What value of $p$ makes the equation true?
$-3 p+\frac{1}{8}=-\frac{1}{4}$
$p=$
The objective is to find the value of the variable \(p\) that satisfies the linear equation:
\[-3p + \frac{1}{8} = -\frac{1}{4}\]
This equation involves a linear term \(-3p\) and two constant fractions. Our goal is to isolate \(p\) by using basic algebraic manipulation, specifically performing operations on both sides of the equation to maintain equality.
To isolate the term with \(p\), we first subtract \(\frac{1}{8}\) from both sides:
\[-3p = -\frac{1}{4} - \frac{1}{8}\]
Next, we combine the fractions on the right side. To subtract them, we need a common denominator, which is 8:
\[-\frac{1}{4} = -\frac{2}{8}\]
Substituting this into the equation, we get:
\[-3p = -\frac{2}{8} - \frac{1}{8} = -\frac{3}{8}\]
Now we have the equation:
\[-3p = -\frac{3}{8}\]
To solve for \(p\), we divide both sides by \(-3\):
\[p = \frac{-\frac{3}{8}}{-3}\]
Simplifying the expression, the negatives cancel out, and we divide the numerator by 3:
\[p = \frac{3}{8 \times 3} = \frac{3}{24} = \frac{1}{8}\]
Thus, the value of \(p\) that makes the equation true is \(\frac{1}{8}\).
This type of problem is fundamental in many real-world scenarios. For example, if you are calculating the interest rate on a loan or adjusting a budget, you often encounter linear equations where you need to isolate a variable to find an unknown quantity, such as a missing fee or a required monthly payment. Solving \(-3p + \frac{1}{8} = -\frac{1}{4}\) is akin to balancing a simple budget where you have recurring deductions and a fixed end balance. Mastering these steps allows you to solve more complex financial or scientific problems efficiently.