Question
Finish the equation so that the equation that has zero solutions
5
(
6
x
+
7
)
−
6
=
5(6x+7)−6=
x
+
x+
5
(
6
x
+
7
)
−
6
=
5(6x+7)−6=
x
+
x+
Answers
GPT-4o mini
To create an equation with zero solutions, we can start with the equation given:
\[
5(6x + 7) - 6 = x + x + C
\]
Here, \(C\) is a constant that we will choose in such a way that the equation has no solutions.
First, simplify the left-hand side:
\[
5(6x + 7) - 6 = 30x + 35 - 6 = 30x + 29
\]
Now, the left-hand side becomes:
\[
30x + 29 = x + x + C
\]
The right-hand side simplifies to:
\[
30x + 29 = 2x + C
\]
Next, let's rearrange this equation:
\[
30x + 29 - 2x = C
\]
This simplifies to:
\[
28x + 29 = C
\]
To ensure there are zero solutions, we can set \(C\) to a value that contradicts the equation for any \(x\). For example, if we set \(C = 29\), we would have:
\[
28x + 29 = 29
\]
Subtract \(29\) from both sides:
\[
28x = 0
\]
This would imply \(x = 0\), leading to one solution. To create a condition with no solutions, we can set \(C\) to a value that does not match. For instance, if we set \(C = 30\):
We would have:
\[
28x + 29 = 30
\]
Subtracting \(29\) from both sides gives:
\[
28x = 1
\]
This suggests \(x = \frac{1}{28}\), indicating one solution again.
To further ensure zero solutions, we can set \(C > 29\):
For example, if we set \(C = 31\):
Now we have:
\[
28x + 29 = 31
\]
Subtracting \(29\) gives:
\[
28x = 2
\]
Again, this leads to a single solution.
To achieve an equation that truly has zero solutions, set it such that the coefficients are contradictory.
Let’s take another approach:
Instead of \(C = 31\), let’s set our equation with contradictory constants. Thus, if we were to write:
\[
28x + 29 = 28x + 31
\]
Subtract \(28x\) from both sides:
\[
29 = 31
\]
This is clearly a contradiction, and there are no values for \(x\) that satisfy the equation.
Thus, the final equation for "zero solutions" is:
\[
5(6x + 7) - 6 = 28x + 31
\]
This creates a contradiction:
\[
29 = 31
\]
So, the completed equation is:
\[
5(6x + 7) - 6 = x + x + 31
\]
\[
5(6x + 7) - 6 = x + x + C
\]
Here, \(C\) is a constant that we will choose in such a way that the equation has no solutions.
First, simplify the left-hand side:
\[
5(6x + 7) - 6 = 30x + 35 - 6 = 30x + 29
\]
Now, the left-hand side becomes:
\[
30x + 29 = x + x + C
\]
The right-hand side simplifies to:
\[
30x + 29 = 2x + C
\]
Next, let's rearrange this equation:
\[
30x + 29 - 2x = C
\]
This simplifies to:
\[
28x + 29 = C
\]
To ensure there are zero solutions, we can set \(C\) to a value that contradicts the equation for any \(x\). For example, if we set \(C = 29\), we would have:
\[
28x + 29 = 29
\]
Subtract \(29\) from both sides:
\[
28x = 0
\]
This would imply \(x = 0\), leading to one solution. To create a condition with no solutions, we can set \(C\) to a value that does not match. For instance, if we set \(C = 30\):
We would have:
\[
28x + 29 = 30
\]
Subtracting \(29\) from both sides gives:
\[
28x = 1
\]
This suggests \(x = \frac{1}{28}\), indicating one solution again.
To further ensure zero solutions, we can set \(C > 29\):
For example, if we set \(C = 31\):
Now we have:
\[
28x + 29 = 31
\]
Subtracting \(29\) gives:
\[
28x = 2
\]
Again, this leads to a single solution.
To achieve an equation that truly has zero solutions, set it such that the coefficients are contradictory.
Let’s take another approach:
Instead of \(C = 31\), let’s set our equation with contradictory constants. Thus, if we were to write:
\[
28x + 29 = 28x + 31
\]
Subtract \(28x\) from both sides:
\[
29 = 31
\]
This is clearly a contradiction, and there are no values for \(x\) that satisfy the equation.
Thus, the final equation for "zero solutions" is:
\[
5(6x + 7) - 6 = 28x + 31
\]
This creates a contradiction:
\[
29 = 31
\]
So, the completed equation is:
\[
5(6x + 7) - 6 = x + x + 31
\]