# Step-by-Step Solution Let's determine the force in each cable needed to support the **20-kg flowerpot**. ## **Step 1: Draw a Free-Body Diagram** Identify all forces acting at point A (where the cables meet and the flowerpot is attached): - Tension in cable \( AB \) (\( T_{AB} \)) - Tension in cable \( AC \) (\( T_{AC} \)) - Tension in cable \( AD \) (\( T_{AD} \)) - Weight of the flowerpot (\( W_{pot} \)) --- ## **Step 2: Write Position Vectors** Let’s assign coordinates as given: - \( A = (0, 0, 4) \) - \( B = (0, -3, 10) \) - \( C = (3, 4, 0) \) - \( D = (-2, 2, 0) \) --- ## **Step 3: Find Unit Vectors for Each Cable** ### **For \( AB \):** \[ \overrightarrow{AB} = B - A = (0, -3, 10) - (0, 0, 4) = (0, -3, 6) \] \[ |\overrightarrow{AB}| = \sqrt{0^2 + (-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 6.71 \] \[ \mathbf{u}_{AB} = \frac{1}{6.71}(0\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = (0, -0.447, 0.894) \] --- ### **For \( AC \):** \[ \overrightarrow{AC} = (3, 4, 0) - (0, 0, 4) = (3, 4, -4) \] \[ |\overrightarrow{AC}| = \sqrt{3^2 + 4^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41} = 6.40 \] \[ \mathbf{u}_{AC} = \frac{1}{6.40}(3\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}) = (0.469, 0.625, -0.625) \] --- ### **For \( AD \):** \[ \overrightarrow{AD} = (-2, 2, 0) - (0, 0, 4) = (-2, 2, -4) \] \[ |\overrightarrow{AD}| = \sqrt{(-2)^2 + 2^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 4.90 \] \[ \mathbf{u}_{AD} = \frac{1}{4.90}(-2\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}) = (-0.408, 0.408, -0.816) \] --- ### **For the Pot Weight:** \[ \mathbf{u}_{pot} = -\mathbf{k} \] --- ## **Step 4: Write the Force Equilibrium Equations** The sum of forces at point A must be zero: \[ \sum \vec{F} = 0 \] \[ T_{AB}\mathbf{u}_{AB} + T_{AC}\mathbf{u}_{AC} + T_{AD}\mathbf{u}_{AD} + W_{pot}\mathbf{u}_{pot} = 0 \] --- ## **Step 5: Write in Component Form** ### Let \( W_{pot} = mg = 20 \times 9.81 = 196.2\, N \) Sum the forces in \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) directions: #### **\(\mathbf{i}\):** \[ 0 \cdot T_{AB} + 0.469 T_{AC} - 0.408 T_{AD} = 0 \] #### **\(\mathbf{j}\):** \[ -0.447 T_{AB} + 0.625 T_{AC} + 0.408 T_{AD} = 0 \] #### **\(\mathbf{k}\):** \[ 0.894 T_{AB} - 0.625 T_{AC} - 0.816 T_{AD} - 196.2 = 0 \] --- ## **Step 6: Solve the System of Equations** ### From \(\mathbf{i}\): \[ 0.469 T_{AC} = 0.408 T_{AD} \implies T_{AC} = \frac{0.408}{0.469} T_{AD} \approx 0.870 T_{AD} \] ### Substitute \( T_{AC} \) into \(\mathbf{j}\) equation: \[ -0.447 T_{AB} + 0.625 T_{AC} + 0.408 T_{AD} = 0 \] \[ -0.447 T_{AB} + 0.625 (0.870 T_{AD}) + 0.408 T_{AD} = 0 \] \[ -0.447 T_{AB} + 0.544 T_{AD} + 0.408 T_{AD} = 0 \implies -0.447 T_{AB} + 0.952 T_{AD} = 0 \implies T_{AB} = \frac{0.952}{0.447} T_{AD} \approx 2.13 T_{AD} \] --- ### Substitute \(T_{AB}\) and \(T_{AC}\) into \(\mathbf{k}\) equation: \[ 0.894 T_{AB} - 0.625 T_{AC} - 0.816 T_{AD} = 196.2 \] \[ 0.894 (2.13 T_{AD}) - 0.625 (0.87 T_{AD}) - 0.816 T_{AD} = 196.2 \] \[ 1.905 T_{AD} - 0.544 T_{AD} - 0.816 T_{AD} = 196.2 \] \[ (1.905 - 0.544 - 0.816) T_{AD} = 196.2 \] \[ 0.545 T_{AD} = 196.2 \] \[ T_{AD} = \frac{196.2}{0.545} = 360\, N \] --- ### Back-solve for \(T_{AB}\) and \(T_{AC}\): \[ T_{AB} = 2.13 \times 360 = 766.8\, N \] \[ T_{AC} = 0.87 \times 360 = 313.2\, N \] --- ## **Step 7: Final Answers** \[ \boxed{ \begin{aligned} T_{AB} & = 767\, \text{N} \\ T_{AC} & = 313\, \text{N} \\ T_{AD} & = 360\, \text{N} \end{aligned} } \] --- ### **Formulas Used** 1. **Unit Vector:** \( \mathbf{u} = \frac{\overrightarrow{r}}{|\overrightarrow{r}|} \) 2. **Force Equilibrium:** \( \sum \vec{F} = 0 \) 3. **Weight:** \( W = mg \) --- **Let me know if you need a breakdown for a specific cable or direction!**

# Step-by-Step Solution We will determine the force in each cable needed to support the **20-kg flowerpot** using the provided coordinates. ## **Step 1: Draw a Free-Body Diagram** Identify the forces acting at point A: - Tension in cable \( AB \) (\( T_{AB} \)) - Tension in cable \( AC \) (\( T_{AC} \)) - Tension in cable \( AD \) (\( T_{AD} \)) - Weight of the flowerpot (\( W_{pot} \)) ## **Step 2: Given Coordinates** Assign coordinates as follows: - \( A = (0, -4, 0) \) - \( B = (0, -7, 6) \) - \( C = (2, 0, 0) \) - \( D = (-2, 0, 0) \) ## **Step 3: Calculate Unit Vectors for Each Cable** ### **For \( AB \):** \[ \overrightarrow{AB} = B - A = (0, -7, 6) - (0, -4, 0) = (0, -3, 6) \] \[ |\overrightarrow{AB}| = \sqrt{0^2 + (-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.71 \] \[ \mathbf{u}_{AB} = \frac{1}{6.71}(0\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = (0, -0.447, 0.894) \] ### **For \( AC \):** \[ \overrightarrow{AC} = C - A = (2, 0, 0) - (0, -4, 0) = (2, 4, 0) \] \[ |\overrightarrow{AC}| = \sqrt{2^2 + 4^2 + 0^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \] \[ \mathbf{u}_{AC} = \frac{1}{4.47}(2\mathbf{i} + 4\mathbf{j}) = (0.447, 0.894, 0) \] ### **For \( AD \):** \[ \overrightarrow{AD} = D - A = (-2, 0, 0) - (0, -4, 0) = (-2, 4, 0) \] \[ |\overrightarrow{AD}| = \sqrt{(-2)^2 + 4^2 + 0^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \] \[ \mathbf{u}_{AD} = \frac{1}{4.47}(-2\mathbf{i} + 4\mathbf{j}) = (-0.447, 0.894, 0) \] ## **Step 4: Calculate Weight of the Pot** The weight of the flowerpot is: \[ W_{pot} = mg = 20 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 196.2 \, \text{N} \] ## **Step 5: Write the Force Equilibrium Equations** The sum of forces at point A must be zero: \[ \sum \vec{F} = 0 \] Thus: \[ T_{AB} \mathbf{u}_{AB} + T_{AC} \mathbf{u}_{AC} + T_{AD} \mathbf{u}_{AD} - W_{pot} \mathbf{k} = 0 \] ### Write in Component Form #### **\(\mathbf{i}\):** \[ 0 \cdot T_{AB} + 0.447 