Problem 52
Question
In Exercises \(45-54\), evaluate the algebraic expression for the given values of the variables. If it is not possible, state the reason. \(\frac{2 x-y}{y^{2}+1}\) (a) \(x=1, y=2\) (b) \(x=1, y=3\)
Step-by-Step Solution
Verified Answer
Hence, for \(x=1, y=2\), the solution is \(0\) and for \(x=1, y=3\), the solution is \(-0.1\).
1Step 1: Substitute for \(x=1, y=2\)
First, replace x with 1 and y with 2 in the expression, so that it becomes \(\frac{2*1 - 2}{2^{2} + 1}\).
2Step 2: Simplify for \(x=1, y=2\)
Simplify the expression to get the answer. In the numerator, \(2*1 - 2= 0\). In the denominator, \(2^{2} + 1= 5\). Therefore, the fraction simplifies to \(\frac{0}{5}=0\)
3Step 3: Substitute for \(x=1, y=3\)
Next, replace x with 1 and y with 3 in the original expression \(\frac{2 x-y}{y^{2}+1}\), so it becomes \(\frac{2*1 - 3}{3^{2} + 1}\).
4Step 4: Simplify for \(x=1, y=3\)
Simplify the expression to get the answer. In the numerator, \(2*1 - 3= -1\). In the denominator, \(3^{2} + 1=10\). Therefore, the fraction simplifies to \(\frac{-1}{10}=-0.1\)
Key Concepts
Variable SubstitutionExpression SimplificationEvaluating Expressions
Variable Substitution
Imagine algebraic expressions as recipes and variables as ingredients that can vary. When we want to evaluate such an expression, our first task is to substitute the given values of variables into the expression. Think of it like swapping out generic ingredients with real ones, say flour for "two cups of sugar". This step of replacement is called **Variable Substitution**.
For example, take the expression \(\frac{2x-y}{y^2+1}\). If we're given that \(x=1\) and \(y=2\), we substitute these values directly into the expression:
This substitution method sets the stage for the next critical step: simplifying the expression. But remember, accurate substitution is essential for obtaining meaningful results. Any error in substituting can lead to incorrect outcomes.
For example, take the expression \(\frac{2x-y}{y^2+1}\). If we're given that \(x=1\) and \(y=2\), we substitute these values directly into the expression:
- Replace \(x\) with 1, resulting in \(2 \cdot 1\).
- Replace \(y\) with 2, transforming the expression to \(\frac{2 \cdot 1 - 2}{2^2 + 1}\).
This substitution method sets the stage for the next critical step: simplifying the expression. But remember, accurate substitution is essential for obtaining meaningful results. Any error in substituting can lead to incorrect outcomes.
Expression Simplification
After variable substitution, it's time to simplify the expression—a crucial part of evaluating any algebraic expression. Think of **Expression Simplification** like tidying up your workspace by combining like terms and performing basic operations. The goal is to make the expression as simple as possible so it can be easily evaluated.
Consider our example where the expression has been modified to \(\frac{2 \cdot 1 - 2}{2^2 + 1}\) from the earlier substitution:
This step of simplification allows us to clearly see the final result of the expression under the specified values. It’s like having each item in your workspace neatly organized, allowing you to focus on what's important. Simplification is often the step where students make mistakes, so practice is key.
Consider our example where the expression has been modified to \(\frac{2 \cdot 1 - 2}{2^2 + 1}\) from the earlier substitution:
- Start with the numerator: Calculate \(2 \cdot 1 - 2\) which simplifies to 0.
- Move to the denominator: Calculate \(2^2 + 1\) which results in 5.
This step of simplification allows us to clearly see the final result of the expression under the specified values. It’s like having each item in your workspace neatly organized, allowing you to focus on what's important. Simplification is often the step where students make mistakes, so practice is key.
Evaluating Expressions
Once you've substituted the variables and simplified the expression, the final step is **Evaluating Expressions**. This step involves calculating the precise value that the simplified expression represents under the given conditions.
For instance, taking the previously simplified expression \(\frac{0}{5}\), we can see the value is simply 0. Another example from our exercise is substituting \(x=1\) and \(y=3\), which eventually simplifies to \(\frac{-1}{10}\). Here, you go through the following steps:
For instance, taking the previously simplified expression \(\frac{0}{5}\), we can see the value is simply 0. Another example from our exercise is substituting \(x=1\) and \(y=3\), which eventually simplifies to \(\frac{-1}{10}\). Here, you go through the following steps:
- Evaluate the numerator: \(2 \cdot 1 - 3 = -1\).
- Evaluate the denominator: \(3^2 + 1 = 10\).
- Combine them for the final value: \(\frac{-1}{10} = -0.1\).
Other exercises in this chapter
Problem 52
The dimensions of a rectangular lawn are 150 feet by 250 feet. The property owner buys a rectangular strip of land \(x\) feet wide along one 250 -foot side of t
View solution Problem 52
In Exercises \(47-66\), simplify the expression by removing symbols of grouping and combining like terms. $$ 8 l-(3 l-7) $$
View solution Problem 53
In Exercises 51-54, determine whether each value of \(x\) is a solution of the equation. \(\frac{5}{x-1}+\frac{1}{x}=5\) (a) \(x=3\) (b) \(x=\frac{1}{6}\)
View solution Problem 53
A bubble rises through water at a rate of about \(1.15\) feet per second. How far will the bubble rise in 5 seconds?
View solution