Problem 46 - Algebra for College Students

Question

Solve each equation for the indicated variable. \((y-2)(a+1)=x \quad\) for \(y\)

Step-by-Step Solution

Answer

\(y = \frac{x + 2a + 2}{a+1}\)

Step 1: Distribute and Expand

The given equation is \((y-2)(a+1) = x\). To solve for \(y\), first expand the left-hand side by applying the distributive property: \((y-2)(a+1) = y(a+1) - 2(a+1)\). This becomes \((ya + y) - (2a + 2)\). So the equation is \(ya + y - 2a - 2 = x\).

Step 2: Isolate Terms with y

Group all terms containing \(y\) on one side and other terms on the opposite side. Rearrange the equation: \(ya + y = x + 2a + 2\). Factor out \(y\) from the left side: \(y(a + 1) = x + 2a + 2\).

Step 3: Solve for y

Now, divide both sides of the equation by \(a+1\) to isolate \(y\). Thus, \(y = \frac{x + 2a + 2}{a+1}\).

Key Concepts

Distributive Property

The Distributive Property is one of the fundamental principles in algebra. It allows you to multiply a sum by a value by distributing the multiplication over each term inside the parentheses. In simple terms, if you have an expression like

• (a+b)c, you can rewrite it as (ac + bc).

In the given exercise, applying the distributive property helps break down the expression

• (y-2)(a+1) = y(a+1) - 2(a+1).

This step simplifies the expression, making it easier to manage, understand, and solve for the variable. By breaking down complex expressions into simpler parts, we can find a path to solve for the desired variable.

Solving for a Variable

When solving an equation for a specific variable, the goal is to manipulate the equation so that the variable of interest is isolated on one side of the equation. In this problem, our aim is to find what the variable y equals. We start with the equation

• ya + y - 2a - 2 = x.

Here, we need to make sure that y is by itself on one side. This process involves:

• Rearranging terms to group all y-related terms together.
• Carefully moving other terms to the opposite side of the equation.

By focusing on these techniques, solving for y becomes a clear and systematic process that involves restructuring the equation step by step.

Factoring

Factoring is another essential algebraic technique used to simplify equations. It involves breaking down an expression into simpler "factors" that, when multiplied together, give back the original expression. In our algebra equation, once we grouped terms containing y on one side,

• ya + y = x + 2a + 2,

we factored out y from the left side. This process can be seen as:

• Taking out y to transform the expression into y(a + 1).

Factoring is crucial because it consolidates terms, allowing easier manipulation which leads us closer to isolating the variable and finding its value. It turns a complex problem into a straightforward path to follow.

Equation Manipulation

Equation manipulation is the practice of applying different algebraic techniques to rearrange and simplify equations. In this exercise, after applying the distributive property and factoring, we reached an equation where

• y(a + 1) = x + 2a + 2.

The final manipulation involves dividing both sides by (a+1) to solve for y.

• y = \(\frac{x + 2a + 2}{a+1}\)

This step is crucial as it reflects the final stretch of solving the equation. Equation manipulation often requires combining various algebraic strategies to move terms around correctly. This systematic process allows you to solve for any given variable using known values on the other side of the equation.