Problem 55 - Elementary Algebra

Question

For the following problems, solve the literal equations for the indicated variable. When directed, find the value of that variable for the given values of the other variables. Solve \(z=\frac{x-\bar{x}}{s}\) for \(x\). Find the value of \(x\) when \(z=1.96, s=2.5,\) and \(\bar{x}=15\).

Step-by-Step Solution

Answer

Question: Given the literal equation \(z=\frac{x-\bar{x}}{s}\), find the value of \(x\) when \(z=1.96\), \(s=2.5\), and \(\bar{x}=15\). Answer: 19.9

Step 1: Solve the equation for x

We are given the equation \(z=\frac{x-\bar{x}}{s}\). Our goal is to isolate \(x\) on one side of the equation. Start by multiplying both sides of the equation by \(s\): \(z * s = \frac{x-\bar{x}}{s} * s\) This simplifies to: \(zs = x - \bar{x}\) Now, add \(\bar{x}\) to both sides of the equation: \(zs + \bar{x} = x\) We have now solved the equation for \(x\).

Step 2: Substitute the given values

Now that we have an expression for \(x\), we can substitute the given values for \(z\), \(s\), and \(\bar{x}\): \(x = 1.96 * 2.5 + 15\)

Step 3: Calculate the value of x

Now, we can perform the calculations to find the value of \(x\): \(x = 4.9 + 15\) \(x = 19.9\) Thus, when \(z=1.96\), \(s=2.5\), and \(\bar{x}=15\), the value of \(x\) is \(19.9\).

Key Concepts

Algebraic Manipulation

Algebraic manipulation is the process of rearranging, simplifying, or rewriting an algebraic expression or equation to reach a desired form or solve for a specific variable. It involves a variety of operations such as adding, subtracting, multiplying, and dividing both sides of an equation in order to isolate the variable you're solving for.

For example, consider the literal equation from the exercise:
\(z=\frac{x-\bar{x}}{s}\).
In this scenario, we aim to solve for \(x\). Algebraic manipulation comes into play when we multiply both sides of the equation by \(s\) to eliminate the fraction and isolate terms containing \(x\) on one side. Once that's achieved, we further manipulate the equation by adding \(\bar{x}\) to both sides, to solve for \(x\). The careful application of algebraic manipulation leads us to the final expression where \(x\) stands alone:
\(zs + \bar{x} = x\).

The key to algebraic manipulation is understanding the properties of equality and operations. These manipulations maintain the balance of the equation, meaning what you do to one side, you must do to the other to keep the equation equal. Mastery over algebraic manipulation is fundamental in solving literal equations effectively.

Substitution Method

The substitution method is a technique used in algebra to solve for a variable by replacing given values into the equation. Once you have manipulated the literal equation into the desired form, you can substitute the known values to evaluate the unknown.

Looking back at our exercise, after solving the literal equation \(z=\frac{x-\bar{x}}{s}\) for \(x\), the resulting equation we obtained is:
\(x = zs + \bar{x}\).
With the substitution method, you take the given numbers for \(z\), \(s\), and \(\bar{x}\) and put them in place of the corresponding variables. In this case:
\(x = (1.96 * 2.5) + 15\).

Substituting the values correctly is crucial as it sets the stage for finding the precise value of \(x\). This method is incredibly useful when dealing with equations that have multiple variables — it allows you to find the value of one variable when the others are known.

Literal Equation Solution

A literal equation solution means solving an equation for one variable in terms of other variables. Unlike numerical equations where you obtain a numerical answer, a literal equation involves variables representing known and unknown quantities.

For instance, the given exercise requires us to solve the equation for \(x\), which is originally part of an algebraic formula involving other variables, namely \(z\) and \(s\), as well as the constant \(\bar{x}\). When we say 'solve for \(x\)', we rearrange the equation to express \(x\) as a function of the remaining variables and constants:
\(x = zs + \bar{x}\).
This new equation is the solution to our literal equation. It is now a straightforward formula that we can use to calculate the value of \(x\) for any values of \(z\) and \(s\), as we did in the exercise by substituting \(z=1.96\), \(s=2.5\), and \(\bar{x}=15\) to find \(x=19.9\).

A correct literal equation solution is a powerful tool in algebraic analysis and applied mathematics, as it can be used to understand relationships between variables and to solve a broad range of practical problems.