Question

What solutions do these equations have?
3(2i-7)=6i-21
2i+3=2i-5
3i+2i-4=18i

Answer 1

Let's solve each equation one by one.

1. **Equation**: \( 3(2i - 7) = 6i - 21 \)

Start by distributing the left side:
\[
3 \cdot 2i - 3 \cdot 7 = 6i - 21
\]
which simplifies to:
\[
6i - 21 = 6i - 21
\]

Both sides of the equation are equal, which means this equation is true for all values of \( i \). Therefore, the solution is:
\[
\text{All values of } i
\]

2. **Equation**: \( 2i + 3 = 2i - 5 \)

Next, we subtract \( 2i \) from both sides:
\[
3 = -5
\]

This is a contradiction, which means there are no values of \( i \) that can satisfy this equation. Therefore, the solution is:
\[
\text{No solution}
\]

3. **Equation**: \( 3i + 2i - 4 = 18i \)

First, combine the like terms on the left side:
\[
(3i + 2i) - 4 = 18i
\]
which simplifies to:
\[
5i - 4 = 18i
\]

Next, we can isolate \( i \) by subtracting \( 5i \) from both sides:
\[
-4 = 13i
\]

Now, divide both sides by \( 13 \):
\[
i = -\frac{4}{13}
\]

Therefore, the solution is:
\[
i = -\frac{4}{13}
\]

### Summary of Solutions:
1. \( 3(2i - 7) = 6i - 21 \): **All values of \( i \)**
2. \( 2i + 3 = 2i - 5 \): **No solution**
3. \( 3i + 2i - 4 = 18i \): **\( i = -\frac{4}{13} \)**