A school bought 28 bags of maize and 15 bags of beans for a total of sh.263400. In the following term, the trader increased the cost of bag of maize by 20% and reduced the cost of a bag of beans by 10%,the school was then able to buy 24 bags of maize and 12 bags of beans for a total of sh.231120. Taking x to represent cost of bag of maize and y to represent cost of bag of beans; Form two simplified equations using the information above (ii)Use matrix method to find the cost of 1 bag of maize and 1 bag of beans bought from the trader initially (iii) Another school bought 30 bags of maize and 20 bags of beans after the price adjustments. Use matrix method to determine the total amount of money the school spent.

Shule ilinunua mifuko 28 ya mahindi kwa bei ya $x$ na mifuko 15 ya maharagwe kwa bei ya $y$, kwa jumla ya Sh. 263400.

Bei ya mahindi iliongezeka kwa 20%, hivyo bei mpya ni $1.2x$.
Bei ya maharagwe ilipungua kwa 10%, hivyo bei mpya ni $0.9y$.
Shule ilinunua mifuko 24 ya mahindi na mifuko 12 ya maharagwe kwa jumla ya Sh. 231120.

Ili kuondoa desimali, zidisha mlinganyo mzima kwa 10:

Rahisisha mlinganyo kwa kugawanya kwa 36 (GCD ya 288, 108, na 2311200):

Milinyango iliyorahisishwa ni:

Milinganyo ni:

Katika mfumo wa matriki:

Wacha $A = \begin{bmatrix} 28 & 15 \\ 8 & 3 \end{bmatrix}$.
Kiamua cha $A$, $det(A) = (28 \times 3) - (15 \times 8) = 84 - 120 = -36$.
Kinyume cha $A$, $A^{-1} = \frac{1}{-36} \begin{bmatrix} 3 & -15 \\ -8 & 28 \end{bmatrix}$.

Gharama ya awali ya mfuko 1 wa mahindi ni Sh. 4800 na mfuko 1 wa maharagwe ni Sh. 8600.

Bei mpya ya mahindi ($x'$): $1.2 \times 4800 = 5760$ Sh.
Bei mpya ya maharagwe ($y'$): $0.9 \times 8600 = 7740$ Sh.

Shule nyingine ilinunua mifuko 30 ya mahindi na mifuko 20 ya maharagwe.
Wacha $Q = \begin{bmatrix} 30 & 20 \end{bmatrix}$ iwe matriki ya idadi.
Wacha $C' = \begin{bmatrix} 5760 \\ 7740 \end{bmatrix}$ iwe matriki ya bei zilizorekebishwa.
Jumla ya pesa iliyotumiwa ($T$) ni $Q \cdot C'$.

Jumla ya pesa iliyotumiwa na shule nyingine ni Sh. 327600.