Problem 57 - Algebra for College Students

Question

Evaluate the algebraic expressions in Problems 35-57 for the given values of the variables. \(5(x-2 y)-3(2 x+y)-2(x-y), \quad x=\frac{1}{3}\) and \(y=-\frac{3}{4}\)

Step-by-Step Solution

Answer

The evaluated expression is \(\frac{29}{4}\).

Step 1: Substitute Values.

First, substitute the given values of the variables into the expression. Replace \(x\) with \(\frac{1}{3}\) and \(y\) with \(-\frac{3}{4}\): \[ 5\left(\frac{1}{3} - 2\left(-\frac{3}{4}\right)\right) - 3\left(2\cdot\frac{1}{3} + \left(-\frac{3}{4}\right)\right) - 2\left(\frac{1}{3} - \left(-\frac{3}{4}\right)\right) \]

Step 2: Simplify Each Term

Calculate each part of the expression separately after substitution. Start with simplifying inside the parentheses. Compute \(\frac{1}{3} - 2(-\frac{3}{4})\), \(2\cdot\frac{1}{3} - \frac{3}{4}\), and \(\frac{1}{3} - \left(-\frac{3}{4}\right)\).\[ \frac{1}{3} + \frac{6}{4} = \frac{1}{3} + \frac{3}{2} = \frac{11}{6} \]\[ \frac{2}{3} - \frac{3}{4} = \frac{8}{12} - \frac{9}{12} = -\frac{1}{12} \]\[ \frac{1}{3} + \frac{3}{4} = \frac{4}{12} + \frac{9}{12} = \frac{13}{12} \]

Step 3: Apply Distributive Property

Distribute the coefficients across the simplified terms. Evaluate following expressions:\[ 5 \times \frac{11}{6},\quad -3 \times -\frac{1}{12},\quad -2 \times \frac{13}{12} \]. Calculate each product.\[ 5 \times \frac{11}{6} = \frac{55}{6} \]\[ -3 \times -\frac{1}{12} = \frac{3}{12} = \frac{1}{4} \]\[ -2 \times \frac{13}{12} = -\frac{26}{12} = -\frac{13}{6} \]

Step 4: Combine Terms

Add the results from the distribution to combine like terms:\[ \frac{55}{6} + \frac{1}{4} - \frac{13}{6} \] Convert all terms to a common denominator before combining. Note that the common denominator for the fractions is 12.\[ \frac{55}{6} = \frac{110}{12}, \quad \frac{1}{4} = \frac{3}{12}, \quad -\frac{13}{6} = -\frac{26}{12} \]\[ \frac{110}{12} + \frac{3}{12} - \frac{26}{12} = \frac{110 + 3 - 26}{12} = \frac{87}{12} \]

Step 5: Final Simplification

Simplify the fraction \(\frac{87}{12}\) by dividing both the numerator and the denominator by their greatest common divisor (GCD), which is 3.\[ \frac{87}{12} = \frac{29}{4} \] This fraction cannot be simplified further.

Key Concepts

Substitution

Substitution is the process of replacing variables with their given values within an algebraic expression. It is a fundamental step in solving equations and evaluating expressions. By doing this, you're transforming an abstract equation into something more tangible:

• If the exercise provides values like \(x = \frac{1}{3}\) and \(y = -\frac{3}{4}\), you replace every occurrence of \(x\) and \(y\) in the expression with these values.
• This is akin to swapping placeholders with actual numbers.

This simple step allows you to evaluate the expression numerically rather than symbolically. For instance, replacing \(x\) and \(y\) in the given expression yields:\[5\left(\frac{1}{3} - 2\left(-\frac{3}{4}\right)\right) - 3\left(2 \cdot \frac{1}{3} + \left(-\frac{3}{4}\right)\right) - 2\left(\frac{1}{3} - \left(-\frac{3}{4}\right)\right)\]Substitution simplifies your task as you shift from working with variables to numbers.

Distributive Property

The distributive property is a useful principle in algebra that connects addition and multiplication. It allows you to distribute or "spread" a multiplication operation across numbers inside parentheses:

• Mathematically, it is expressed as \(a(b + c) = ab + ac\).
• This property helps to simplify expressions where a number or variable is multiplied by a sum or difference.

Once you've substituted the values into the expression, apply the distributive property to simplify further. Take for example:\[5 \times \frac{11}{6},\quad -3 \times -\frac{1}{12},\quad -2 \times \frac{13}{12}\]These computations are simplified by immediately distributing the multiplication across the fractional results from substitution.

Fraction Simplification

Fraction simplification, or reducing fractions, involves adjusting the fraction to its simplest form. This means making the numerator and the denominator as small as possible by using their greatest common divisor (GCD):

• To simplify \(\frac{65}{15}\), find the GCD, which is 5.
• Divide both the numerator and denominator by 5 to get \(\frac{13}{3}\).

Apply this step in solving the given exercise where necessary, ensuring all parts of your expression are manageable. For instance, when simplifying the product \(-\frac{26}{12}\), you divide both by their GCD, which is 2, resulting in \(-\frac{13}{6}\). This method helps to navigate and simplify through complex algebraic expressions more effectively.

Common Denominator

A common denominator is a shared multiple of the denominators of several fractions. Before adding or subtracting fractions, they must have the same denominator:

• Common multiple helps in the alignment of fractions for operations.
• For example, in the given exercise, the common denominator for \(\frac{55}{6}, \frac{1}{4}, \text{ and } -\frac{13}{6}\) is 12.

Converting each fraction so that they share a common denominator allows you to perform addition or subtraction accurately. For instance:\[\frac{55}{6} = \frac{110}{12},\quad \frac{1}{4} = \frac{3}{12},\quad -\frac{13}{6} = -\frac{26}{12}\]This adjustment is crucial to combine fractions correctly, leading to the accurate computation of the final result, in this case \(\frac{87}{12}\), which you then simplify further.