1. Problem: Find the price of 1 book and 1 notebook given two purchases: Let $x$ be the price of 1 book and $y$ be the price of 1 notebook. Equations: $$3x + 4y = 19500$$ $$5x + 3y = 23000$$ 2. Solve one of the equations for $x$ or $y$; from the first: $$3x = 19500 - 4y \\ x = \frac{19500 - 4y}{3}$$ 3. Substitute $x$ in the second equation: $$5 \times \frac{19500 - 4y}{3} + 3y = 23000$$ Multiply both sides by 3: $$5(19500 - 4y) + 9y = 69000$$ $$97500 - 20y + 9y = 69000$$ $$-11y = 69000 - 97500 = -28500$$ $$y = \frac{28500}{11} = 2590.91$$ 4. Substitute $y$ back to find $x$: $$x = \frac{19500 - 4 \times 2590.91}{3} = \frac{19500 - 10363.64}{3} = \frac{9136.36}{3} = 3045.45$$ --- 2. Problem: Determine the price difference between mango and orange from: $$4j + 7m = 66500$$ $$9j + m = 63700$$ Let $j$ be price of 1 orange, $m$ price of 1 mango. 1. Solve second for $m$: $$m = 63700 - 9j$$ 2. Substitute in first: $$4j + 7(63700 - 9j) = 66500$$ $$4j + 445900 - 63j = 66500$$ $$-59j = 66500 - 445900 = -379400$$ $$j = \frac{379400}{59} = 6433.90$$ 3. Find $m$: $$m = 63700 - 9 \times 6433.90 = 63700 - 57805 = 5895$$ 4. Price difference: $$|m - j| = |5895 - 6433.90| = 538.90$$ --- 3. Problem: Find prices of drink bottle ($x$), bread ($y$), chocolate ($z$) from: $$3x + 2y + z = 37000$$ $$2x + 3y + 2z = 39000$$ $$x + y + 3z = 38000$$ 1. Solve system by elimination or substitution. Multiply third by 2: $$2x + 2y + 6z = 76000$$ Subtract second from this: $$(2x + 2y + 6z) - (2x + 3y + 2z) = 76000 - 39000$$ $$-y + 4z = 37000$$ 2. From this: $$y = 4z - 37000$$ 3. Substitute $y$ in first equation: $$3x + 2(4z - 37000) + z = 37000$$ $$3x + 8z - 74000 + z = 37000$$ $$3x + 9z = 111000$$ 4. Substitute $y$ in third: $$x + (4z - 37000) + 3z = 38000$$ $$x + 7z = 75000$$ 5. From above, express $x$: $$x = 75000 - 7z$$ 6. Substitute $x$ in equation with $3x + 9z = 111000$: $$3(75000 - 7z) + 9z = 111000$$ $$225000 - 21z + 9z = 111000$$ $$-12z = 111000 - 225000 = -114000$$ $$z = \frac{114000}{12} = 9500$$ 7. Find $x$: $$x = 75000 - 7 \times 9500 = 75000 - 66500 = 8500$$ 8. Find $y$: $$y = 4 \times 9500 - 37000 = 38000 - 37000 = 1000$$ --- 4. Problem parts: a) Weight of fertilizers A, B, C: Equations: $$3a + 2b + c = 50$$ $$a + 3b + 2c = 54$$ 1. Solve first for $c$: $$c = 50 - 3a - 2b$$ 2. Substitute $c$ in second: $$a + 3b + 2(50 - 3a - 2b) = 54$$ $$a + 3b + 100 - 6a - 4b = 54$$ $$-5a - b = -46$$ $$b = -5a + 46$$ 3. Substitute back to $c$: $$c = 50 - 3a - 2(-5a + 46) = 50 -3a + 10a - 92 = 7a - 42$$ b) If prices per kg are $p_a, p_b, p_c$, total price sum given is $p_a + p_b + p_c$ (not provided actual prices so cannot calculate). c) Discount 25% on price of C means price of C after discount is $0.75 p_c$. --- 5. Problem: Find number of 10000 bills each donated: Let $x_n, x_c, x_f$ be the number of 10000 bills from Nadine, Chaca, Floretta respectively. Equations given: Nadine: $$2\times 