Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Help me solve for x for these I'm confused
![### Algebraic Expressions and Equations
#### Basic Algebraic Manipulation
In the first part, we have the equation \(x + 6 = 23FG - 6\).
To isolate \(x\), we subtract 6 from both sides:
\[
x + 6 - 6 = 23FG - 6 - 6
\]
This simplifies to:
\[
x = 23FG - 6
\]
This demonstrates how to solve for \(x\) when it is in a simple linear equation.
#### Complex Algebraic Expression
Below the simplified equation, there is a more complex algebraic notation:
\[
\frac{C^2 D^5 x}{E^4 R} = \frac{T^3 K}{Y}
\]
This equation involves multiple variables, and both sides of the equation include fractions and exponents.
##### Explanation:
- \(C^2\): This denotes \(C\) squared.
- \(D^5 x\): This denotes \(D\) raised to the 5th power, multiplied by \(x\).
- \(E^4 R\): This denotes \(E\) raised to the 4th power, multiplied by \(R\).
- \(T^3 K\): This denotes \(T\) cubed, multiplied by \(K\).
- \(Y\): The variable in the denominator on the right side of the equation.
The entire expression can be read as:
"The product of \(x\), \(C\) squared, and \(D\) to the 5th power all divided by the product of \(E\) to the 4th power and \(R\), is equal to the product of \(T\) cubed and \(K\) divided by \(Y\)."
This kind of equation might come up in advanced algebra, physics, or engineering problems where multiple variables and operations are involved.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F51435bf2-32bb-499c-aa57-a85dc35bd95d%2F2277879a-a2d1-4e30-95f0-6130856a2e51%2Fpycekfj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Algebraic Expressions and Equations
#### Basic Algebraic Manipulation
In the first part, we have the equation \(x + 6 = 23FG - 6\).
To isolate \(x\), we subtract 6 from both sides:
\[
x + 6 - 6 = 23FG - 6 - 6
\]
This simplifies to:
\[
x = 23FG - 6
\]
This demonstrates how to solve for \(x\) when it is in a simple linear equation.
#### Complex Algebraic Expression
Below the simplified equation, there is a more complex algebraic notation:
\[
\frac{C^2 D^5 x}{E^4 R} = \frac{T^3 K}{Y}
\]
This equation involves multiple variables, and both sides of the equation include fractions and exponents.
##### Explanation:
- \(C^2\): This denotes \(C\) squared.
- \(D^5 x\): This denotes \(D\) raised to the 5th power, multiplied by \(x\).
- \(E^4 R\): This denotes \(E\) raised to the 4th power, multiplied by \(R\).
- \(T^3 K\): This denotes \(T\) cubed, multiplied by \(K\).
- \(Y\): The variable in the denominator on the right side of the equation.
The entire expression can be read as:
"The product of \(x\), \(C\) squared, and \(D\) to the 5th power all divided by the product of \(E\) to the 4th power and \(R\), is equal to the product of \(T\) cubed and \(K\) divided by \(Y\)."
This kind of equation might come up in advanced algebra, physics, or engineering problems where multiple variables and operations are involved.
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