Problem 22 - Prealgebra

Question

Solve each equation. Check your solution. $$8 y-5=5(y-1)+3 y$$

Step-by-Step Solution

Answer

The equation holds for all real numbers, so it has infinitely many solutions.

Step 1: Expand and Simplify the Equation

First, we need to expand the right-hand side of the equation: \(5(y-1) + 3y\). Distribute the 5: \(5 \times y - 5 \times 1\), which results in \(5y - 5\). Now the equation becomes: \[ 8y - 5 = 5y - 5 + 3y \] which simplifies further to \[ 8y - 5 = 8y - 5 \].

Step 2: Simplify Both Sides

Now we rearrange both sides to investigate the variable terms. Since both sides are equal \(8y - 5 = 8y - 5\), we have identical terms on both sides. We can deduct \(8y\) from both sides: \(8y - 8y = -5 + 5\), which simplifies to \(0 = 0\).

Step 3: Check the Consistency

Since we end up with the statement \(0 = 0\), which is always true, this indicates the equation holds for any real number \(y\). This implies that every \(y\) is a solution, confirming the equation has infinitely many solutions.

Key Concepts

Equation Solving

Equation solving is a fundamental skill in algebra. It involves finding the value of the variable that makes the equation true. Understanding the steps to solve an equation can simplify what seems complex at first. Let's start by examining the given equation: \[ 8y - 5 = 5(y - 1) + 3y \]To solve, we first expand the terms, particularly the expression \(5(y - 1)\). Utilizing distribution, multiply 5 by each term inside the parenthesis: \(5 \times y\) and \(5 \times (-1)\), resulting in \(5y - 5\). This change turns the equation into:\[ 8y - 5 = 5y - 5 + 3y \]With further simplification, we consolidate like terms. \(5y + 3y\) combines into \(8y\), resulting in both sides of the equation being identical: \[ 8y - 5 = 8y - 5 \]Sometimes, simplifying equations leads to a straightforward solution, or as in this exercise, a realization about the nature of solutions. By understanding these steps, equation solving becomes a less daunting task, letting you break down problems with confidence.

Variables in Algebra

Variables are symbols used to represent unknown values in equations. In algebra, they function much like placeholders. In our exercise, \(y\) is the variable we're working with.When solving for a variable, the goal is to isolate it on one side of the equation. However, as we saw in our step-by-step solution, sometimes equations reflect that any number could satisfy the variable.Consider this nature of variables:

• They allow the equation to take on multiple forms and have flexibility.
• Used effectively, they help express real-world situations in mathematical terms.
• Isolating variables is key when aiming to find specific solutions.

In our case, simplifying left us with an equation \(8y - 5 = 8y - 5\), which already represents how variables can reinforce balance.Overall, variables not only define an equation's structure but also serve as the core of problem-solving mechanics in algebra.

Infinitely Many Solutions

Sometimes, solving an equation doesn't lead to a single solution but rather infinitely many solutions. This happens when both sides of the equation reflect the same mathematical expression.In our example, simplifying the given equation led to:\[ 8y - 5 = 8y - 5 \]Reducing further by eliminating identical terms on both sides results in:\\[ 0 = 0 \]This is a statement that is inherently true, suggesting that any value of \(y\) will satisfy the equation. Therefore, instead of narrowing down to one solution, you conclude that there are an infinite number of solutions.Such equations are called identities and are defined by their self-consistent, always-true nature. It is important to recognize this when solving equations to understand the scope of possible solutions. Infinite solutions can, at first, seem counterintuitive but they highlight the elegant balance and symmetry in algebra.