Problem 34 - Algebra 1

Question

Rewrite the equation in function form. \(\frac{1}{2} x+\frac{5}{2} y=1\)

Step-by-Step Solution

Answer

The equation rewritten in function form is \(y = -\frac{1}{5}x + \frac{2}{5}\)

Step 1: Isolate x terms on one side of the equation

To begin, isolate the \(x\) term, this is done by subtracting \(\frac{1}{2} x\) from both sides of the equation. This results in: \(\frac{5}{2}y = -\frac{1}{2}x + 1\)

Step 2: Isolate y on one side to get function

To obtain the function form, isolate \(y\) by dividing the whole equation by \(\frac{5}{2}\). As dividing by a fraction is the same as multiplying by its reciprocal, this is equivalent to multiplying the equation by \(\frac{2}{5}\). This leads to: \(y = -\frac{1}{5}x + \frac{2}{5}\)

Step 3: Simplify function if possible

In this case, the function is already in its simplest form, so no further steps are necessary.

Key Concepts

Algebra

Algebra is a vital part of mathematics where we work with symbols and letters to represent numbers and quantities in equations and expressions. It allows us to create a general formula from a specific problem, making it easier to solve various problems with similar structures. In algebra, variables are often used to stand for unknown values, and equations are made up of expressions that show relationships between these variables.

When working with algebra, there are different operations such as addition, subtraction, multiplication, and division that we can use on variables and constants (known numbers). This creates new expressions or modifies existing ones. For example, in the given equation \(rac{1}{2}x + \frac{5}{2}y = 1\), we see variables \(x\) and \(y\) being part of an algebraic expression.

• Variables: Letters representing numbers whose values can change.
• Constants: Fixed numbers in expressions or equations.
• Expressions: Combinations of variables and constants using operations.

Understanding how to manipulate these components is key to solving equations and expressing them in different forms.

Isolating Variables

Isolating variables is an essential technique in algebra that helps us express one variable in terms of another. This is particularly useful when rewriting equations in function form, as it allows us to see the relationship between variables more clearly.

In our problem, the goal was to isolate \(y\). We started with the equation \(\frac{1}{2}x + \frac{5}{2}y = 1\).

• First, we moved the \(x\) term to the other side by subtracting \(\frac{1}{2}x\) from both sides, resulting in \(\frac{5}{2}y = -\frac{1}{2}x + 1\).
• Next, we needed \(y\) alone on one side. To achieve this, we divided every term by \(\frac{5}{2}\), which effectively means multiplying by its reciprocal, \(\frac{2}{5}\), to get \(y = -\frac{1}{5}x + \frac{2}{5}\).

This technique of isolating variables is crucial for understanding how changes in one variable affect another, thereby providing a clear view of the functional relationship between them.

Linear Equations

Linear equations are mathematical statements of equality that graph as straight lines when plotted on a coordinate plane. They usually take the form \(y = mx + b\), where \(m\) represents the slope and \(b\) the y-intercept.

In our solved problem, transforming the original equation into function form means rewriting it as a linear equation. We have achieved this with \(y = -\frac{1}{5}x + \frac{2}{5}\). In this form:

• The slope \(-\frac{1}{5}\) tells us the direction and steepness of the line. A negative slope means the line slopes downward from left to right.
• The y-intercept \(\frac{2}{5}\) is the point where the line crosses the y-axis.

Linear equations are incredibly useful as they model real-world phenomena across various disciplines, such as economics and physics, allowing us to predict one variable based on the linear relationship with another.