To determine the mission values that would make the equation \( 3(6x+5)=2x+_____x+______ \) have an infinite number of solutions, we need to rewrite the equation in a way that facilitates analysis.
Let's analyze the left-hand side first: \[ 3(6x + 5) = 18x + 15 \]
Now let’s set the right-hand side: \[ 2x + A x + B \] where \( A \) and \( B \) are the mission values that we will determine.
Now, we need to combine the terms on the right-hand side: \[ (2 + A)x + B \]
Setting the left-hand side equal to the right-hand side gives us: \[ 18x + 15 = (2 + A)x + B \]
For this equation to have an infinite number of solutions, the coefficients of \( x \) must be equal, and the constant terms must also be equal. Thus, we can form two equations from this comparison:
- \( 18 = 2 + A \)
- \( 15 = B \)
From the first equation: \[ A = 18 - 2 = 16 \]
From the second equation: \[ B = 15 \]
So, the mission values that would make the given equation have an infinite amount of solutions are: \[ A = 16 \quad \text{and} \quad B = 15 \]
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