Problem 5 - Beginning and Intermediate Algebra

Question

Evaluate (if possible) for a) \(x=2\) and b) \(x=-1\). $$\frac{7 x+1}{3 x-1}$$

Step-by-Step Solution

Answer

For the given values of x: a) x = 2: The expression evaluates to 3. b) x = -1: The expression evaluates to \(\frac{3}{2}\).

Step 1: Write the given expression

The given expression is: \[\frac{7x+1}{3x-1}\]

Step 2: Evaluate the expression for x = 2

Now, we will substitute x with 2 and simplify the expression: \[\frac{7(2)+1}{3(2)-1} = \frac{14+1}{6-1} = \frac{15}{5}\] Now, we will simplify the fraction: \[\frac{15}{5} = 3\] So, the expression evaluates to 3 when x=2.

Step 3: Evaluate the expression for x = -1

Now, we will substitute x with -1 and simplify the expression: \[\frac{7(-1)+1}{3(-1)-1} = \frac{-7+1}{-3-1} = \frac{-6}{-4}\] Now, we will simplify the fraction: \[\frac{-6}{-4} = \frac{3}{2}\] So, the expression evaluates to \(\frac{3}{2}\) when x=-1.

Step 4: Write down the final answer

For the given values of x: a) x = 2: The expression evaluates to 3. b) x = -1: The expression evaluates to \(\frac{3}{2}\).

Key Concepts

Evaluating Expressions

When it comes to evaluating expressions in algebra, the key is to substitute the chosen number for each occurrence of the variable, and then carefully perform arithmetic operations. This has nothing to do with solving equations, but simply finding what the expression equals to for certain values of variables. Think of it like following a recipe where each variable has a specific value as an ingredient.

For example, let's evaluate the expression \( \frac{7x+1}{3x-1} \) for \( x=2 \). You replace every 'x' in the expression with 2. It would look like this:

• Substitute 2 into the expression: \( \frac{7(2)+1}{3(2)-1} \)
• Calculate the numerator: \( 7 \times 2 + 1 = 15 \)
• Calculate the denominator: \( 3 \times 2 - 1 = 5 \)
• Resulting fraction: \( \frac{15}{5} \)

The final evaluated result is simply the number 3. These simple substitutions and calculations can turn complex-looking expressions into straightforward problems to solve.

Substitution Method

The substitution method is a powerful tool in algebra that lets you simplify problems by replacing a variable with a specific number. This technique is widely used not only in simple evaluations but also in solving systems of equations. By using substitution, you can turn abstract symbols into concrete numbers, making calculations more intuitive.

Take, for instance, the expression \( \frac{7x+1}{3x-1} \). If you want to evaluate it for \( x=-1 \), here is how substitution is applied:

• Replace every 'x' with -1: \( \frac{7(-1) + 1}{3(-1) - 1} \)
• Calculate the numerator: \( 7 \times -1 + 1 = -6 \)
• Calculate the denominator: \( 3 \times -1 - 1 = -4 \)
• The expression becomes: \( \frac{-6}{-4} \)

After substitution, you’re left with numbers that can be easily simplified. This method unlocks the solution by solidifying the role of variables and getting rid of abstraction.

Simplifying Fractions

Simplifying fractions is an essential skill in algebra that involves reducing fractions to their simplest form. This process ensures that the fraction is expressed in the lowest terms, making mathematical work easier and more precise. Simplifying a fraction involves both the numerator and the denominator and finding the greatest common divisor (GCD) to divide them.For example, after substituting numbers into the expression \( \frac{7x+1}{3x-1} \), we evaluated it to \( \frac{15}{5} \) when \( x=2 \). We can simplify this fraction as follows:

• Identify the greatest common divisor of 15 and 5, which is 5.
• Divide both the numerator and the denominator by 5: \( \frac{15 \div 5}{5 \div 5} = 3 \).

Similarly, with \( x=-1 \), the fraction \( \frac{-6}{-4} \) can be simplified:

• The GCD of 6 and 4 is 2.
• Divide the numerator and the denominator by 2: \( \frac{-6 \div 2}{-4 \div 2} = \frac{3}{2} \).

Remember, simplifying fraction is all about finding and using the GCD. It's useful across problems in algebra and ensures results are as neat and understandable as possible.