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The equation provided is:

\[
d = b + c + a
\]

We need to solve for \( a \) in terms of \( b \), \( c \), and \( d \).

### Steps:
1. Subtract \( b + c \) from both sides of the equation to isolate \( a \):

\[
a = d - (b + c)
\]

2. Simplify the equation:

\[
a = d - b - c
\]

Thus, the solution is:

\[
a = d - b - c
\]

Would you like further details or have any questions?

Here are 5 related questions to practice:

1. Solve for \( b \) in terms of \( a \), \( c \), and \( d \): \( d = b + c + a \).
2. Solve for \( d \) when \( a = 5 \), \( b = 3 \), and \( c = 4 \).
3. If \( a + b + c = d \), how can you express \( b \) in terms of the others?
4. Given the equation \( d = 2b + a \), solve for \( a \).
5. How would the solution change if \( d = b + c - a \)?

### Tip:
When solving for a variable, try to isolate it on one side by using inverse operations (addition ↔ subtraction, multiplication ↔ division).

Solve for a in terms of b, c, and d. Given the equation: d = b + c + a.

Learn how to solve for a variable in a linear equation with step-by-step instructions. This problem focuses on isolating variables and understanding basic algebraic operations, suitable for students in Grades 6-8.

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