Verified Answer :
- The equation \(3x \pm 5 = \frac{80}{75} - \frac{8}{75}\) yields two solutions: \(x = -\frac{101}{75}\) and \(x = \frac{149}{75}\).
- Using the system of equations \(R = 3M + 1\) and \(M + R = 45\), Ryan volunteered for \(R = 34\) hours.
- A rectangle with width \(w\) has a length of \(2w - 6\), valid for \(w > 3\).
- Final answers: \(x = -\frac{101}{75}, \frac{149}{75}\), \(R = 34\), and \(\text{Dimensions: } w \text{ and } 2w-6\). \(\boxed{x \in \{-\frac{101}{75}, \frac{149}{75}\}, R=34, \text{Width}=w, \text{Length}=2w-6}\)
Explanation
- Solving Equation 1
We begin by solving for \(x\) in the equation \(3x \pm 5 = \frac{80}{75} - \frac{8}{75}\).
First, simplify the right-hand side of the equation:
\[\frac{80}{75} - \frac{8}{75} = \frac{72}{75} = \frac{24}{25} = 0.96.\]
This gives us two separate linear equations to solve:
- \(3x + 5 = 0.96 \implies 3x = 0.96 - 5 = -4.04 \implies x = -\frac{4.04}{3} = -\frac{404}{300} = -\frac{101}{75}\).
- \(3x - 5 = 0.96 \implies 3x = 0.96 + 5 = 5.96 \implies x = \frac{5.96}{3} = \frac{596}{300} = \frac{149}{75}\).
Thus, the solutions are \(x = -\frac{101}{75}\) and \(x = \frac{149}{75}\).
- Solving Problem 2 (Volunteer Hours)
Next, we determine the number of hours Ryan volunteered. Let \(M\) be the number of hours Maddie volunteered and \(R\) be the number of hours Ryan volunteered. We are given the following system of equations:
- \(R = 3M + 1\)
- \(M + R = 45\)
Substitute the first equation into the second:
\[M + (3M + 1) = 45 \implies 4M + 1 = 45 \implies 4M = 44 \implies M = 11.\]
Now, find \(R\) using the first equation:
\[R = 3(11) + 1 = 33 + 1 = 34.\]
Ryan volunteered for 34 hours.
- Solving Problem 3 (Rectangle Dimensions)
Finally, we express the dimensions of a rectangle in terms of its width \(w\).
We are given that the length \(l\) is six inches less than twice its width. Therefore, we have:
- Width = \(w\)
- Length = \(2w - 6\)
Since dimensions must be positive, we have \(w > 0\) and \(2w - 6 > 0\), which implies \(w > 3\). The dimensions are thus \(w\) and \(2w - 6\) for \(w > 3\).
Examples
These problems illustrate how algebra helps in real-life decision-making. For instance, solving for volunteer hours allows organizations to track contributions fairly, while defining dimensions in terms of a variable is essential in fields like architecture or manufacturing, where designs need to be scalable. By expressing unknown quantities through variables and setting up equations, we can solve complex logistical challenges efficiently.