Solve the following equation for $b$.
$-m+\frac{2}{5} b=d$
We start with the given equation:
\[-m + \frac{2}{5}b = d\]
Our objective is to isolate the variable \(b\) in terms of the variables \(m\) and \(d\). This involves performing basic algebraic operations to rearrange the equation such that \(b\) stands alone on one side.
To begin isolating \(b\), we need to remove the term \(-m\) from the left-hand side of the equation. We do this by adding \(m\) to both sides:
\[-m + \frac{2}{5}b + m = d + m\]
Simplifying the left-hand side, we obtain:
\[\frac{2}{5}b = d + m\]
Now, to solve for \(b\), we need to eliminate the fraction \(\frac{2}{5}\) multiplying \(b\). We do this by multiplying both sides of the equation by the reciprocal of \(\frac{2}{5}\), which is \(\frac{5}{2}\):
\[\frac{5}{2} \cdot \left(\frac{2}{5}b\right) = \frac{5}{2} \cdot (d + m)\]
This leaves us with:
\[b = \frac{5}{2}(d + m)\]
Or, distributing the \(\frac{5}{2}\), we can write the expression as:
\[b = \frac{5}{2}d + \frac{5}{2}m\]
Both forms are correct representations for \(b\).
Algebraic isolation is like managing a budget. Suppose you are calculating the total amount spent on a project, where 'b' represents the base cost of materials, 'm' represents a discount, and 'd' represents the delivery fee. By rearranging your formulas, you can easily determine how the base cost 'b' changes when your discount or delivery fees fluctuate. Mastering these manipulations allows you to adapt your calculations to different situations quickly and efficiently.