Question

write the other side of this equation so that it is true for no values of x.\
\\(\frac{1}{2}(6x - 10) - x = \underline{\quad}\\)

Explanation:

Step1: Simplify the left side

First, simplify \(\frac{1}{2}(6x - 10)-x\). Distribute the \(\frac{1}{2}\): \(\frac{1}{2}\times6x - \frac{1}{2}\times10 - x = 3x - 5 - x\). Combine like terms: \(3x - x - 5 = 2x - 5\).

Step2: Create a contradiction

For the equation to be true for no values of \(x\), the right side should be a linear expression with the same coefficient of \(x\) but a different constant term. So, if we set the right side as \(2x + c\) where \(c
eq - 5\). Let's choose \(c = 0\) (or any other number except -5). So we can write \(2x + 1\) (or any expression like \(2x + 2\), \(2x - 4\) etc., as long as the coefficient of \(x\) is 2 and the constant term is not -5). Let's take \(2x + 1\) for example.

Answer:

\(2x + 1\) (any expression of the form \(2x + c\) where \(c
eq - 5\) is correct, e.g., \(2x + 2\), \(2x - 4\) etc.)