We are tasked with solving a system of equations. Let’s organize the equations for clarity:

1. $4a + 2c - 4d = 61$
2. $a + 7c + 2d = 0$
3. $a + 8c - 6d = -32$
4. $-9a + 8c + 9d = 68$
5. $a - 9c + 9d = -33$

To solve this system of equations, we can use either substitution or elimination. Let me solve this step by step.

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Step 1: Express variables in terms of others

Choose one equation to express a variable (e.g., $a$) in terms of $c$ and $d$. Then substitute into the other equations.

From equation (2): $a = -7c - 2d$

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Step 2: Substitute $a = -7c - 2d$ into other equations

Substitute this expression into equations (1), (3), (4), and (5).

Equation (1):

$4(-7c - 2d) + 2c - 4d = 61$ $-28c - 8d + 2c - 4d = 61$ $-26c - 12d = 61 \quad \text{(Equation A)}$

Equation (3):

$(-7c - 2d) + 8c - 6d = -32$ $-7c + 8c - 2d - 6d = -32$ $c - 8d = -32 \quad \text{(Equation B)}$

Equation (4):

$-9(-7c - 2d) + 8c + 9d = 68$ $63c + 18d + 8c + 9d = 68$ $71c + 27d = 68 \quad \text{(Equation C)}$

Equation (5):

$(-7c - 2d) - 9c + 9d = -33$ $-7c - 9c - 2d + 9d = -33$ $-16c + 7d = -33 \quad \text{(Equation D)}$

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Step 3: Solve the simplified system

We now have four equations in two variables ($c$ and $d$):

1. $-26c - 12d = 61$ (A)
2. $c - 8d = -32$ (B)
3. $71c + 27d = 68$ (C)
4. $-16c + 7d = -33$ (D)

Let’s solve this system systematically.

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Step 4: Solve for $c$ and $d$

From Equation (B): $c = -32 + 8d$

Substitute $c = -32 + 8d$ into Equation (A): $-26(-32 + 8d) - 12d = 61$ $832 - 208d - 12d = 61$ $832 - 220d = 61$ $-220d = 61 - 832$ $-220d = -771$ $d = \frac{771}{220} \approx 3.505$

Substitute $d = 3.505$ back into $c = -32 + 8d$: $c = -32 + 8(3.505)$ $c \approx -32 + 28.04$ $c \approx -3.96$

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Step 5: Solve for $a$

Using $a = -7c - 2d$: $a = -7(-3.96) - 2(3.505)$ $a \approx 27.72 - 7.01$ $a \approx 20.71$

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Final Solution:

$a \approx 20.71, \, c \approx -3.96, \, d \approx 3.505$

Would you like a detailed breakdown of any step?

Here are some follow-up questions to deepen understanding:

1. What methods can simplify solving such systems of linear equations?
2. How does substitution compare to elimination for solving linear systems?
3. What are the implications if the system had no solution?
4. How can we verify the solution's accuracy?
5. What tools or software can assist in solving such equations efficiently?

Tip: Always check your solutions by substituting them back into the original equations to verify accuracy!