T_{AC} - 0.447 T_{AD} = 0 \] #### **\(\mathbf{j}\):** \[ -0.447 T_{AB} + 0.894 T_{AC} + 0.894 T_{AD} = 0 \] #### **\(\mathbf{k}\):** \[ 0.894 T_{AB} + 0 \cdot T_{AC} + 0 \cdot T_{AD} - 196.2 = 0 \] ## **Step 6: Solve the System of Equations** ### From \(\mathbf{i}\): \[ 0.447 T_{AC} = 0.447 T_{AD} \implies T_{AC} = T_{AD} \] ### Substitute \(T_{AC}\) into \(\mathbf{j}\) equation: \[ -0.447 T_{AB} + 0.894 T_{AD} + 0.894 T_{AD} = 0 \] \[ -0.447 T_{AB} + 1.788 T_{AD} = 0 \implies T_{AB} = \frac{1.788}{0.447} T_{AD} \approx 4.00 T_{AD} \] ### Substitute \(T_{AB}\) into \(\mathbf{k}\) equation: \[ 0.894 (4.00 T_{AD}) = 196.2 \] \[ 3.576 T_{AD} = 196.2 \implies T_{AD} = \frac{196.2}{3.576} \approx 54.83 \, \text{N} \] ### Back-solve for \(T_{AB}\) and \(T_{AC}\): \[ T_{AC} = T_{AD} \approx 54.83 \, \text{N} \] \[ T_{AB} = 4.00 \times 54.83 \approx 219.32 \, \text{N} \] ## **Step 7: Final Answers** \[ \boxed{ \begin{aligned} T_{AB} & \approx 219.32 \, \text{N} \\ T_{AC} & \approx 54.83 \, \text{N} \\ T_{AD} & \approx 54.83 \, \text{N} \end{aligned} } \] ### **Formulas Used** 1. **Unit Vector:** \( \mathbf{u} = \frac{\overrightarrow{r}}{|\overrightarrow{r}|} \) 2. **Force Equilibrium:** \( \sum \vec{F} = 0 \) 3. **Weight:** \( W = mg \) Let me know if you need further clarification on any part of the solution!

# Step-by-Step Solution for a 22.1 kg Flowerpot We will determine the force in each cable needed to support a **22.1 kg flowerpot** using the provided coordinates. ## **Step 1: Draw a Free-Body Diagram** Identify the forces acting at point A: - Tension in cable \( AB \) (\( T_{AB} \)) - Tension in cable \( AC \) (\( T_{AC} \)) - Tension in cable \( AD \) (\( T_{AD} \)) - Weight of the flowerpot (\( W_{pot} \)) ## **Step 2: Given Coordinates** Assign coordinates as follows: - \( A = (0, -4, 0) \) - \( B = (0, -7, 6) \) - \( C = (2, 0, 0) \) - \( D = (-2, 0, 0) \) ## **Step 3: Calculate the Weight of the Pot** The weight of the flowerpot is: \[ W_{pot} = mg = 22.1 \, \text{kg} \times 9.81 \, \text{m/s}^2 \approx 216.3 \, \text{N} \] ## **Step 4: Calculate Unit Vectors for Each Cable** ### **For \( AB \):** \[ \overrightarrow{AB} = B - A = (0, -7, 6) - (0, -4, 0) = (0, -3, 6) \] \[ |\overrightarrow{AB}| = \sqrt{0^2 + (-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.71 \] \[ \mathbf{u}_{AB} = \frac{1}{6.71}(0\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = (0, -0.447, 0.894) \] ### **For \( AC \):** \[ \overrightarrow{AC} = C - A = (2, 0, 0) - (0, -4, 0) = (2, 4, 0) \] \[ |\overrightarrow{AC}| = \sqrt{2^2 + 4^2 + 0^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \] \[ \mathbf{u}_{AC} = \frac{1}{4.47}(2\mathbf{i} + 4\mathbf{j}) = (0.447, 0.894, 0) \] ### **For \( AD \):** \[ \overrightarrow{AD} = D - A = (-2, 0, 0) - (0, -4, 0) = (-2, 4, 0) \] \[ |\overrightarrow{AD}| = \sqrt{(-2)^2 + 4^2 + 