https://file.aiquickdraw.com/mathbot/file/1726567378006y0io8k4z.jpg

The system displays a set of simultaneous equations:

\[
\begin{cases}
-13 = -4c + 5d \\
3c + 4d - 2 = 0
\end{cases}
\]

Let's solve this system of linear equations step by step.

### Equation 1:
\[
-13 = -4c + 5d \quad \text{(Rearrange for } c\text{)}
\]
\[
4c = 5d + 13 \quad \text{(Divide by 4)}
\]
\[
c = \frac{5d + 13}{4}
\]

### Equation 2:
\[
3c + 4d - 2 = 0 \quad \text{(Substitute } c = \frac{5d + 13}{4}\text{ into this equation)}
\]
\[
3\left(\frac{5d + 13}{4}\right) + 4d - 2 = 0
\]
\[
\frac{3(5d + 13)}{4} + 4d - 2 = 0
\]
Multiply through by 4 to eliminate the fraction:
\[
3(5d + 13) + 16d - 8 = 0
\]
\[
15d + 39 + 16d - 8 = 0
\]
\[
31d + 31 = 0
\]
\[
31d = -31
\]
\[
d = -1
\]

### Solving for \( c \):
Now substitute \( d = -1 \) back into the equation for \( c \):
\[
c = \frac{5(-1) + 13}{4} = \frac{-5 + 13}{4} = \frac{8}{4} = 2
\]

### Solution:
\[
c = 2, \quad d = -1
\]

Let me know if you want more details or have questions!

Here are 5 related questions:
1. How do you solve systems of equations using substitution?
2. What happens if both equations give the same value for \( c \) and \( d \)?
3. Can you solve simultaneous equations using matrices?
4. How would graphing the two equations help visualize the solution?
5. How does solving for one variable simplify solving systems of equations?

**Tip:** Always check your solution by plugging values back into the original equations!

Solve the following simultaneous equations: -13 = -4c + 5d and 3c + 4d - 2 = 0. Find the values of c and d.

Learn how to solve a set of simultaneous linear equations, including -13 = -4c + 5d and 3c + 4d - 2 = 0. The solution involves rearranging the equations and using the substitution method to find the values of c and d. Ideal for students in Grades 9-12.

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