Problem 60 - College Algebra

Question

find all values of \(x\) such that \(y=0\) $$ y=\frac{1}{5 x+5}-\frac{3}{x+1}+\frac{7}{5} $$

Step-by-Step Solution

Answer

The possible values for \(x\) are \(x=1\) and \(x=-1\)

Step 1: Simplify the Equation

First, find a common denominator for the fractions, which in this case is \(5x+5\) and \(x+1\), so the common denominator will be \((5x+5)(x+1)\). Rewrite each fraction in the equation so they each have the common denominator: \[y= \frac{(x+1)-(15x+15)}{(5x+5)(x+1)}+\frac{7}{5}\] Then simplify the fraction and the equation becomes: \[y= \frac{-14x-14}{5x^2+10x+5}+\frac{7}{5}\]

Step 2: Reduce the Equation further

Now, multiply the entire equation by 5 to get rid of the fraction on the right side, and which also makes the equation a little cleaner: \[5y= \frac{-14x-14}{x^2+2x+1}+7\] Continue to reduce the equation by multiplying both sides by the denominator \((x^2+2x+1)\): \[5y(x^2+2x+1) = -14x-14+7(x^2+2x+1)\]

Step 3: Solve for \(x\)

Now the equation is much simpler. The equation can now be solved for \(x\) by expanding the expressions, gathering like terms, setting the expression equal to zero, and then solving for \(x\) using the quadratic formula: \[5yx^2+10yx+5y=-14x-14+7x^2+14x+7\] \[5yx^2-7x^2+10yx-14x+5y+7=0\] \[(5y-7)x^2+(10y-14)x+(5y+7)=0\] This is a quadratic equation in the form \(ax^2+bx+c=0\). The solutions for \(x\) are given by: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Step 4: Obtain the solutions

Now replace \(a\), \(b\) and \(c\) with the corrected values, and solve the equation: \(a=5y-7\), \(b=10y-14\), \(c=5y+7\). Note that since \(y=0\), the equation becomes: \[x = \frac{14 \pm \sqrt{(14)^2-4*7*(7)}}{2*7}\] The solutions are found to be \(x=1\), \(x=-1\), after simplification.

Key Concepts

Rational Expressions

Rational expressions are fractions in which the numerator and/or the denominator are polynomials. In the context of a quadratic equation like this one, you often encounter them as terms in an expression. Recognizing and working with rational expressions is crucial in algebra, especially when they appear as part of solving more complex equations.
To begin with, it's important to understand that each rational expression must be simplified or rewritten to facilitate further calculations. In our original problem, we have fractions like \(\frac{1}{5x+5}\) and \(\frac{3}{x+1}\). Here, the denominators contain variables, which is characteristic of rational expressions.
We approach solving problems with rational expressions by finding a common denominator, allowing us to combine or subtract these expressions more easily. Ultimately, working with rational expressions becomes a bit routine once you understand how to manipulate the polynomials effectively.

Common Denominator

When dealing with rational expressions, it's often necessary to find a common denominator. A common denominator allows you to combine fractions, making it easier to solve the equation. This is essential when you're faced with expressions like in the problem we're solving.
In our exercise, we need to combine \(\frac{1}{5x+5}\) and \(\frac{3}{x+1}\). The least common denominator (LCD) here would be \((5x+5)(x+1)\). Why choose this? Because multiplying these two gives us a shared baseline for each term in the equation. This shared baseline is crucial for simplifying and solving the equation.
By rewriting each term of the equation using this common denominator, we ultimately simplify the mathematical manipulations needed to arrive at a solution. It converts a complicated equation into a more manageable form, which paves the way for the next steps in solving our quadratic equation.

Solving for x

Once you simplify the rational expressions and have a common denominator, you may proceed to solve for \(x\). Here, solving for \(x\) often involves setting the equation to zero and solving the resulting quadratic equation using various algebraic methods.
In our example, we simplify the equation to a standard form of a quadratic equation \((ax^2 + bx + c = 0)\). This eventually allows us to apply the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

This formula is particularly useful when factoring is not straightforward or possible. It provides a systematic way to find solutions, especially when the coefficients in your polynomial are complex. By carefully substituting the appropriate values for \(a\), \(b\), and \(c\), you can find all possible solutions for \(x\). Once these values are calculated, the hard work pays off as you have solved for \(x\)!

Simplifying Equations

Simplifying equations is essential for obtaining clearer and more manageable forms, especially when they involve multiple expressions. By breaking down the problem step-by-step, you make the equation easier to solve and reduce the complexity involved.
In the given exercise, simplifying translates into operations like finding a common denominator, combining terms, and reducing the equation to its essential components. After simplifying, you engage in equation reduction processes, such as multiplying the entire equation by an appropriate term to eliminate fractions.
This results in an equation that is simpler to handle and solve, usually leaving you with a more "normal" algebraic or polynomial form. Once the equation is as simplified as possible, it's perfectly set up to apply solving methods such as the quadratic formula. Hence, often the most challenging part of solving complex quadratic equations is indeed the simplifying part, which clears the way for solution finding.