1. **Problem i:** Solve the system $$\begin{cases} x_1 - 5x_2 = -85 \\ 2x_1 + 4x_2 = 40 \end{cases}$$ Step 1: Multiply the first equation by 2 to align $x_1$ coefficients: $$2x_1 - 10x_2 = -170$$ Step 2: Subtract the second equation from this: $$ (2x_1 - 10x_2) - (2x_1 + 4x_2) = -170 - 40 $$ $$ -14x_2 = -210 $$ Step 3: Solve for $x_2$: $$ x_2 = \frac{-210}{-14} = 15 $$ Step 4: Substitute $x_2=15$ into first original equation: $$ x_1 - 5(15) = -85 $$ $$ x_1 - 75 = -85 $$ $$ x_1 = -85 + 75 = -10 $$ Answer: $x_1 = -10$, $x_2 = 15$. 2. **Problem ii:** Solve $$\begin{cases} 3x - 2y = -13 \\ 4x + 6y = 0 \end{cases}$$ Multiply first equation by 3: $$9x - 6y = -39$$ Add to second: $$ (9x - 6y) + (4x + 6y) = -39 + 0 $$ $$ 13x = -39 $$ $$ x = -3 $$ Substitute $x=-3$ into first: $$ 3(-3) - 2y = -13 $$ $$ -9 - 2y = -13 $$ $$ -2y = -4 $$ $$ y = 2 $$ Answer: $x=-3$, $y=2$. 3. **Problem iii:** Solve $$\begin{cases} 10x + 20y = 278 \\ 20x + 30y = 473 \end{cases}$$ Multiply first equation by 2: $$ 20x + 40y = 556 $$ Subtract second equation: $$ (20x + 40y) - (20x + 30y) = 556 - 473 $$ $$ 10y = 83 $$ $$ y = 8.3 $$ Substitute $y=8.3$ into first: $$ 10x + 20(8.3) = 278 $$ $$ 10x + 166 = 278 $$ $$ 10x = 112 $$ $$ x = 11.2 $$ Answer: $x=11.2$, $y=8.3$. 4. **Problem iv:** Solve $$\begin{cases} x + y = 42 \\ 0.24x + 0.18y = 8.4 \end{cases}$$ Multiply second by 100 for easier numbers: $$ 24x + 18y = 840 $$ Multiply first by 18: $$ 18x + 18y = 756 $$ Subtract: $$ (24x + 18y) - (18x + 18y) = 840 - 756 $$ $$ 6x = 84 $$ $$ x = 14 $$ Then $y = 42 - 14 = 28$ Answer: $x=14$, $y=28$. 5. **Problem v:** Solve $$\begin{cases} 3x + 5z = 14 \\ x + 4y - 2z = -10 \\ x + y + z = 2 \end{cases}$$ Step 1: From third, express $y$: $$ y = 2 - x - z $$ Substitute $y$ into second: $$ x + 4(2 - x - z) - 2z = -10 $$ $$ x + 8 - 4x - 4z - 2z = -10 $$ $$ -3x - 6z = -18 $$ $$ 3x + 6z = 18 $$ Step 2: System now: $$\begin{cases} 3x + 5z = 14 \\ 3x + 6z = 18 \end{cases}$$ Subtract first from second: $$ (3x + 6z) - (3x + 5z) = 18 - 14 $$ $$ z = 4 $$ Step 3: Substitute $z=4$ into first: $$ 3x + 5(4) = 14 $$ $$ 3x + 20 = 14 $$ $$ 3x = -6 $$ $$ x = -2 $$ Step 4: Calculate $y$: $$ y = 2 - (-2) - 4 = 2 + 2 - 4 = 0 $$ Answer: $x = -2$, $y=0$, $z=4$. 6. **Problem vi:** Solve $$\begin{cases} x + y + z = 2 \\ 2x - y + z = 9 \\ 2x + 3y + 4z = 4 \end{cases}$$ Step 1: Add first and second: $$ (x + y + z) + (2x - y + z) = 2 + 9 $$ $$ 3x + 2z = 11 $$ Step 2: Use first for $x$: $$ x = 2 - y - z $$ Substitute into third: $$ 2(2 - y - z) + 3y + 4z = 4 $$ $$ 4 - 2y - 2z + 3y + 4z = 4 $$ $$ 4 + y + 2z = 4 $$ $$ y + 2z = 0 $$ Step 3: From Step 1: $$ 3x + 2z = 11 $$ Substitute $x = 2 - y - z$: $$ 3(2 - y - z) + 2z = 11 $$ $$ 6 - 3y - 3z + 2z = 11 $$ $$ 6 - 3y - z = 11 $$ $$ -3y - z = 5 $$ Step 4: From Step 2, $y = -2z$; substitute into above: $$ -3(-2z) - z = 5 $$ $$ 6z - z = 5 $$ $$ 5z = 5 $$ $$ z = 1 $$ Step 5: Find $y$: $$ y = -2(1) = -2 $$ Step 6: Find $x$: $$ x = 2 - (-2) - 1 = 2 + 2 - 1 = 3 $$ Answer: $x=3$, $y=-2$, $z=1$. 7. **Problem vii:** Solve $$\begin{cases} 2x + 3y + 4z = 29 \\ x + y + 2z = 13 \\ 3x + 2y + z = 16 \end{cases}$$ Step 1: From second: $$ x = 13 - y - 2z $$ Step 2: Substitute into first: $$ 2(13 - y - 2z) + 3y + 4z = 29 $$ $$ 26 - 2y - 4z + 3y + 4z = 29 $$ $$ 26 + y = 29 $$ $$ y = 3 $$ Step 3: Substitute $x,y$ into third: $$ 3(13 - 3 - 2z) + 2(3) + z = 16 $$ $$ 3(10 - 2z) + 6 + z = 16 $$ $$ 30 - 6z + 6 + z = 16 $$ $$ 36 - 5z = 16 $$ $$ -5z = -20 $$ $$ z = 4 $$ Step 4: Find $x$: $$ x = 13 - 3 - 2(4) = 13 - 3 - 8 = 2 $$ Answer: $x=2$, $y=3$, $z=4$.