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Chapter 8: Problem 342

In the following exercises, solve each equation. $$ 0.1 d+0.25(d+8)=4.1 $$

Short Answer

Expert verified
The value of \(d\) is 6.

Step by step solution

01

Distribute the terms

Multiply the 0.25 by each term inside the parentheses: \(0.25(d + 8) = 0.25d + 2\). The equation becomes: \(0.1d + 0.25d + 2 = 4.1\).
02

Combine like terms

Add the like terms on the left side of the equation: \(0.1d + 0.25d = 0.35d\). Now the equation is: \(0.35d + 2 = 4.1\).
03

Isolate the variable term

Subtract 2 from both sides of the equation to isolate the term with \(d\): \(0.35d = 4.1 - 2\). Simplify to get: \(0.35d = 2.1\).
04

Solve for the variable

Divide both sides of the equation by 0.35 to find the value of \(d\): \(d = \frac{2.1}{0.35}\). Simplify to get \(d = 6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

distributive property
When solving linear equations, the distributive property can be very helpful. This property states that multiplying a number by a sum is the same as doing each multiplication separately and then adding the results.
For example, in the equation 0.1d + 0.25(d + 8) = 4.1, we need to distribute the 0.25 to both terms inside the parentheses.
Here's how it works: 0.25(d + 8) becomes 0.25 * d + 0.25 * 8, which simplifies to 0.25d + 2.
Now, our equation looks like this: 0.1d + 0.25d + 2 = 4.1. This step sets up the equation for further simplification and solving.
combining like terms
After using the distributive property, the next step is to combine like terms. Like terms are terms that contain the same variable raised to the same power. In our example, 0.1d and 0.25d are like terms because they both have the variable d.
We simply add the coefficients (the numerical parts) together: 0.1d + 0.25d = 0.35d.
This gives us a simpler equation: 0.35d + 2 = 4.1.
Combining like terms is important because it simplifies the equation, making it easier to solve.
isolating variables
Isolating the variable means getting the variable by itself on one side of the equation. This is a crucial step in solving linear equations.
In our equation, 0.35d + 2 = 4.1, we need to move the constant term (2) to the other side of the equation. We do this by performing the same operation on both sides.
Subtract 2 from both sides of the equation: 0.35d = 4.1 - 2.
Simplify the right-hand side: 0.35d = 2.1.
This leaves us with 0.35d, where d is now isolated.
prealgebra
Prealgebra involves understanding and manipulating basic arithmetic and introductory algebra concepts.
Skills like using the distributive property, combining like terms, and isolating variables are foundational in prealgebra. They prepare students for solving more complex equations in algebra and other higher-level math courses.
For example, in our problem-solving process, we used the distributive property to break down the terms, combined like terms to simplify, and isolated the variable to solve for d.
These steps form the basis of solving linear equations, a key concept in prealgebra.

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