Question

Solve the equation if possible. $$ \frac{1}{4}(60+16 s)=15+4 s $$

Short answer

The solutions to the equation are infinite. This is because when simplified, the terms on both sides of the equation are identical, confirming that there are an infinite number of solutions.

Step-by-step solution

Interpret the equation

The problem presents the equation \( \frac{1}{4}(60+16s)=15+4s \). We must solve this equation for \(s\) which is the unknown. To tackle this problem, we need to distribute the \( \frac{1}{4} \) across the terms in the parentheses on the left side of the equation and then simplify on both sides.

Distribute and Simplicate

Distribute \( \frac{1}{4} \) across \(60+16s\) to get \(15+4s\). Simplify to get \(15+4s = 15+4s\).

Solve for the variable

We notice that \(4s\) on one side is equal to \(4s\) on the other side, and \(15\)-one side is equal to \(15\)-the other side. Therefore we do not need to do anything more. Because each side of the equation is equal, there are an infinite number of solutions to this equation.

Confirm the result

Check that the equation holds true with some arbitrary values of \(s\). If it does, then the solution is correct. If it doesn't, there's an error somewhere in the steps taken to solve for \(s\).

Key concepts

Equation Simplification

When solving linear equations, equation simplification is a critical step that involves reducing complex expressions to simpler forms. This process makes it easier to understand and solve the equation. For instance, consider the equation

\[ \frac{1}{4}(60+16s)=15+4s \. \]
To simplify, we look for like terms and constants that can be combined. In this example, we would first distribute the fraction across the parenthesis, yielding terms that are easier to compare and manipulate. Simplification may involve combining like terms, factoring, expanding expressions, or canceling out terms on both sides of the equation. As a general rule, the goal of simplification is to isolate the variable, making the solution to the equation apparent. In this case, the simplification reveals that both sides of the equation are, in fact, identical, hinting at the presence of infinite solutions, which is a unique scenario in algebra.

Distributive Property

The distributive property is a foundational algebraic principle used to simplify equations. It states that the product of a number and a sum or difference is the same as the sum or difference of the products. In algebraic terms:

\[ a(b + c) = ab + ac \. \]

Applying this property helps to eliminate parentheses by distributing the coefficient to each term inside the parentheses. In our current example, the equation

\[ \frac{1}{4}(60+16s) \. \]
involves distributing the \( \frac{1}{4} \) across both terms within the parentheses, resulting in \(15 + 4s \) after the calculation. This step is vital as it transforms the equation into a form where like terms on both sides can be identified and solved. Understanding and applying the distributive property are critical when dealing with linear equations and contribute to a solid foundation in algebraic operations.

Infinite Solutions

Linear equations may at times have what is known as infinite solutions. This scenario occurs when the simplification process yields an identity; an equation which is true for all values of the variable. This can be confusing for students, as they are often looking for a single solution. In the given example, after simplifying, we have

\[ 15+4s = 15+4s \. \]
Showing that each side is a mirror of the other, regardless of the value assigned to \(s\text{.}\). This tells us that for any real number substituted for \(s\text{,}\) the equation holds true. So, no unique solution exists—instead, there is an infinite number of solutions. Recognizing infinite solutions is an important part of understanding algebra as it allows us to interpret equations not just for a single value but within the context of the system they represent. An equation with infinite solutions points to an identity or a fundamental truth within that mathematical system.