Problem 56 - Algebra for College Students

Question

Set up an algebraic equation and solve each problem. What number must be added to the numerator and denominator of \(\frac{2}{5}\) to produce a rational number that is equivalent to \(\frac{7}{8} ?\)

Step-by-Step Solution

Answer

The number is 19.

Step 1: Identify the rational expression

The original fraction is \( \frac{2}{5} \). We need to alter it by adding a certain number to both the numerator and denominator.

Step 2: Write the equation for the new fraction

Let the number to be added be \( x \). The expression becomes \( \frac{2+x}{5+x} \). This needs to equal the fraction \( \frac{7}{8} \).

Step 3: Set up the equation

We equate the new fraction to the given fraction: \( \frac{2+x}{5+x} = \frac{7}{8} \).

Step 4: Cross-multiply to clear the fractions

Cross-multiply to eliminate the fractions: \( 8(2+x) = 7(5+x) \).

Step 5: Distribute and simplify each side

Expand both sides: \( 16 + 8x = 35 + 7x \).

Step 6: Solve for x

Move all terms involving \( x \) to one side: \( 8x - 7x = 35 - 16 \), resulting in \( x = 19 \).

Step 7: Verify the solution

Substitute \( x = 19 \) back into the new fraction: \( \frac{2+19}{5+19} = \frac{21}{24} \). Simplifying, \( \frac{21}{24} \) can be reduced to \( \frac{7}{8} \), confirming the solution is correct.

Key Concepts

Rational Expression

A rational expression is essentially a fraction, but with polynomials as the numerator and/or denominator. In simpler terms, it is an expression made by dividing one algebraic expression by another. Just like regular fractions, rational expressions have a numerator (top part) and a denominator (bottom part). Understanding how to work with these forms is fundamental in algebra.

For instance, in the problem provided, the rational expression starts as \( \frac{2}{5} \). We are looking to find a number \( x \) that when added to both 2 (numerator) and 5 (denominator), turns the expression into \( \frac{7}{8} \).

When working with rational expressions, keep in mind:

• You can perform operations such as addition, subtraction, multiplication, and division.
• Always aim to simplify the expression to its lowest terms, if it’s possible.
• While solving equations involving rational expressions, ensure the denominator never equals zero as it makes the expression undefined.

By understanding these basics, handling rational expressions in algebra becomes much easier.

Cross-Multiplication

Cross-multiplication is a handy method to solve equations involving two fractions set equal to each other. It essentially means multiplying the numerator of each fraction by the denominator of the other. This helps in clearing the fractions, making the equation simpler to solve.

In our example, we have the equation \( \frac{2 + x}{5 + x} = \frac{7}{8} \). Here’s how you would use cross-multiplication:

• Multiply the numerator of the left fraction by the denominator of the right: \( 8(2 + x) \).
• Multiply the numerator of the right fraction by the denominator of the left: \( 7(5 + x) \).
• Set the products equal: \( 8(2 + x) = 7(5 + x) \).

This forms a new equation without fractions, easing the path to solving for \( x \). Remember, cross-multiplication only works if the equation is in the form of a single fraction on each side.

Fraction Simplification

Simplification is reducing a fraction to its simplest form. This means making the numerator and denominator as small as possible while keeping their ratio identical. Simplification is crucial not only for clarity but also to verify the accuracy of solutions.

After solving the equation for \( x \) in the previous steps, we substitute it back to check our work: \( \frac{2+19}{5+19} = \frac{21}{24} \). Now, we simplify this fraction:

• Identify the greatest common divisor (GCD) of the numerator and denominator. For 21 and 24, the GCD is 3.
• Divide both the numerator and the denominator by their GCD: \( \frac{21 \div 3}{24 \div 3} = \frac{7}{8} \).

The simplified fraction \( \frac{7}{8} \) confirms that our previous steps were correct. Simplifying fractions ensures that you have the most straightforward version of the solution possible.