Problem 51 - Introductory and Intermediate Algebra for College Students 4th

Question

The average, or mean, \(A,\) of three exam grades, \(x, y,\) and \(z,\) is given by the formula $$ A=\frac{x+y+z}{3} $$ a. Solve the formula for \(z\). b. Use the formula in part (a) to solve this problem. On your first two exams, your grades are \(86 \%\) and \(88 \%\) : \(x=86\) and \(y=88 .\) What must you get on the third exam to have an average of \(90 \% ?\)

Step-by-Step Solution

Answer

To have an average of \(90\%\), you must score \(96\%\) on the third exam.

Step 1: Rewrite the Average Formula

In order to solve for \(z\), rearrange the equation \(A=\frac{x+y+z}{3}\) to give \(z\) in terms of \(x\), \(y\) and \(A\). This can be done by multiplying both sides by 3 and then subtracting \(x\) and \(y\) from both sides: \[3A = x + y + z\] \[3A - x - y = z\]

Step 2: Substitute Given Values

Substitute \(x=86\%\) and \(y=88\%\) into the rewritten equation \[z = 3A - x - y\] to get \[z = 3 * 90 - 86 - 88\]

Step 3: Calculate z

Calculate the value of \(z\): \[z = 3 * 90 - 86 - 88 = 270 - 86 - 88 = 96\]. This means that you should score \(96\%\) on the third exam to have an average of \(90\%\).

Key Concepts

Mean Calculation in Algebra

Understanding how to calculate an average—or mean—is essential in algebra, particularly when dealing with sets of numbers such as exam grades. The mean is a type of average found by adding up all the numbers and then dividing this total by the count of the numbers. In algebra, this concept is often represented with variables and equations. In the context of exam grades, if a student receives three exam grades represented by the variables \( x, y, \) and \( z \), the mean \( A \) can be expressed as: \[ A = \frac{x + y + z}{3} \]

When expressed algebraically, the formula allows us to solve for any unknown variable, assuming we have the required data. For instance, if a student knows two exam grades and the average they desire, they can rearrange the formula to solve for the third exam grade needed to achieve that average. The ability to manipulate this formula provides a powerful tool in planning and goal setting for students seeking to reach a specific academic target.

Solving for a Variable

When faced with an algebraic equation like the mean grade formula, we often need to isolate one variable to find its value. This process is known as 'solving for a variable'. It involves performing operations that will rearrange the equation, so the variable of interest is by itself on one side of the equation. For example, to solve the average grade formula for \( z \), we start with the equation \( A = \frac{x + y + z}{3} \) and by carrying out the appropriate algebraic steps—multiplying each side by 3 and then subtracting \( x \) and \( y \) from both sides—we isolate \( z \):

\[ 3A = x + y + z \]
\[ z = 3A - x - y \]

This manipulation demonstrates a key aspect of algebra: understanding the 'balance' in an equation. When we perform an operation on one side, we must do the same to the other side in order to maintain equality. Mastering this principle is fundamental to algebra and crucial to solving for variables in more complex equations.

Substituting Values in Equations

Substituting values into an algebraic equation is a straightforward yet crucial skill in solving math problems. Once we’ve correctly arranged the equation to solve for the variable we’re interested in, the next step is to replace the variables with the given numeric values. This process requires careful attention to ensure that each variable is substituted correctly and that operations are performed accurately.

In our example, we substituted \( x = 86 \) and \( y = 88 \) into the equation we derived for \( z \): \[ z = 3A - x - y \]. With the average \( A \) we desire to achieve, say \( 90 \), we replace the variables with these numbers: \[ z = 3 \times 90 - 86 - 88 \]. The resulting calculation yields the value needed for \( z \), or in one's actual scenario, the grade needed on the third exam to achieve the target average. Substitution not only helps in finding specific numerical answers but also in checking one's work for accuracy and understanding the relationship between variables in the equation.