Solve for $x$. Express your answer as a whole number, a decimal, or a fraction in simplest form.
$-6-(-4 x+2)-x=-(5 x-2)$
The goal is to solve for the variable \(x\) in the equation:
\[-6 - (-4x + 2) - x = -(5x - 2).\]
This is a linear equation in one variable. Our approach will be to expand the expressions on both sides by distributing the negative signs, combine like terms, and finally isolate \(x\) to find its value.
First, we simplify the left side of the equation:
\[-6 - (-4x + 2) - x.\]
Distributing the negative sign through the parentheses \((-4x + 2)\) gives:
\[-6 + 4x - 2 - x.\]
Now, we group the variable terms and the constant terms:
\[(4x - x) + (-6 - 2) = 3x - 8.\]
Thus, the left side simplifies to \(3x - 8\).
Next, we simplify the right side of the equation:
\[-(5x - 2).\]
Distributing the negative sign through the parentheses gives:
\[-5x + 2.\]
So, the equation now reads:
\[3x - 8 = -5x + 2.\]
Now we move all terms involving \(x\) to one side and the constant terms to the other. Adding \(5x\) to both sides, we get:
\[3x + 5x - 8 = 2,\]
\[8x - 8 = 2.\]
Next, adding 8 to both sides gives:
\[8x = 2 + 8,\]
\[8x = 10.\]
Finally, dividing both sides by 8, we find:
\[x = \frac{10}{8}.\]
Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, 2, we obtain:
\[x = \frac{5}{4}.\]
Linear equations like this are foundational for solving real-world problems involving relationships between quantities. For instance, if you are comparing two different pricing plans, say Plan A costs a fixed fee plus a variable rate, and Plan B has a different structure, you can set their cost expressions equal to find the "break-even point"—the quantity at which the cost of both plans is exactly the same. Mastering the technique of isolating variables ensures you can find these critical values efficiently in economics, engineering, and daily budgeting.