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Magazine Chapter 6: Problem 6
30 x y^{3}-y^{8}=15 \text { between } y=1 \text { and } y=3
Short Answer
Expert verified
For \( y = 1, x = \frac{8}{15} \); for \( y = 2, x = \frac{271}{240} \); for \( y = 3, x = \frac{4}{5} \).
Step by step solution
01
Rewrite the Equation
Start by rewriting the equation: \( 30xy^3 - y^8 = 15 \). In this problem, we are interested in checking specific values of \( y \).
02
Check Values of y
You are given a range for \( y \) between 1 and 3. Check the values for \( y = 1, \; y = 2, \; y = 3 \). This means you need to substitute these values into the left side of the equation.
03
Substituting y=1
Substituting \( y = 1 \) into the equation gives: \( 30x(1)^3 - (1)^8 = 15 \). Simplify to get \( 30x - 1 = 15 \), which simplifies further to \( 30x = 16 \). Solve for \( x \) to get \( x = \frac{16}{30} = \frac{8}{15} \).
04
Substituting y=2
Substitute \( y = 2 \) into the equation: \( 30x(2)^3 - (2)^8 = 15 \). Simplify to get \( 30x imes 8 - 256 = 15 \), which simplifies to \( 240x = 271 \). Solve for \( x \) to get \( x = \frac{271}{240} \).
05
Substituting y=3
Substitute \( y = 3 \) into the equation: \( 30x(3)^3 - (3)^8 = 15 \). Simplify to get \( 30x imes 27 - 6561 = 15 \), which simplifies to \( 810x = 6576 \). Solve for \( x \) to get \( x = \frac{6576}{810} = \frac{12}{15} = \frac{4}{5} \).
06
Review Results
After calculating \( x \) for each value of \( y \), you have \( x = \frac{8}{15} \) for \( y = 1 \), \( x = \frac{271}{240} \) for \( y = 2 \), and \( x = \frac{4}{5} \) for \( y = 3 \). Check these calculations for arithmetic accuracy.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Algebraic manipulation
Algebraic manipulation is a technique used to transform equations and expressions into a form that makes them easier to solve or analyze. In the context of our exercise, we are given the equation: \( 30xy^3 - y^8 = 15 \). Our goal is to simplify this expression by isolating variables or simplifying constants.
To manipulate this equation:
To manipulate this equation:
- Start by substituting specific values for a variable, like substituting values of \( y \) from the given range.
- This is done to test if simpler calculations can isolate another variable (in this case, \( x \)).
- Rearrange the equation to make one of the variables the subject. This helps us see the relationship between the variables clearly.
Equation solving
Equation solving is a fundamental skill in algebra and calculus that involves finding unknown values that satisfy a given equation. In our exercise, the equation \( 30xy^3 - y^8 = 15 \) needed to be solved for \( x \) by plugging in different values of \( y \).
Here's how we tackled this:
Here's how we tackled this:
- For each trial value of \( y \), substitute it into the equation.
- Simplify the equation by performing necessary arithmetic operations like addition, multiplication, or exponentiation of constants and variables.
- Isolate the variable \( x \) (if it's still present) on one side by rearranging terms.
- Use inverse operations to solve for \( x \), which may involve division or the use of fractions to make calculations precise.
Substitution method
The substitution method is a strategy used to solve equations involving more than one variable by replacing a variable with a specific value or another expression. This method is extremely helpful when direct simplification is challenging.
In the exercise, using substitution was crucial:
In the exercise, using substitution was crucial:
- Given the equation \( 30xy^3 - y^8 = 15 \), we selected specific values for \( y \) (like 1, 2, and 3).
- These values were plugged into the equation in place of \( y \), transforming the original equation into one that can be solved for \( x \).
- This results in a new equation each time, making it possible to calculate a specific solution for \( x \) instead of dealing with a general case.
