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Chapter 2: Problem 35

Solve each formula for the indicated letter. $$ 2 x+5 y=10, \text { for } y $$

Short Answer

Expert verified
y = \frac{10 - 2x}{5}

Step by step solution

01

Isolate terms involving y

Start by moving the term involving y to one side of the equation. Subtract 2x from both sides of the equation to isolate the term with y. \[ 2x + 5y - 2x = 10 - 2x \] Simplify this to get: \[ 5y = 10 - 2x \]
02

Solve for y

Divide both sides of the equation by 5 to solve for y. \[ y = \frac{10 - 2x}{5} \]

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Key Concepts

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

Linear Equations
A linear equation is an equation where each term is either a constant or the product of a constant and a single variable. These equations form a straight line when graphed on a coordinate plane. In simpler terms, linear equations are used to describe relationships where one variable changes in direct proportion to another. The general form of a linear equation is written as \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. Here, our given equation is \(2x + 5y = 10\).
Isolation Technique
The isolation technique involves moving terms to isolate the variable you want to solve for. This step-by-step method ensures clarity and accuracy. In our problem, we're solving for \(y\). First, move the term containing \(x\) to the other side of the equation. Subtract \(2x\) from both sides:
\[2x + 5y - 2x = 10 - 2x\]
This simplifies to: \[5y = 10 - 2x\]
By isolating the \(y\) term, it becomes easier to solve for it in the next steps.
Algebraic Manipulation
Algebraic manipulation involves applying arithmetic operations to both sides of an equation to simplify or rearrange it. Here, we divide both sides by 5 to isolate \(y\).
\[y = \frac{10 - 2x}{5}\]
This division evenly distributes \(10\) and \(-2x\) by 5, resulting in:
\[y = 2 - \frac{2x}{5}\]
Such manipulation is crucial for solving complex problems efficiently. It involves various arithmetic operations like addition, subtraction, multiplication, and division while maintaining the equation's balance.

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