x Which equation has NO solution? A 15z+12 3 = 5x + 4 B 8-4z 10 = 4+ 2x C 43-20 2 = 2x-10 D 15+5 3 = 5x + 6

1 answer

To determine which equation has no solution, we can solve each equation step-by-step to see if we arrive at a contradiction.

Let's evaluate each equation:

Option A:

\[ 15z + 12 = 5x + 4 \]

Rearranging, we get:

\[ 15z - 5x = -8 \]

This can potentially have many solutions depending on the values of \(z\) and \(x\).


Option B:

\[ 8 - 4z = 10 - 4 + 2x \]

This simplifies to:

\[ 8 - 4z = 6 + 2x \]

Rearranging gives:

\[ -4z - 2x = -2 \]

Which can again have solutions depending on \(z\) and \(x\).


Option C:

\[ 43 - 20 = 2x - 10 \]

This simplifies to:

\[ 23 = 2x - 10 \]

Adding 10 to both sides:

\[ 33 = 2x \]

Dividing by 2, we get:

\[ x = 16.5 \]

This has one solution.


Option D:

\[ 15 + 5 = 5x + 6 \]

This simplifies to:

\[ 20 = 5x + 6 \]

Rearranging gives:

\[ 14 = 5x \]

Dividing by 5, we get:

\[ x = \frac{14}{5} \]

This also has a solution.


Next, let's look for contradictions or cases where the equations lead to impossible statements.

In options A, B, C, and D, we do not find any of them leading directly to a contradiction but we still need to make sure about each:

  1. Option A and Option B: Both have the form where they relate variables and could have solutions.
  2. Option C gives a specific solution.
  3. Option D also leads to a specific solution.

Upon re-evaluation, it seems all provided equations could have solutions. However, if any of these could be contrived to show no solutions through contradictory values, then there may be an oversight.

Conclusion:

If looking specifically for contradictions in terms of values such as equal values leading to false statements such as \(z = z + c\) where \(c \neq 0\), none are immediately visible.

Could you re-check if there may be more specific errors in how the original equations were constructed or their forms? Currently from the analysis, none provide a definitive fail case without further context on variable constraints.