50000 + 1\times 20000 + x_n \times 10000 = 120000$$ $$100000 + 20000 + 10000 x_n = 120000$$ $$10000 x_n = 0$$ $$x_n = 0$$ Chaca: $$50000 + 2\times 20000 + 10000 x_c = 120000$$ $$50000 + 40000 + 10000 x_c = 120000$$ $$10000 x_c = 120000 - 90000 = 30000$$ $$x_c = 3$$ Floretta: $$50000 + 20000 + 10000 x_f = 90000$$ $$70000 + 10000 x_f = 90000$$ $$10000 x_f = 20000$$ $$x_f = 2$$ --- 6. Problem: Solve system $$2x + 3y = 13$$ $$3x - y = 7$$ 1. Solve second for $y$: $$y = 3x - 7$$ 2. Substitute in first: $$2x + 3(3x - 7) = 13$$ $$2x + 9x - 21 = 13$$ $$11x = 34$$ $$x = \frac{34}{11}$$ 3. Find $y$: $$y = 3 \times \frac{34}{11} - 7 = \frac{102}{11} - 7 = \frac{102 - 77}{11} = \frac{25}{11}$$ --- 7. Problem: Solve $$x + y + z = 6$$ $$2x - y + 3z = 14$$ $$x + 2y - z = 3$$ 1. From first: $$z = 6 - x - y$$ 2. Substitute in second: $$2x - y + 3(6 - x - y) = 14$$ $$2x - y + 18 - 3x - 3y = 14$$ $$-x - 4y = -4$$ $$x + 4y = 4$$ 3. Substitute $z$ in third: $$x + 2y - (6 - x - y) = 3$$ $$x + 2y - 6 + x + y = 3$$ $$2x + 3y = 9$$ 4. Solve system: $$\begin{cases} x + 4y = 4 \\ 2x + 3y = 9 \end{cases}$$ Multiply first by 2: $$2x + 8y = 8$$ Subtract second: $$(2x + 8y) - (2x + 3y) = 8 - 9$$ $$5y = -1$$ $$y = -\frac{1}{5}$$ 5. Find $x$: $$x = 4 - 4y = 4 - 4(-\frac{1}{5}) = 4 + \frac{4}{5} = \frac{24}{5}$$ 6. Find $z$: $$z = 6 - x - y = 6 - \frac{24}{5} + \frac{1}{5} = \frac{30 - 24 + 1}{5} = \frac{7}{5}$$ --- 8. Problem: $$2a + 3b + c = 12$$ $$a - 2b + 2c = 3$$ $$3a + b - c = 8$$ 1. Express $c$ from first: $$c = 12 - 2a - 3b$$ 2. Substitute $c$ into second: $$a - 2b + 2(12 - 2a - 3b) = 3$$ $$a - 2b + 24 - 4a - 6b = 3$$ $$-3a - 8b = -21$$ $$3a + 8b = 21$$ 3. Substitute $c$ into third: $$3a + b - (12 - 2a - 3b) = 8$$ $$3a + b - 12 + 2a + 3b = 8$$ $$5a + 4b = 20$$ 4. Solve system: $$\begin{cases} 3a + 8b = 21 \\ 5a + 4b = 20 \end{cases}$$ Multiply second by 2: $$10a + 8b = 40$$ Subtract first multiplied by 1: $$(10a + 8b) - (3a + 8b) = 40 - 21$$ $$7a = 19$$ $$a = \frac{19}{7}$$ 5. Find $b$: $$3 \times \frac{19}{7} + 8b = 21$$ $$\frac{57}{7} + 8b = 21$$ $$8b = 21 - \frac{57}{7} = \frac{147 - 57}{7} = \frac{90}{7}$$ $$b = \frac{90}{56} = \frac{45}{28}$$ 6. Find $c$: $$c = 12 - 2\times \frac{19}{7} - 3 \times \frac{45}{28} = 12 - \frac{38}{7} - \frac{135}{28}$$ Convert to 28 denominator: $$12 = \frac{336}{28}, \frac{38}{7} = \frac{152}{28}$$ $$c = \frac{336}{28} - \frac{152}{28} - \frac{135}{28} = \frac{49}{28} = \frac{7}{4}$$ 7. Part b: $$a + b + c = \frac{19}{7} + \frac{45}{28} + \frac{7}{4} = \frac{76}{28} + \frac{45}{28} + \frac{49}{28} = \frac{170}{28} = \frac{85}{14}$$ 8. Part c: $$a^2 + 3b = \left( \frac{19}{7} \right)^2 + 3 \times \frac{45}{28} = \frac{361}{49} + \frac{135}{28}$$ Convert to common denominator 196: $$\frac{361}{49} = \frac{1444}{196}$$ $$\frac{135}{28} = \frac{945}{196}$$ Sum: $$\frac{1444 + 945}{196} = \frac{2389}{196}$$ Final answers incorporated in steps.