0^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \] \[ \mathbf{u}_{AD} = \frac{1}{4.47}(-2\mathbf{i} + 4\mathbf{j}) = (-0.447, 0.894, 0) \] ## **Step 5: Write the Force Equilibrium Equations** The sum of forces at point A must be zero: \[ \sum \vec{F} = 0 \] Thus: \[ T_{AB} \mathbf{u}_{AB} + T_{AC} \mathbf{u}_{AC} + T_{AD} \mathbf{u}_{AD} - W_{pot} \mathbf{k} = 0 \] ### Write in Component Form #### **\(\mathbf{i}\):** \[ 0 \cdot T_{AB} + 0.447 T_{AC} - 0.447 T_{AD} = 0 \] #### **\(\mathbf{j}\):** \[ -0.447 T_{AB} + 0.894 T_{AC} + 0.894 T_{AD} = 0 \] #### **\(\mathbf{k}\):** \[ 0.894 T_{AB} + 0 \cdot T_{AC} + 0 \cdot T_{AD} - 216.3 = 0 \] ## **Step 6: Solve the System of Equations** ### From \(\mathbf{i}\): \[ 0.447 T_{AC} = 0.447 T_{AD} \implies T_{AC} = T_{AD} \] ### Substitute \(T_{AC}\) into \(\mathbf{j}\) equation: \[ -0.447 T_{AB} + 0.894 T_{AD} + 0.894 T_{AD} = 0 \] \[ -0.447 T_{AB} + 1.788 T_{AD} = 0 \implies T_{AB} = \frac{1.788}{0.447} T_{AD} \approx 4.00 T_{AD} \] ### Substitute \(T_{AB}\) into \(\mathbf{k}\) equation: \[ 0.894 (4.00 T_{AD}) = 216.3 \] \[ 3.576 T_{AD} = 216.3 \implies T_{AD} = \frac{216.3}{3.576} \approx 60.50 \, \text{N} \] ### Back-solve for \(T_{AB}\) and \(T_{AC}\): \[ T_{AC} = T_{AD} \approx 60.50 \, \text{N} \] \[ T_{AB} = 4.00 \times 60.50 \approx 243.00 \, \text{N} \] ## **Step 7: Final Answers** \[ \boxed{ \begin{aligned} T_{AB} & \approx 243.00 \, \text{N} \\ T_{AC} & \approx 60.50 \, \text{N} \\ T_{AD} & \approx 60.50 \, \text{N} \end{aligned} } \] ### **Formulas Used** 1. **Unit Vector:** \( \mathbf{u} = \frac{\overrightarrow{r}}{|\overrightarrow{r}|} \) 2. **Force Equilibrium:** \( \sum \vec{F} = 0 \) 3. **Weight:** \( W = mg \) If you need further clarification or adjustments, feel free to ask!

# Step-by-Step Solution for a 22.1 kg Flowerpot We will determine the force in each cable needed to support a **22.1 kg flowerpot** using the provided coordinates and correcting the weight calculation. ## **Step 1: Draw a Free-Body Diagram** Identify the forces acting at point A: - Tension in cable \( AB \) (\( T_{AB} \)) - Tension in cable \( AC \) (\( T_{AC} \)) - Tension in cable \( AD \) (\( T_{AD} \)) - Weight of the flowerpot (\( W_{pot} \)) ## **Step 2: Given Coordinates** Assign coordinates as follows: - \( A = (0, -4, 0) \) - \( B = (0, -7, 6) \) - \( C = (2, 0, 0) \) - \( D = (-2, 0, 0) \) ## **Step 3: Calculate the Weight of the Pot** The weight of the flowerpot is: \[ W_{pot} = mg = 22.1 \, \text{kg} \times 9.81 \, \text{m/s}^2 \approx 216.801 \, \text{N} \] ## **Step 4: Calculate Unit Vectors for Each Cable** ### **For \( AB \):** \[ \overrightarrow{AB} = B - A = (0, -7, 6) - (0, -4, 0) = (0, -3, 6) \] \[ |\overrightarrow{AB}| = \sqrt{0^2 + (-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.71 \] \[ \mathbf{u}_{AB} = \frac{1}{6.71}(0\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = (0, -0.447, 