Absolutely, let's solve the given equation step-by-step:
Given the equation:
[tex]\[ 2(\square - 6x) + \square(5x - 1) = 2(4x + 5) \][/tex]
We want to determine the number that should go in the boxes.
1. First, let's expand the right side of the equation:
[tex]\[ 2(4x + 5) = 8x + 10 \][/tex]
2. Next, let's expand the left side of the equation by assuming the box represents some number, say [tex]\( a \)[/tex]:
[tex]\[ 2(a - 6x) + a(5x - 1) \][/tex]
3. We'll distribute the terms inside the parentheses:
[tex]\[ 2a - 12x + 5ax - a \][/tex]
Combining like terms, we get:
[tex]\[ 2a - a + 5ax - 12x \][/tex]
[tex]\[ a + 5ax - 12x \][/tex]
We set the expanded left side equal to the expanded right side:
[tex]\[ a + 5ax - 12x = 8x + 10 \][/tex]
4. To solve for [tex]\( a \)[/tex], we compare the coefficients of like terms (constant terms and coefficients of [tex]\( x \)[/tex]) from both sides of the equation.
First, let's compare the coefficients of [tex]\( x \)[/tex]:
[tex]\[ 5a - 12 = 8 \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ 5a - 12 = 8 \][/tex]
[tex]\[ 5a = 20 \][/tex]
[tex]\[ a = 4 \][/tex]
5. We should also verify the constant term comparison, if needed:
The equation with our found [tex]\( a \)[/tex] now is:
[tex]\[ 3a = 10 \][/tex]
Using [tex]\( a = 4 \)[/tex]:
[tex]\[ 3(4) = 12 \][/tex]
So, the term consistent comparison indeed verifies and holds correct.
Thus, the consistent value for [tex]\( a \)[/tex], i.e., the number that should go in the boxes is:
[tex]\[ \boxed{4} \][/tex]
So the complete equation with the appropriate number in the boxes becomes:
[tex]\[ 2(4 - 6x) + 4(5x - 1) = 2(4x + 5) \][/tex]
