Problem 41
Question
Evaluate each expression for \(a=3\) and \(b=-5\). \(\frac{a-b}{2 a}\)
Step-by-Step Solution
Verified Answer
\(\frac{4}{3}\)
1Step 1: Substitute the given values into the expression
Replace the variables with their given values in the expression. In this case, substitute the value of 3 for a and the value of -5 for b into the expression \(\frac{a-b}{2a}\).
2Step 2: Perform the subtraction in the numerator
Calculate the value of the numerator by subtracting b from a. With the substituted values, this becomes \(\frac{3 - (-5)}{2 \cdot 3}\).
3Step 3: Simplify the expression
Add the values in the numerator and then multiply the denominator according to the order of operations. This results in \(\frac{3 + 5}{6}\), which simplifies to \(\frac{8}{6}\).
4Step 4: Reduce the fraction to lowest terms
Divide both the numerator and the denominator by their greatest common divisor (GCD) to reduce the fraction to its lowest terms. In this case, 8 and 6 are both divisible by 2, so the simplified expression is \(\frac{8/2}{6/2} = \frac{4}{3}\).
Key Concepts
Substitution Method in AlgebraSimplifying Algebraic FractionsOrder of Operations in Algebra
Substitution Method in Algebra
The substitution method in algebra is a fundamental technique used to evaluate expressions, especially when specific values are assigned to variables. It involves replacing variables with their given numerical values and then performing arithmetic operations to find the result.
For example, if you have an expression such as \( \frac{a-b}{2a} \) and you want to evaluate it for \( a=3 \) and \( b=-5 \) as given in the exercise, you start by substituting the values directly into the expression. This looks like \( \frac{3 - (-5)}{2 \cdot 3} \), effectively turning the algebraic expression into a simple arithmetic problem.
It's crucial, when using this method, to substitute accurately and to include parentheses where necessary to maintain the integrity of the original expression. After substitution, the next steps involve simplifying the expression, which leads us to another important topic: simplifying algebraic fractions.
For example, if you have an expression such as \( \frac{a-b}{2a} \) and you want to evaluate it for \( a=3 \) and \( b=-5 \) as given in the exercise, you start by substituting the values directly into the expression. This looks like \( \frac{3 - (-5)}{2 \cdot 3} \), effectively turning the algebraic expression into a simple arithmetic problem.
It's crucial, when using this method, to substitute accurately and to include parentheses where necessary to maintain the integrity of the original expression. After substitution, the next steps involve simplifying the expression, which leads us to another important topic: simplifying algebraic fractions.
Simplifying Algebraic Fractions
When dealing with algebraic fractions, being able to simplify them is quite useful. Simplifying fractions is the process of reducing them to their simplest form, so that the numerator and the denominator are as small as possible. This is done by dividing both by their greatest common divisor (GCD).
Starting from the substituted expression in the exercise, \( \frac{3 + 5}{6} \), we combine the numbers in the numerator to get \( \frac{8}{6} \). Even though this fraction is correct, it's not in its simplest form. The numbers 8 and 6 share a GCD of 2, which means we can divide both the numerator and the denominator by 2 to simplify the fraction to \( \frac{4}{3} \).
Simplifying algebraic fractions not only makes expressions more aesthetically pleasing, but also easier to work with, especially in further calculations or algebraic manipulations.
Starting from the substituted expression in the exercise, \( \frac{3 + 5}{6} \), we combine the numbers in the numerator to get \( \frac{8}{6} \). Even though this fraction is correct, it's not in its simplest form. The numbers 8 and 6 share a GCD of 2, which means we can divide both the numerator and the denominator by 2 to simplify the fraction to \( \frac{4}{3} \).
Simplifying algebraic fractions not only makes expressions more aesthetically pleasing, but also easier to work with, especially in further calculations or algebraic manipulations.
Order of Operations in Algebra
The order of operations, also known as PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction), is a set of rules used to clarify which procedures to perform first in a given mathematical expression.
When evaluating the expression \( \frac{a-b}{2a} \) after substitution, it's important to follow these rules to get the correct result. In our example, we must add the values within the parentheses \(3 - (-5)\), and then multiply the denominator \(2 \cdot 3\). The multiplication in the denominator occurs before dividing the numerator by this product. Understanding and following the order of operations ensures that every algebraic expression is simplified correctly and consistently.
When evaluating the expression \( \frac{a-b}{2a} \) after substitution, it's important to follow these rules to get the correct result. In our example, we must add the values within the parentheses \(3 - (-5)\), and then multiply the denominator \(2 \cdot 3\). The multiplication in the denominator occurs before dividing the numerator by this product. Understanding and following the order of operations ensures that every algebraic expression is simplified correctly and consistently.
Common Mistakes to Avoid
One pitfall in this process is neglecting to add or subtract before multiplying or dividing. For instance, if you multiply 2 by 3 before adding 3 and -5 in our initial example, you would get the wrong result. Always perform operations inside parentheses first and proceed with multiplication or division before moving on to addition or subtraction.Other exercises in this chapter
Problem 40
Classify each system without graphing. $$ \left\\{\begin{array}{l}{3 a+6 b=14} \\ {-a+2 b=3}\end{array}\right. $$
View solution Problem 41
Open-Ended Write and graph a system of inequalities for which the solution is bounded by a dashed vertical line and a solid horizontal line.
View solution Problem 41
Solve each system by elimination. \(\left\\{\begin{array}{l}{3 m+4 n=-13} \\ {5 m+6 n=-19}\end{array}\right.\)
View solution Problem 41
Classify each system without graphing. $$ \left\\{\begin{array}{l}{3 m=-5 n+4} \\ {n-\frac{6}{5}=-\frac{3}{5} m}\end{array}\right. $$
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