0.894) \] ### **For \( AC \):** \[ \overrightarrow{AC} = C - A = (2, 0, 0) - (0, -4, 0) = (2, 4, 0) \] \[ |\overrightarrow{AC}| = \sqrt{2^2 + 4^2 + 0^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \] \[ \mathbf{u}_{AC} = \frac{1}{4.47}(2\mathbf{i} + 4\mathbf{j}) = (0.447, 0.894, 0) \] ### **For \( AD \):** \[ \overrightarrow{AD} = D - A = (-2, 0, 0) - (0, -4, 0) = (-2, 4, 0) \] \[ |\overrightarrow{AD}| = \sqrt{(-2)^2 + 4^2 + 0^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \] \[ \mathbf{u}_{AD} = \frac{1}{4.47}(-2\mathbf{i} + 4\mathbf{j}) = (-0.447, 0.894, 0) \] ## **Step 5: Write the Force Equilibrium Equations** The sum of forces at point A must be zero: \[ \sum \vec{F} = 0 \] Thus: \[ T_{AB} \mathbf{u}_{AB} + T_{AC} \mathbf{u}_{AC} + T_{AD} \mathbf{u}_{AD} - W_{pot} \mathbf{k} = 0 \] ### Write in Component Form #### **\(\mathbf{i}\):** \[ 0 \cdot T_{AB} + 0.447 T_{AC} - 0.447 T_{AD} = 0 \] #### **\(\mathbf{j}\):** \[ -0.447 T_{AB} + 0.894 T_{AC} + 0.894 T_{AD} = 0 \] #### **\(\mathbf{k}\):** \[ 0.894 T_{AB} + 0 \cdot T_{AC} + 0 \cdot T_{AD} - 216.801 = 0 \] ## **Step 6: Solve the System of Equations** ### From \(\mathbf{i}\): \[ 0.447 T_{AC} = 0.447 T_{AD} \implies T_{AC} = T_{AD} \] ### Substitute \(T_{AC}\) into \(\mathbf{j}\) equation: \[ -0.447 T_{AB} + 0.894 T_{AD} + 0.894 T_{AD} = 0 \] \[ -0.447 T_{AB} + 1.788 T_{AD} = 0 \implies T_{AB} = \frac{1.788}{0.447} T_{AD} \approx 4.00 T_{AD} \] ### Substitute \(T_{AB}\) into \(\mathbf{k}\) equation: \[ 0.894 (4.00 T_{AD}) = 216.801 \] \[ 3.576 T_{AD} = 216.801 \implies T_{AD} = \frac{216.801}{3.576} \approx 60.50 \, \text{N} \] ### Back-solve for \(T_{AB}\) and \(T_{AC}\): \[ T_{AC} = T_{AD} \approx 60.50 \, \text{N} \] \[ T_{AB} = 4.00 \times 60.50 \approx 243.00 \, \text{N} \] ## **Step 7: Final Answers** \[ \boxed{ \begin{aligned} T_{AB} & \approx 243.00 \, \text{N} \\ T_{AC} & \approx 60.50 \, \text{N} \\ T_{AD} & \approx 60.50 \, \text{N} \end{aligned} } \] ### **Formulas Used** 1. **Unit Vector:** \( \mathbf{u} = \frac{\overrightarrow{r}}{|\overrightarrow{r}|} \) 2. **Force Equilibrium:** \( \sum \vec{F} = 0 \) 3. **Weight:** \( W = mg \) If you have further questions or need additional assistance, feel free to ask!

# Step-by-Step Solution for the System of Equations We will solve the given system of equations involving coefficients \( C_1, C_2, C_3, C_4, C_5, C_6 \). ## **Given Equations:** 1. \[ -\frac{1}{3}(3^4) + 3C_1 + C_2 = 0 \] 2. \[ \frac{22}{3}(3^3) - 84(3^2) + 3C_3 + C_4 = 0 \] 3. \[ -\frac{4}{3}(3^3) + C_1 = 22(3^2) - 168(3) + C_3 \] 4. \[ -\frac{8}{3}(12^3) + 96(12^2) + 12C_5 + C_6 = 0 \] 5. \[ \frac{22}{3}(6^3) - 84(6^2) + 6C_3 + C_4 = -\frac{8}{3}(6^3) + 96(6^2) + 6C_5 + C_6 \] 6. \[ 22(6^2) - 168(6) + C_3 = -8(6^2) + 192(6) + C_5 \] ## **Step 1: Simplify Each Equation** ### 1. Simplifying