https://file.aiquickdraw.com/mathbot/file/1725403502199gk2t8uma.jpg

To solve the simultaneous equations provided in the image:

\[
\begin{aligned}
3c + 4d &= 30 \quad \text{(Equation 1)} \\
3d + 2c &= 21 \quad \text{(Equation 2)}
\end{aligned}
\]

### Step 1: Express one variable in terms of the other.
Let's solve Equation 2 for \(c\):

\[
3d + 2c = 21
\]

Subtract \(3d\) from both sides:

\[
2c = 21 - 3d
\]

Now, divide by 2:

\[
c = \frac{21 - 3d}{2} \quad \text{(Equation 3)}
\]

### Step 2: Substitute Equation 3 into Equation 1.
Now that we have \(c\) in terms of \(d\), substitute it into Equation 1:

\[
3\left(\frac{21 - 3d}{2}\right) + 4d = 30
\]

Multiply the 3 inside the parentheses:

\[
\frac{3(21 - 3d)}{2} + 4d = 30
\]

This simplifies to:

\[
\frac{63 - 9d}{2} + 4d = 30
\]

### Step 3: Clear the fraction.
Multiply every term by 2 to eliminate the fraction:

\[
63 - 9d + 8d = 60
\]

Simplify the equation:

\[
63 - d = 60
\]

### Step 4: Solve for \(d\).
Subtract 63 from both sides:

\[
-d = -3
\]

So,

\[
d = 3
\]

### Step 5: Substitute \(d = 3\) back into Equation 3 to find \(c\).
Substitute \(d = 3\) into Equation 3:

\[
c = \frac{21 - 3(3)}{2}
\]

Simplify:

\[
c = \frac{21 - 9}{2} = \frac{12}{2} = 6
\]

### Final Answer:
The solution to the system of equations is:

\[
c = 6, \quad d = 3
\]

Would you like further details or have any questions?

### Related Questions:
1. What is the graphical interpretation of solving simultaneous equations?
2. How can substitution and elimination methods differ in solving simultaneous equations?
3. Can these equations be solved using matrices? If so, how?
4. What are some common mistakes to avoid when solving simultaneous equations?
5. How does the nature of coefficients affect the solution of simultaneous equations?

**Tip:** Always check your solutions by substituting them back into the original equations to verify correctness.

Solve these simultaneous equations: 3c + 4d = 30 and 3d + 2c = 21

This guide explains how to solve the simultaneous linear equations 3c + 4d = 30 and 3d + 2c = 21 using the substitution method. The solution covers key algebraic concepts suitable for students in Grades 8-10.

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