To create an equation with one solution, we need to manipulate the equation such that both sides can be simplified to yield a single value for \( x \).
Let's start with the equation you provided:
\[ 5(3x - 4) = 4 + \text{Choose...}x \]
First, let's simplify the left side:
\[ 5(3x - 4) = 15x - 20 \]
Now, we have:
\[ 15x - 20 = 4 + \text{Choose...}x \]
Next, let's assume "Choose..." is represented by some coefficient \( k \), so we can say:
\[ 15x - 20 = 4 + kx \]
To create an equation with one solution, we can set \( k \) so that the terms involving \( x \) can be isolated properly on one side, leading to a solvable equation.
Let’s isolate \( x \) on one side:
\[ 15x - kx = 24 \] \[ (15 - k)x = 24 \] \[ x = \frac{24}{15 - k} \]
For this to have one solution, \( 15-k \) cannot be zero (so \( k \neq 15 \)).
Now, let’s choose a specific value for \( k \) to ensure we have an equation with one solution. For example:
If we let \( k = 3 \), we have:
\[ 5(3x - 4) = 4 + 3x \]
Now we can check if it has one solution:
\[ 15x - 20 = 4 + 3x \]
Rearranging this gives:
\[ 15x - 3x = 20 + 4 \] \[ 12x = 24 \] \[ x = 2 \]
Thus, the completed equation is:
\[ 5(3x - 4) = 4 + 3x \]
This equation has exactly one solution, \( x = 2 \).
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