Question

Determine the mission values that would make the equation have an infinite amount of solutions
3(6x+5)=2x+________x+_________

Answer 1

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:
1. \( 18 = 2 + A \)
2. \( 15 = B \)

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:
\[
A = 16 \quad \text{and} \quad B = 15
\]