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 - hand side

First, we simplify the left - hand side of the equation \(\frac{1}{2}(6x - 10)-x\).
Using the distributive property \(a(b + c)=ab+ac\), where \(a=\frac{1}{2}\), \(b = 6x\) and \(c=- 10\), we have \(\frac{1}{2}(6x-10)=3x - 5\).
Then, subtract \(x\) from \(3x - 5\): \(3x-5 - x=2x-5\).

Step2: Determine the right - hand side

We want an equation that is true for no values of \(x\). An equation of the form \(ax + b=cx + d\) has no solution when \(a = c\) and \(b
eq d\).
Since the left - hand side simplifies to \(2x-5\), if we set the right - hand side to \(2x + k\) where \(k
eq - 5\). Let's choose \(k = 0\) (we can choose any number except - 5). So the right - hand side can be \(2x\) (or any expression of the form \(2x + c\) where \(c
eq - 5\)). For example, if we take the right - hand side as \(2x+1\), the equation \(2x - 5=2x + 1\) simplifies to \(-5 = 1\), which is false for all \(x\).

Answer:

\(2x + 1\) (or any expression of the form \(2x + c\) where \(c
eq - 5\), for example, \(2x\), \(2x+2\) etc.)