Problem 56 - Algebra A Combined Function

Question

Solve each equation. See Section \(2.3 .\) \(-y+12\left(\frac{y-1}{4}\right)=3\)

Step-by-Step Solution

Answer

The solution to the equation is \(y = 3\).

Step 1: Distribute the 12

We begin by distributing the 12 across the terms in the parenthetical expression: \(-y + 12\left(\frac{y-1}{4}\right) = 3\).When we distribute 12 to each term inside the parenthesis, we first multiply it by \(\frac{y}{4}\) and then by \(\frac{-1}{4}\):\(-y + 12 \times \frac{y}{4} - 12 \times \frac{1}{4} = 3\).Simplifying this gives us:\(-y + 3y - 3 = 3\).

Step 2: Simplify the equation

Combine like terms in the equation:\(-y + 3y - 3 = 3\).This simplifies to:\(2y - 3 = 3\).

Step 3: Isolate the variable term

To isolate the term with \(y\), add 3 to both sides of the equation:\(2y - 3 + 3 = 3 + 3\).This simplifies to:\(2y = 6\).

Step 4: Solve for y

Divide both sides by 2 to solve for \(y\):\(\frac{2y}{2} = \frac{6}{2}\).This simplifies to:\(y = 3\).

Key Concepts

Linear Equations

Linear equations are a fundamental part of algebra that help us understand relationships between variables. These equations take the form of a constant plus a variable term equal to another constant, like:

• Single-variable linear equation: \( ax + b = c \)
• Two-variable linear equation: \( ax + by = c \)

The main idea is to find the value of the variable that makes the equation true. In the equation given in the exercise, \(-y + 12\left(\frac{y-1}{4}\right)=3\), we have one variable, \(y\), which we need to solve for. The challenge is to manipulate the equation to isolate and solve for \(y\), ultimately finding its specific value. Understanding linear equations is crucial as they lay the groundwork for solving more complex mathematical problems.

Distributive Property

The distributive property is a critical algebraic property used to solve equations like the one in the exercise. It states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products. This property is represented as:

• \(a(b + c) = ab + ac\)

In the equation from the exercise, \(-y + 12\left(\frac{y-1}{4}\right) = 3\), we distribute 12 across the terms inside the parentheses. This distribution transforms the expression into:

• \(-y + 3y - 3 = 3\)

This step is crucial because it simplifies the equation, making it easier to combine like terms and progress towards solving for \(y\). The distributive property helps to break down complex expressions and is a tool used frequently in algebra to simplify and solve equations.

Solving Equations

Solving equations involves a systematic process of finding the value of the unknown variable that makes the equation true. After using the distributive property, the equation becomes simpler, as in:

• \(2y - 3 = 3\)

From here, the goal is to simplify the equation further and isolate the variable. This involves several steps such as:

• Combining like terms: Grouping constants and variable terms together.
• Transposing terms: Moving terms across the equal sign to group similar math operations.

Each step involves arithmetic operations and logical reasoning to ensure each transformation keeps the equation balanced while iterating closer to the solution. This skill of solving equations is fundamental, not only in mathematics but also in various fields including physics, engineering, and economics.

Isolation of Variables

Isolating variables is a key step in solving linear equations. The objective is to get the variable alone on one side of the equation, enabling us to see its value. This can be achieved with the following steps:

• Eliminating constant terms: Add or subtract constants from both sides.
• Removing coefficients: Divide or multiply both sides by a number to "free" the variable.

In the exercise, after combining terms, you isolate \(y\) by first adding 3 to both sides, yielding:

• \(2y = 6\)

Then, you divide each side by 2 resulting in:

• \(y = 3\)

Isolating variables is important because it allows us to directly determine the value of the unknown, providing clarity and solution to the problem at hand. Mastery of this concept builds confidence and proficiency in managing algebraic equations.