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Chapter 2: Problem 46

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

Short Answer

Expert verified
The solution for \(y\) is \(y = \frac{x}{a+1} + 2\).

Step by step solution

01

Understand the Equation Structure

The given equation is \((y-2)(a+1)=x\). We need to solve this equation for \(y\). The equation is in the form of a product on the left equal to \(x\).
02

Isolate the Term Containing y

To isolate the term containing \(y\), divide both sides of the equation by \((a+1)\), assuming \(a+1eq 0\): \(y-2 = \frac{x}{a+1}\).
03

Solve for y

Add 2 to both sides to solve for \(y\):\[y = \frac{x}{a+1} + 2\]
04

Verify the Expression

Ensure that \(y\) is correctly isolated. The solution obtained, \(y = \frac{x}{a+1} + 2\), expresses \(y\) in terms of the other variables.

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Key Concepts

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

Equation Solving
When solving an equation, our primary aim is to find the unknown value that makes the equation true. In the given exercise, we have the equation \((y-2)(a+1) = x\). To solve it for a specific variable, such as \(y\), we need to change the equation so that \(y\) is by itself on one side. This process involves systematically performing operations that maintain the equality of the equation.
  • Identify the part of the equation containing the variable you're solving for, in this case, \(y-2\).
  • We have a product \((y-2)(a+1)\), so the goal is to manipulate this expression to isolate \(y\).
Being diligent in each step ensures the balance of the equation is maintained, leading to an accurate solution. Always check your final answer to verify its correctness by plugging it back into the original equation.
Variable Isolation
Isolating a variable is the process of manipulating an equation so the desired variable appears alone on one side of the equation. For our example, we aim to isolate \(y\) in the equation \((y-2)(a+1)=x\). Here's how:
  • First, divide each side of the equation by \((a+1)\) to remove the multiplication with \((y-2)\). This gives us the equation: \(y-2 = \frac{x}{a+1}\).
  • The next step is to eliminate the subtraction of 2 by adding 2 to both sides. This results in \(y = \frac{x}{a+1} + 2\).
By systematically performing these operations, \(y\) becomes the subject of the formula. Isolating the variable this way ensures you are able to express \(y\) clearly in terms of \(x\) and \(a\). This method is crucial in algebra for finding the value of unknowns.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions to solve equations or find expressions for specific variables. In the process of solving the equation \((y-2)(a+1)=x\), several algebraic manipulations were performed:
  • Dividing both sides by \((a+1)\) simplified the equation to \(y-2 = \frac{x}{a+1}\).
  • Adding 2 to both sides allowed us to further simplify it to \(y = \frac{x}{a+1} + 2\).
These operations ensured that we maintained equality while transforming the equation into a simpler form. Algebraic manipulation is a fundamental skill in mathematics, enabling us to change equations into forms that make it easier to derive unknown values. Always perform one operation at a time to keep track of changes and verify accuracy.

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Most popular questions from this chapter

For Problems \(35-44\), solve each compound inequality and graph the solution sets. Express the solution sets in interval notation. $$ x+1>0 \quad \text { and } \quad 3 x-4<0 $$

Solve each inequality. $$ |x-2|+4 \geq 10 $$

For Problems \(1-16\), solve each equation. $$ |3 x-1|-1=9 $$

Solve each inequality. $$ |-2 x+7| \leq 13 $$

For Problems \(17-30\), solve each inequality and graph the solution. $$ |x-3| \geq 2 $$
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