Question

Isaac wants the equation below to have no solution when the missing number is placed in the box.

Box (x + 2) + 2 x = 2 (x + 6) + 4 x

Which number should he place in the box?
2
4
6
8

Answers

GPT-4o mini
To determine which number Isaac should place in the box so that the equation has no solution, we first need to simplify the given equation.

The equation is:

\[
\text{Box} (x + 2) + 2x = 2(x + 6) + 4x
\]

Let’s denote the number in the box as \( a \). The equation can be rewritten as:

\[
a(x + 2) + 2x = 2(x + 6) + 4x
\]

Now, simplify both sides of the equation.

### Left Side:
\[
a(x + 2) + 2x = ax + 2a + 2x
\]

### Right Side:
\[
2(x + 6) + 4x = 2x + 12 + 4x = 6x + 12
\]

Now, the equation becomes:

\[
ax + 2a + 2x = 6x + 12
\]

Next, combine like terms on the left side:
\[
(ax + 2x) + 2a = 6x + 12
\]
\[
(a + 2)x + 2a = 6x + 12
\]

Now, we can set coefficients equal to each other. The equation can be split into two equations:

1. Coefficient of \( x \): \( a + 2 = 6 \)
2. Constant term: \( 2a = 12 \)

From the first equation \( a + 2 = 6 \):

\[
a = 6 - 2 = 4
\]

From the second equation \( 2a = 12 \):

\[
a = \frac{12}{2} = 6
\]

So we have two potential values from these equations: 4 and 6. However, we want to find the value of \( a \) such that the two equations contradict each other (i.e., no solution).

Thus, if \( a + 2 \neq 6 \) and \( 2a \neq 12 \) simultaneously, we can potentially find the answer.

If we take \( a = 8 \):

### Checking:
1. \( 8 + 2 = 10 \neq 6 \) (coefficient of \( x \))
2. \( 2(8) = 16 \neq 12 \) (constant term)

Both conditions are not satisfied, which means that:

If \( a = 8 \), then the equation becomes:
\[
(8 + 2)x + 2(8) = 6x + 12
\]
\[
10x + 16 = 6x + 12
\]
\[
10x - 6x + 16 - 12 = 0
\]
\[
4x + 4 = 0
\]
\[
4x = -4
\]
\[
x = -1
\]

However, this has a solution; we diverged with the sign.

### Hence:
The value of \( a \) that leads to no solution would be the only possible value \( a = 4 \) since both parts would not lead to consistency since it leads to 0=12, and we have inconsistency if we lead \( a \neq 4 \).

### Conclusion:

Therefore, he should place \( \text{6} \) in the box so that the equation has no solutions.