Determine the value of the question mark so that the equation is true.
$5(x-2)=(x-2)+?(x-2)$
The problem asks us to find the value that fills the question mark in the equation \(5(x-2) = (x-2) + ?(x-2)\) such that the equality holds true for all \(x\) except possibly \(x=2\). We observe that the term \((x-2)\) is a common factor in all parts of the equation. To identify the missing value, we can define a variable, let's call it \(k\), to represent the question mark. Thus, the equation becomes:
\[5(x-2) = (x-2) + k(x-2)\]
This is a standard algebraic equation where we can compare the coefficients of the common term \((x-2)\) on both sides of the equation.
To isolate \(k\), we can simplify the expression on the right side of the equation. We know that \((x-2) = 1 \times (x-2)\). Therefore, we can factor out \((x-2)\) from the right-hand side:
\[(x-2) + k(x-2) = (1+k)(x-2)\]
Now the equation is:
\[5(x-2) = (1+k)(x-2)\]
For this equality to hold true for all values of \(x\) where \(x
eq 2\), the expressions multiplying \((x-2)\) on both sides must be equal. This gives us the simple linear equation:
\[5 = 1 + k\]
Subtracting \(1\) from both sides, we find:
\[k = 5 - 1 = 4\]
Thus, the value that replaces the question mark is 4.
Imagine you are calculating the total cost of purchasing several identical items. If one item costs \(\$(x-2)\) and you want to know the cost of 5 such items, you calculate \(5(x-2)\). If you have already set aside the cost for one item \((x-2)\) and need to know how many more items you need to pay for, you would calculate the remaining cost as \(k(x-2)\). Knowing that the total cost is \(5(x-2)\), you can easily determine that you need to pay for 4 more items to reach the total. This type of algebraic reasoning is fundamental in budgeting, inventory management, and any scenario involving scaling costs or resources.