the first equation: \[ -\frac{1}{3}(81) + 3C_1 + C_2 = 0 \] \[ -27 + 3C_1 + C_2 = 0 \implies 3C_1 + C_2 = 27 \tag{Eq. 1} \] ### 2. Simplifying the second equation: \[ \frac{22}{3}(27) - 84(9) + 3C_3 + C_4 = 0 \] \[ \frac{594}{3} - 756 + 3C_3 + C_4 = 0 \implies 198 - 756 + 3C_3 + C_4 = 0 \] \[ 3C_3 + C_4 = 558 \tag{Eq. 2} \] ### 3. Simplifying the third equation: \[ -\frac{4}{3}(27) + C_1 = 22(9) - 168(3) + C_3 \] \[ -\frac{108}{3} + C_1 = 198 - 504 + C_3 \] \[ -36 + C_1 = -306 + C_3 \implies C_1 - C_3 = -270 \tag{Eq. 3} \] ### 4. Simplifying the fourth equation: \[ -\frac{8}{3}(1728) + 96(144) + 12C_5 + C_6 = 0 \] \[ -4608 + 13824 + 12C_5 + C_6 = 0 \implies 12C_5 + C_6 = -9216 \tag{Eq. 4} \] ### 5. Simplifying the fifth equation: \[ \frac{22}{3}(216) - 84(36) + 6C_3 + C_4 = -\frac{8}{3}(216) + 96(36) + 6C_5 + C_6 \] Calculating both sides: \[ \frac{4752}{3} - 3024 + 6C_3 + C_4 = -\frac{1728}{3} + 3456 + 6C_5 + C_6 \] \[ 1584 - 3024 + 6C_3 + C_4 = -576 + 3456 + 6C_5 + C_6 \] \[ 6C_3 + C_4 = 576 + 6C_5 + C_6 - 1440 \implies 6C_3 + C_4 = 4032 + 6C_5 + C_6 \tag{Eq. 5} \] ### 6. Simplifying the sixth equation: \[ 22(36) - 168(6) + C_3 = -8(36) + 192(6) + C_5 \] Calculating: \[ 792 - 1008 + C_3 = -288 + 1152 + C_5 \] \[ C_3 - C_5 = 288 \tag{Eq. 6} \] ## **Step 2: Solve the System of Equations** ### From Eq. 1: \[ C_2 = 27 - 3C_1 \tag{A} \] ### From Eq. 3: \[ C_3 = C_1 + 270 \tag{B} \] ### Substitute Eq. B into Eq. 2: \[ 3(C_1 + 270) + C_4 = 558 \] \[ 3C_1 + 810 + C_4 = 558 \implies C_4 = -252 - 3C_1 \tag{C} \] ### Substitute Eq. A and C into Eq. 5: \[ 6(C_1 + 270) + (-252 - 3C_1) = 4032 + 6C_5 + C_6 \] \[ 6C_1 + 1620 - 252 - 3C_1 = 4032 + 6C_5 + C_6 \] \[ 3C_1 + 1368 = 4032 + 6C_5 + C_6 \tag{D} \] ### From Eq. 4: \[ C_6 = -9216 - 12C_5 \tag{E} \] ### Substitute Eq. E into Eq. D: \[ 3C_1 + 1368 = 4032 + 6C_5 - 9216 - 12C_5 \] \[ 3C_1 + 1368 = 4032 - 9216 - 6C_5 \] \[ 3C_1 + 1368 + 6C_5 = -5184 \] \[ 3C_1 + 6C_5 = -6552 \tag{F} \] ### From Eq. 6: Substituting for \( C_5 \) gives: \[ C_5 = C_3 - 288 \implies C_5 = (C_1 + 270) - 288 \implies C_5 = C_1 - 18 \tag{G} \] ### Substitute Eq. G into Eq. F: \[ 3C_1 + 6(C_1 - 18) = -6552 \] \[ 3C_1 + 6C_1 - 108 = -6552 \] \[ 9C_1 - 108 = -6552 \implies 9C_1 = -6444 \implies C_1 = -716 \] ### Back-substitute to find other variables: Using \( C_1 = -716 \): - From Eq. A: \[ C_2 = 27 - 3(-716) = 27 + 2148 = 2175 \] - From Eq. B: \[ C_3 = -716 + 270 = -446 \] - From Eq. C: \[ C_4 = -252 - 3(-716) = -252 + 2148 = 1896 \] - From Eq. G: \[ C_5 = -716 - 18 = -734 \] - From Eq. E: \[ C_6 = -9216 - 12(-734) = -9216 + 8808 = -408 \] ## **Final Values** The solution for the system is: \[ \boxed{ \begin{aligned} C_1 & = -716 \\ C_2 & = 2175 \\ C_3 & = -446 \\ C_4 & = 1896 \\ C_5 & = -734 \\ C_6 & = -408 \end{aligned} } \] If you require further assistance or verification, feel free to ask!