Question

solve for x.
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$\frac{1}{4}x + 2 = \frac{3}{4}x$
$x + 2 \cdot 4 = 3x$
add to both sides :add what to both sides?
$x + 8 + \square = 3x + \square$

Explanation:

Step1: Analyze the equation

We have the equation \(x + 8=3x\) (after simplifying \(x + 2\cdot4=3x\) to \(x + 8 = 3x\)). To solve for \(x\), we want to get all the \(x\)-terms on one side. So we can add \(-x\) (or subtract \(x\)) to both sides.

Step2: Determine what to add

If we add \(-x\) to both sides of the equation \(x+8 = 3x\), we get \(x + 8+(-x)=3x+(-x)\). Simplifying the left side: \(x - x+8=8\), and the right side: \(3x - x = 2x\). But in the given form \(x + 8+\square=3x+\square\), we need to add \(-x\) (or \(-x\) is equivalent to subtracting \(x\)) to both sides. So we add \(-x\) to both sides.

Answer:

We add \(-x\) to both sides. So the boxes should be filled with \(-x\) and \(-x\) respectively. That is, \(x + 8+(-x)=3x+(-x)\)