Question

You are loading a large pile of newspapers onto a truck. You divide the pile into four equal-size bundles. You find that one bundle weighs 37 pounds. You want to find the weight \(x\) of the original pile. Which equation models the situation? Solve the correct equation. $$A \cdot \frac{x}{4}=37$$ $$B \cdot 4 x=37$$$

Short answer

The weight of the original pile of newspapers is 148 pounds.

Step-by-step solution

Understand the problem

We want to find the weight of the original pile of newspapers, lets denote it as \(x\). The stack was divided into 4 equal piles which means each pile weighs \(x/4\) pounds. We know that one bundle is 37 pounds.

Formulate the equation

Since one bunlde weighs \(x/4\) and the actual weight of the bundle is 37 pounds, we can form the equation as: \(\frac{x}{4} = 37\)

Solve the equation

To solve for \(x\), we can multiply both sides of the equation by 4: \(x = 37 * 4\). This will give us the weight of the original pile.

Compute the answer

After performing the multiplication we get: \(x = 148\), so the weight of the original pile is 148 pounds.

Key concepts

Solving Equations

When we talk about solving equations, we refer to the process of finding the value of a variable that makes the equation true. An equation is a statement that two expressions are equal, and it often contains one or more variables. To solve an equation, you need to isolate the variable on one side, which sometimes involves performing the same mathematical operation on both sides of the equal sign.

Just like in our exercise where we wanted to find the weight of the original pile of newspapers represented by the variable x, we begin by formulating the equation based on the given conditions: one of the four equal bundles weighs 37 pounds. If we set this up as \(\frac{x}{4} = 37\), our next step is to perform the operation that will isolate x. Multiply both sides by 4 gives us \(x = 37 \times 4\), simplifying it to \(x = 148\).

Here, it's critical to apply inverse operations to maintain the balance of the equation. If an equation is multiplied by a number, we divide; if it's added, we subtract, and so on. This foundational process helps ensure that when students plug the solution back into the original equation, both sides remain equal, confirming that their solution is correct.

Linear Equations

Linear equations are equations of the first degree, which means that any variables in the equation are to the first power and do not multiply each other. These equations form a straight line when graphed on a coordinate plane, hence the name 'linear'. They can have one or more variables, but the most basic form in two dimensions is \(y = mx + b\), where m is the slope and b is the y-intercept.

In our textbook example, the equation \(\frac{x}{4} = 37\) is a linear equation. It has just one variable and is of the first degree. When we rearrange it as \(x = 148\), it matches the standard form \(Ax + B = 0\), with A being 1 (since 1*x = x) and B being -148. It's simple but fundamental to understand that regardless of how a linear equation may be presented or disguised in a problem, the objective remains to isolate the variable to find its value.

Variables in Algebra

Variables are symbols, often letters like x or y, that represent numbers. In algebra, variables are placeholders for values that can change or that we do not yet know. When working with variables, it's important to remember they follow the same arithmetic rules as constants. You can add, subtract, multiply, or divide them, just as you would with numbers.

Returning to our newspaper bundle problem, the variable x stands for the unknown total weight of the newspapers. The equation \(\frac{x}{4} = 37\) tells us that if we divide the total weight by 4, we should get 37. Knowing the role of variables helps you construct and deconstruct algebraic expressions and equations. As students practice, they begin to see variables not as mysterious symbols, but as keys that unlock the solutions to various problems.