Problem 116 - College Algebra

Question

A piece of wire is 8 inches long. The wire is cut into two pieces and then each piece is bent into a square. Find the length of each piece if the sum of the areas of these squares is to be 2 square inches. (image cant copy)

Step-by-Step Solution

Answer

The lengths of the two pieces of wire that will form squares with a combined area of 2 square inches are \(x = 4\sqrt{2}\) inches and \(y = 8 - 4\sqrt{2}\) inches.

Step 1: Express in terms of variables

Let's define \(x\) and \(y\) as the lengths of the two pieces of wire, with \(x > y\). The total length of the wire is given so we can write this relation as: \(x + y = 8\). The squares formed from these wires have perimeters equal to \(x\) and \(y\). Therefore, their sides are \(x/4\) and \(y/4\) respectively.

Step 2: Express the area condition

The sum of the areas of the squares can be expressed as: \(Area1 + Area2 = 2\). Since the side of a square is half the wire length, and the area of a square is \(side^2\), we can express this condition as: \((x/4)^2 + (y/4)^2 = 2\).

Step 3: Solve the system of equations

Solving our system of equations, we isolate \(y\) in the first equation, getting \(y = 8 - x\), then substitute this value in the second equation. This will give us a quadratic equation for \(x\). After solving for \(x\), we substitute its value back in our original equation to get \(y\).

Step 4: Conclusion

After substituting and solving the quadratic equation, suitable values are \(x = 4\sqrt{2}\) and \(y = 8 - 4\sqrt{2}\)

Key Concepts

Algebra

Algebra is the branch of mathematics dealing with symbols and the rules for manipulating these symbols. It is about finding the unknown or putting real-life variables into equations and then solving them. In our wire problem sequence, we start with defining variables to represent the length of the wire pieces. Let's say we have two pieces, one is length \( x \) and the other \( y \). You can write the relationship between them as \( x + y = 8 \). This equation represents the simple algebraic manipulation showing how the total length of the wire is distributed into two parts.

In algebra, expressing unknown quantities with variables allows us to form equations which we can mathematically solve. The power of algebra is that it can turn a word problem into an equation that can be systematically solved. Here, when we know the relationship \( x + y = 8 \) and also that certain conditions must be met, we use these conditions to find exact values for \( x \) and \( y \). This steps us into the realm of creating simultaneous equations.

System of Equations

A system of equations involves two or more equations that have common variables. The goal is to find values for the variables that satisfy all equations simultaneously. In our problem, we start with the equation \( x + y = 8 \) as one equation. To implement the area condition of the bent wires, we develop a second equation:

\[ \left( \frac{x}{4} \right)^2 + \left( \frac{y}{4} \right)^2 = 2 \].

Now, the problem transforms into solving these two equations where \( x \) and \( y \) must satisfy both equations. By isolating \( y \) with \( y = 8 - x \) and substituting it into the second equation, you reduce the system into a single equation that can be solved using quadratic methods.

Solving systems of equations is crucial in mathematics as it allows us to address complex problems where several conditions are interdependent. It requires methods such as substitution or elimination to isolate each variable and find a mutual solution set.

Geometry in Algebra

Geometry in algebra allows us to explore shapes using algebraic equations. By converting geometric properties into algebraic form, we address spatial problems with numerical solutions. In our wire problem, each wire bent into a square forms a geometric shape. The algebraic element comes from expressing the perimeter of the wire as the side lengths of the squares. Each wire piece perimeter initially, \( x \) and \( y \), becomes the side lengths \( \frac{x}{4} \) and \( \frac{y}{4} \) respectively, based on the formula for a square's perimeter, \( 4s \), where \( s \) is the side.

This understanding allows us to then use the formula for the area of a square, \( s^2 \). So, each piece of wire turns into a geometric algebra problem where the sum of squares' areas leads to our desired equation \( \left( \frac{x}{4} \right)^2 + \left( \frac{y}{4} \right)^2 = 2 \). Thus, transforming the problem from purely visual geometric shapes into solvable algebraic equations. This proficient blend of geometry and algebra is vital in solving real-world problems where spatial forms need quantifiable solutions.