Problem 120 - College Algebra Essentials

Question

In Exercises \(115-122,\) find all values of \(x\) satisfying the given conditions. $$ y_{1}=\frac{3}{x-1}, y_{2}=\frac{8}{x}, \text { and } y_{1}+y_{2}=3 $$

Step-by-Step Solution

Answer

The solution for \(x\) is \(x = 2\) and \(x = \frac{4}{3}\)

Step 1: Set up the equation

Combine the equations given: \(y_{1}+y_{2}=\frac{3}{x-1}+\frac{8}{x}\) which we know is equal to 3, thus making the equation \(\frac{3}{x-1}+\frac{8}{x}=3\).

Step 2: Simplify the equation

To simplify the equation, multiply every term by \(x(x-1)\) to clear the fractions. Therefore, the equation becomes: \(3x+8(x-1)=3x(x-1)\).

Step 3: Further simplify the equation

distribute the right hand side to get: \(3x+8x-8=3x^{2}-3x\), Combine like terms to get: \(11x-8=3x^{2}-3x\).

Step 4: Rearrange the equation

Rearrange the equation to bring all terms to one side yielding a quadratic equation: \(3x^{2} - 14x + 8 = 0\)

Step 5: Solve the quadratic equation

Use the quadratic formula \(-b \pm \sqrt{{b^{2}-4ac}} / 2a\) to solve for \(x\). Therefore, \(x\) equals \( \frac{14 \pm \sqrt{(14^2 - 4 * 3 * 8)}}{2*3} \)

Step 6: Simplify to find the values of \(x\)

Solving the equations gives two possible values of \(x\), which are \(2\) and \(frac{4}{3}\)

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in algebra for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. This formula provides a method to find the solutions to any quadratic equation.
The formula is \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). With this tool, you can determine the roots of the quadratic equation regardless of whether they are real or complex numbers.
Let’s look at how we applied the quadratic formula in the exercise provided. After arranging our equation into the standard quadratic form \(3x^2 - 14x + 8 = 0\), we can identify \(a=3\), \(b=-14\), and \(c=8\). Substituting these values into the quadratic formula, we find the two possible solutions for \(x\), which helps us gain a complete understanding of all potential values of \(x\) that satisfy the given conditions.

Algebraic Fractions

Algebraic fractions are just like regular fractions but include variables in their numerators, denominators, or both. Working with algebraic fractions generally involves finding a common denominator to combine them or to resolve an equation.
In our exercise, we had two algebraic fractions, \(\frac{3}{x-1}\) and \(\frac{8}{x}\) that we needed to add together. To do this effectively, we first aimed to find a common denominator, which would be the product of \(x\) and \(x-1\), leading us to multiply each fraction by a form of 1 to achieve equivalent fractions with the common denominator. Clear communication of the steps to handle these additions simplifies the concept for students.

Quadratic Equations

Quadratic equations, generally in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are known values, make up a fundamental class of polynomial equations. These equations are solved for the variable \(x\), which may yield zero, one, or two solutions.
In our exercise, after clearing fractions and combining like terms, we brought all terms to one side to get \(3x^2 - 14x + 8 = 0\), a standard quadratic equation. Recognizing these forms and knowing the methods to solve them, such as factoring, completing the square, or using the quadratic formula, is crucial in algebra. Each method has its place, but the quadratic formula is the most comprehensive, which is why we used it in our solution.

Clearing Fractions

When solving equations that contain fractions, one helpful strategy is to clear the fractions to ease the process of finding a solution. This technique involves multiplying each term of the equation by the least common denominator (LCD) to get rid of the fractions.
In our original problem, we started with the equation \(\frac{3}{x-1} + \frac{8}{x} = 3\). The LCD of \(x\) and \(x-1\) is their product, \(x(x-1)\). We cleared the fractions by multiplying each term by this LCD, simplifying the equation to a more manageable form without fractions. Clearing fractions is an invaluable step because it transforms the problem into a more familiar and solvable quadratic equation.