Problem 64 - Algebra 1

Question

The time \(t\) (in hours) needed to produce \(x\) units of a product is modeled by \(t=p x+s .\) If it takes 265 hours to produce 200 units and 390 hours to produce 300 units, what is the value of \(s ?\) (A) 1.25 (B) 15 (C) 100 (D) 125

Step-by-Step Solution

Answer

The correct option is (B) 15

Step 1: Set up the equations

The first step is to translate the given conditions into mathematical equations using the model \(t = px + s\). We have two conditions: \n1. \(t = 265\) when \(x = 200\), which gives the equation \(265 = 200p + s\)\n2. \(t = 390\) when \(x = 300\), which gives the equation \(390 = 300p + s\)

Step 2: Solve the system of equations

After setting up the equations, we can write them in matrix form and solve for \(p\) and \(s\):\n\[\begin{cases} 200p + s = 265 \ 300p + s = 390 \end{cases}\]\nBy subtracting the first equation from the second, we find: \(100p = 125\). This gives us \(p = 1.25\). Now, we can replace \(p\) in the first equation with \(1.25\) to obtain: \(265 = 200(1.25) + s\), which upon simplification gives \(s = 15\).

Step 3: Verify

To ensure the solution is correct, we can substitute \(p = 1.25\) and \(s = 15\) into the second condition and check if the left hand side equals the right hand side. This gives: \(390 = 300(1.25) + 15\), which simplifies to \(390 = 375 + 15\), which is true.

Key Concepts

Linear Equations

Linear equations are one of the basic building blocks of algebra. They are equations that form a straight line when graphed. A simple linear equation is usually written in the form \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. In essence, a linear equation shows how two quantities change at a constant rate relative to each other.

• The equation \(t = px + s\) from the original exercise is a linear equation in terms of \(x\).
• Here, \(t\) represents the total time, \(p\) is the production rate (or slope), and \(s\) is the initial setup time (or y-intercept).

Understanding linear equations is crucial because they allow for the modeling of real-world situations, like production times, expenses, or distance traveled, based on constant rates of change. Additionally, the solutions to these equations can often be visually interpreted as points on a line in a coordinate system, making them not only simple but also intuitive.

Solving Systems of Equations

A system of equations refers to a set of two or more equations that share the same set of unknowns. Solving a system of equations means finding a single set of values for the unknowns that satisfy all equations simultaneously. In our exercise, we have two such simultaneous equations determined by the two given conditions. These are:

• \(265 = 200p + s\)
• \(390 = 300p + s\)

There are several methods to solve systems of equations, including substitution, elimination, and graphical methods. In this scenario, the elimination method helps to find the values of \(p\) and \(s\):

• By subtracting the first equation from the second, you remove \(s\) and isolate \(p\).
• This gives \(100p = 125\), and so, \(p = 1.25\).

Substituting \(pg1.25\) back into the first equation allows us to solve for \(s\), confirming that \(s = 15\). Solving equations in this way requires careful step-by-step manipulation to ensure accuracy.

Matrix Method

The matrix method transforms systems of linear equations into matrices, making them simpler to visualize and solve, especially for larger systems. A matrix is a rectangular array of numbers arranged in rows and columns.
In the exercise, if we express the system

• \(200p + s = 265\)
• \(300p + s = 390\)

as a matrix, it comes down to:\[\begin{bmatrix}200 & 1 \300 & 1\end{bmatrix}\begin{bmatrix}p \s\end{bmatrix} = \begin{bmatrix}265 \390\end{bmatrix}\]This representation allows us to use matrix operations to solve for \(p\) and \(s\). Although for simple systems with only two equations, the direct methods might be more efficient, matrices shine when handling large systems with many variables.
By applying techniques like Gaussian elimination or computing the inverse of matrices, one can solve such systems methodically and efficiently. Understanding matrices not only aids in solving equations but also enhances comprehension of linear transformations and vector spaces in higher mathematics.