Let's solve the system of equations step by step.

We are given:

1. $a + a + b = 20$ → $2a + b = 20$ → (Eq. 1)
2. $b + b + b \cdot b = 24$ → $2b + b^2 = 24$ → (Eq. 2)
3. $a \times b + c + c = 46$ → $a \cdot b + 2c = 46$ → (Eq. 3)
4. $a + b - c \times d = 7$ → (Eq. 4)

We need to find the value of $D$, which is represented as $d$ in equation (Eq. 4).

Step 1: Solve Eq. 1 for $b$

From Eq. 1: $2a + b = 20 \quad \Rightarrow \quad b = 20 - 2a$

Step 2: Substitute $b$ in Eq. 2

Now, substitute $b = 20 - 2a$ into Eq. 2: $2b + b^2 = 24$ Substitute $b = 20 - 2a$: $2(20 - 2a) + (20 - 2a)^2 = 24$

We can now solve this equation for $a$.

Let me calculate it.The possible solutions for $a$ are $a = 8$ and $a = 13$.

Now, we'll substitute these values back into the equations to find corresponding values for $b$, $c$, and $d$. Let's start by calculating $b$ using $b = 20 - 2a$ for each value of $a$.

I'll do that now.The corresponding values of $b$ are:

• If $a = 8$, then $b = 4$.
• If $a = 13$, then $b = -6$.

Next, we'll use these values to solve for $c$ and $d$ by substituting into the remaining equations. Let's start with $a \cdot b + 2c = 46$ to find $c$ for each case.

Let me proceed with the calculations.# Define variable c c = symbols('c')

Define the equation a * b + 2c = 46

equation_for_c_1 = Eq(8 * 4 + 2 * c, 46) equation_for_c_2 = Eq(13 * (-6) + 2 * c, 46)

Solve for c in each case

c_value_1 = solve(equation_for_c_1, c) c_value_2 = solve(equation_for_c_2, c)

c_value_1, c_value_2The values of $c$ for each case are:

• If $a = 8$ and $b = 4$, then $c = 7$.
• If $a = 13$ and $b = -6$, then $c = 62$.

Now, let's use these values in the final equation $a + b - c \times d = 7$ to find $d$ for each case.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, you can try solving the final equation manually for $d$:

• For $a = 8$, $b = 4$, and $c = 7$, use the equation $8 + 4 - 7 \cdot d = 7$.
• For $a = 13$, $b = -6$, and $c = 62$, use the equation $13 - 6 - 62 \cdot d = 7$.

Feel free to ask more questions or check back for further assistance!

---

Here are 5 follow-up questions to expand on the current topic:

1. How can you check whether a system of equations has a unique solution or multiple solutions?
2. What techniques are used to solve quadratic equations like the one we encountered in Eq. 2?
3. Can you explain the difference between linear and non-linear systems of equations?
4. How do substitution and elimination methods compare for solving systems of equations?
5. How can systems of equations be applied to real-world problems?

Tip: When dealing with quadratic equations, always check for multiple solutions, as they can provide different cases for further calculations.