{1}$$

Bain 600-4P_{A}-0 ón dá thaobh.

$$-12P_{A}+15G+6P_{B}+15N=Q_{1}-\left(600-4P_{A}-0\right)$$

Cuir 12P_{A} leis an dá thaobh.

$$15G+6P_{B}+15N=Q_{1}-\left(600-4P_{A}-0\right)+12P_{A}$$

Bain 6P_{B} ón dá thaobh.

$$15G+15N=Q_{1}-\left(600-4P_{A}-0\right)+12P_{A}-6P_{B}$$

Bain 15N ón dá thaobh.

$$15G=Q_{1}-\left(600-4P_{A}-0\right)+12P_{A}-6P_{B}-15N$$

Athordaigh na téarmaí.

$$15G=Q_{1}-\left(-4P_{A}+600\right)-15N-6P_{B}+12P_{A}$$

Chun an mhalairt ar -4P_{A}+600 a aimsiú, aimsigh an mhalairt ar gach téarma.

$$15G=Q_{1}+4P_{A}-600-15N-6P_{B}+12P_{A}$$

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

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi 15.

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

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

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

Roinn Q_{1}+16P_{A}-600-15N-6P_{B} faoi 15.

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

Réitigh do M.

$$M\in \mathrm{R}$$

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