Answer :
Sure, let's solve the given equation step-by-step. The problem gives us a partially filled equation:
[tex]\[ 2 (\square - 6x) + \square \square(5x - 1) = 2(4x + 5) \][/tex]
First, we need to fill in the appropriate numbers in the boxes to complete the calculation. Let's try to complete the equation logically.
Let's fill the first box. We know it should distribute correctly within the parenthesis and be consistent with typical algebraic manipulations.
Consider the form [tex]\( 2 (\square - 6x) \)[/tex]. If we consider common algebraic values, we might place an [tex]\( x \)[/tex] inside to be consistent with typical variable isolations in equations. So the first box should be [tex]\( x \)[/tex]. Now we have:
[tex]\[ 2 (x - 6x) \][/tex]
Simplify [tex]\( x - 6x \)[/tex]:
[tex]\[ x - 6x = -5x \][/tex]
So,
[tex]\[ 2 (x - 6x) = 2 (-5x) = -10x \][/tex]
Next, let's examine the second term, [tex]\( \square \square(5x - 1) \)[/tex]. Normally, it would correspond with the first box [tex]\( x \)[/tex] again in a typical scenario:
[tex]\[ x \cdot x(5x - 1) \][/tex]
However, considering algebraic manipulations, let's confirm it in step by step expressions:
[tex]\[ x \cdot 5x - x\cdot 1 = 5x^2 - x \][/tex]
Thus, our complete left-hand side becomes:
[tex]\[ -10x + 5x^2 - x \][/tex]
Combine like terms:
[tex]\[ -10x - x + 5x^2 = 5x^2 - 11x \][/tex]
Our equation now is:
[tex]\[ 5x^2 - 11x = 2(4x + 5) \][/tex]
Simplify the right-hand side:
[tex]\[ 2(4x + 5) = 8x + 10 \][/tex]
Combining results of full equation:
[tex]\[ 5x^2 - 11x = 8x + 10 \][/tex]
Move all terms to one side to setup for solving quadratically:
[tex]\[ 5x^2 - 11x - 8x - 10 = 0 \][/tex]
Combine like terms:
[tex]\[ 5x^2 - 19x - 10 = 0 \][/tex]
Now solve:
Using the quadratic formula, [tex]\( a = 5 \)[/tex], [tex]\( b = -19 \)[/tex], [tex]\( c = -10 \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{19 \pm \sqrt{(-19)^2 - 4(5)(-10)}}{2(5)} \][/tex]
[tex]\[ x = \frac{19 \pm \sqrt{361 + 200}}{10} \][/tex]
[tex]\[ x = \frac{19 \pm \sqrt{561}}{10} \][/tex]
Thus we have our quadratic solutions as:
[tex]\[ x = \frac{19 + \sqrt{561}}{10} \quad \text{or} \quad x= \frac{19 - \sqrt{561}}{10} \][/tex]
Concluding assigning,
\[ \text{First box} = x, \text{ Second box} = x.
[tex]\[ 2 (\square - 6x) + \square \square(5x - 1) = 2(4x + 5) \][/tex]
First, we need to fill in the appropriate numbers in the boxes to complete the calculation. Let's try to complete the equation logically.
Let's fill the first box. We know it should distribute correctly within the parenthesis and be consistent with typical algebraic manipulations.
Consider the form [tex]\( 2 (\square - 6x) \)[/tex]. If we consider common algebraic values, we might place an [tex]\( x \)[/tex] inside to be consistent with typical variable isolations in equations. So the first box should be [tex]\( x \)[/tex]. Now we have:
[tex]\[ 2 (x - 6x) \][/tex]
Simplify [tex]\( x - 6x \)[/tex]:
[tex]\[ x - 6x = -5x \][/tex]
So,
[tex]\[ 2 (x - 6x) = 2 (-5x) = -10x \][/tex]
Next, let's examine the second term, [tex]\( \square \square(5x - 1) \)[/tex]. Normally, it would correspond with the first box [tex]\( x \)[/tex] again in a typical scenario:
[tex]\[ x \cdot x(5x - 1) \][/tex]
However, considering algebraic manipulations, let's confirm it in step by step expressions:
[tex]\[ x \cdot 5x - x\cdot 1 = 5x^2 - x \][/tex]
Thus, our complete left-hand side becomes:
[tex]\[ -10x + 5x^2 - x \][/tex]
Combine like terms:
[tex]\[ -10x - x + 5x^2 = 5x^2 - 11x \][/tex]
Our equation now is:
[tex]\[ 5x^2 - 11x = 2(4x + 5) \][/tex]
Simplify the right-hand side:
[tex]\[ 2(4x + 5) = 8x + 10 \][/tex]
Combining results of full equation:
[tex]\[ 5x^2 - 11x = 8x + 10 \][/tex]
Move all terms to one side to setup for solving quadratically:
[tex]\[ 5x^2 - 11x - 8x - 10 = 0 \][/tex]
Combine like terms:
[tex]\[ 5x^2 - 19x - 10 = 0 \][/tex]
Now solve:
Using the quadratic formula, [tex]\( a = 5 \)[/tex], [tex]\( b = -19 \)[/tex], [tex]\( c = -10 \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{19 \pm \sqrt{(-19)^2 - 4(5)(-10)}}{2(5)} \][/tex]
[tex]\[ x = \frac{19 \pm \sqrt{361 + 200}}{10} \][/tex]
[tex]\[ x = \frac{19 \pm \sqrt{561}}{10} \][/tex]
Thus we have our quadratic solutions as:
[tex]\[ x = \frac{19 + \sqrt{561}}{10} \quad \text{or} \quad x= \frac{19 - \sqrt{561}}{10} \][/tex]
Concluding assigning,
\[ \text{First box} = x, \text{ Second box} = x.
