Problem 26 - Intermediate Algebra

Question

Solve equation. \(\frac{n}{7}=\frac{n-19}{5 n-45}-\frac{1}{5}\)

Step-by-Step Solution

Answer

Use the quadratic formula on simplified quadratic equation to find the values of \(n\).

Step 1: Clear Denominators

To clear the fractions, we'll find a common denominator for the fractions on the right side of the equation. The common denominator for \(5n - 45\) and 5 is \((5n - 45)\ldots 5\). Multiply each term by this common denominator to eliminate the fractions.

Step 2: Multiply Each Term

Multiply every term by \(5\times(5n - 45)\):\[5(5n - 45) \cdot \frac{n}{7} = (5n - 45) \cdot (n - 19) - 1(5n - 45)\]The goal is to simplify by distributing and eliminating fractions.

Step 3: Simplify Each Side

Now simplify both sides:\[(25n - 225) \cdot \frac{n}{7} = (5n^2 - 95n + 855) - (5n - 45)\]Continue to distribute and combine like terms on each side.

Step 4: Distribute and Simplify

Distribute in the equation:Left Side: \[25n^2 - 225n = 5n^2 - 95n + 855 - 5n + 45\]Right Side simplifies to:\[5n^2 - 100n + 900\]

Step 5: Combine Like Terms

Combine all like terms:\[25n^2 - 225n = 5n^2 - 100n + 900\]

Step 6: Move All Terms to One Side

Move all terms to one side to set the equation to zero:\[0 = 25n^2 - 5n^2 - 225n + 100n + 900\]This simplifies to:\[0 = 20n^2 - 125n + 900\]

Step 7: Factor or Use Quadratic Formula

Notice the equation is quadratic, \[20n^2 - 125n + 900 = 0\]Use the quadratic formula: \[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where (a = 20, b = -125, c = 900).Calculate under the square root and then solve for the two possible values of n.

Step 8: Solve for n

Substitute into the quadratic formula:\[n = \frac{-(-125) \pm \sqrt{(-125)^2 - 4 \cdot 20 \cdot 900}}{2 \cdot 20}\]Solve for both possible values of \(n\).

Key Concepts

Understanding the Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are any polynomials of the form \(ax^2 + bx + c = 0\). Using this formula, you can find the roots of the quadratic equation, which are the values of \(x\) that satisfy the equation. The formula is expressed as follows:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• "\(-b\)": This part reverses the sign of the coefficient \(b\).
• "\(b^2 - 4ac\)": Known as the discriminant, this determines the nature and number of roots. A positive discriminant suggests two real roots, zero means one real root, and a negative means no real roots.
• "\(\pm\)": Indicates that there may be two possible solutions due to the square root.

Applying the quadratic formula starts by identifying \(a\), \(b\), and \(c\) from the equation, substituting these values into the formula, and then calculating the result, taking care to resolve \(\pm\sqrt{b^2 - 4ac}\) to find both solutions.

Factoring as a Method to Solve Equations

Factoring is a method that transforms a polynomial equation into a product of simpler polynomials. This method is particularly useful if the polynomial can be easily broken down into factors, especially binomials. To factor quadratics, look for two numbers that both add up to \(b\) (the middle coefficient) and multiply to \(a \times c\) (product of the leading coefficient \(a\) and the constant term \(c\)).
For instance, consider the expression \(20n^2 - 125n + 900\). While this particular quadratic equation was solved using the quadratic formula, factoring involves:

• Checking if there is a simple common factor among terms.
• Rewriting the equation in terms that can be broken down.
• Factoring by grouping if applicable.

This method is less complex and faster than the quadratic formula when applicable, but not all quadratics can be factored easily, thus making the quadratic formula a crucial alternative.

Common Denominator for Equation Solving

When solving equations that involve fractions, finding a common denominator is key to eliminating them. A common denominator is a shared multiple of the denominators involved, allowing the equation to be simplified to contain whole numbers. In the original exercise, a common denominator for terms \(\frac{n}{7}\) and \(\frac{n-19}{5n-45} - \frac{1}{5}\) was necessary.
Here's how you find and use a common denominator:

• Identify all denominators in the equation.
• Calculate their least common multiple (LCM).
• Multiply every term in the equation by this LCM to clear out the fractions.

By doing this, you transform the equation from a potentially confusing fraction-laden expression into a simpler algebraic expression, which can be further manipulated without the complexity added by fractions.

Distributing in Algebra

Distributing, also known as the distributive property, is used to simplify expressions. It involves multiplying a single term by each term in a parenthesis. In algebraic terms, it states that \(a(b + c) = ab + ac\). This property is useful for clearing terms marked by parentheses, particularly in solving equations or simplifying expressions.
Consider this example: In the equation solving step, you multiply everything by the common denominator. This includes phrases such as \(5(5n - 45) \cdot \frac{n}{7}\), where you apply the distributive property:

• Multiply \(5n - 45\) across the terms inside the parenthesis.
• Simplify the expanded terms, making it easier to combine like terms and solve.

This process streamlines equations, especially those involving multiple fractions or parenthesis, guiding them towards a form that can be solved using standard algebraic methods like the quadratic formula or factoring.