/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Rewrite each equation in \(Y=\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Rewrite each equation in \(Y=\) form. a. \(x y=15\) b. \(x y=35\) c. \(x y=3\)

Short Answer

Expert verified
a. \(y = \frac{15}{x}\) b. \(y = \frac{35}{x}\) c. \(y = \frac{3}{x}\)

Step by step solution

01

Step 1a: Isolate Y in Equation 'a'

Given the equation, \(x y = 15\), we need to solve for \(Y\). To do this, divide both sides of the equation by \(x\) to isolate \(Y\): \(y = \frac{15}{x}\).
02

Step 1b: Isolate Y in Equation 'b'

For the equation \(x y = 35\), to express it in \(Y\)-form, divide both sides by \(x\): \(y = \frac{35}{x}\).
03

Step 1c: Isolate Y in Equation 'c'

Starting with the equation \(x y = 3\), divide both sides by \(x\) to solve for \(Y\): \(y = \frac{3}{x}\).

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Key Concepts

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

Solving for Y
When faced with an algebraic equation like those given in the exercise, the goal is often to rearrange or 'solve' for a specific variable. In this context, we are focusing on rewriting the equations such that we express the variable \(y\) by itself. This is commonly known as solving for \(Y\).

To achieve this, we need to manipulate the equation algebraically until \(y\) is isolated on one side of the equation. This generally involves using operations such as addition, subtraction, multiplication, or division to "undo" the operations surrounding \(y\). In our specific examples, since both sides of the equation involve multiplication by \(x\), division becomes our key operation.

This step-by-step approach not only helps in understanding the relationships between variables but also strengthens problem-solving skills in algebra. By consistently practicing solving for \(y\), students become proficient in viewing and manipulating equations from different perspectives.
Equations in Y-form
Equations are often rewritten into a "form" where one variable is expressed in terms of others. In the context of the exercise, "equations in \(Y\)-form" specifically implies that we want \(y\) as the subject of our equation, typically written as \(y = \text{expression}\).

The standard objective is to express mathematical relationships clearly and concisely where \(y\) is isolated. This formulation is particularly useful to:
  • Assess how changes in other variables, such as \(x\), affect \(y\).
  • Graph equations, making it easier to plot and analyze relationships in the coordinate plane.
  • Simplify complex equations, breaking down components for practical applications.
For instance, by transforming \(x y = 15\) to \(y = \frac{15}{x}\), we clearly see how \(y\) varies inversely with \(x\). Such an approach allows deeper insights into problem-solving, whether in mathematics or applied fields.
Isolating Variables
Isolating variables is a fundamental technique in algebra where the desired variable is left alone on one side of the equation. This allows for direct interpretation or substitution into other equations, making problem-solving more straightforward.

The process often includes the following steps:
  • Identify the variable to isolate, in this case, \(y\).
  • Utilize operations that systematically untwine \(y\) from other parts of the equation, such as using division to "cancel" the multiplication by \(x\).
  • Ensure operations are performed symmetrically on both sides to maintain equality.
In our examples, solving equations such as \(x y = 35\) involves dividing both sides by \(x\) so that \(y\) stands alone as \(y = \frac{35}{x}\).

With practice, isolating variables becomes intuitive, aiding in tackling more advanced topics like functions and systems of equations. Having mastery over this skill simplifies expressions and unlocks the ability to explore and model real-world phenomena efficiently.

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The equation \(\frac{5(\mathrm{~F}-32)}{9}=C\) can be used to change temperatures in Fahrenheit to the Celsius scale. a. What is the first step when converting a temperature in \({ }^{\circ} \mathrm{F}\) to \({ }^{\circ} \mathrm{C}\) ? (A) b. What is the last step when converting a temperature in \({ }^{\circ} \mathrm{F}\) to \({ }^{\circ} \mathrm{C}\) ? c. What is the first step in undoing a temperature in \({ }^{\circ} \mathrm{C}\) to find the temp in \({ }^{\circ} \mathrm{F}\) ? d. What is the last step in undoing a temperature in \({ }^{\circ} \mathrm{C}\) to find the temp in \({ }^{\circ} \mathrm{F}\) ?

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