Problem 26
Question
Solve each rational equation. $$\frac{x-1}{2 x+3}=\frac{6}{x-2}$$
Step-by-Step Solution
Verified Answer
The solutions of the equation are \(x = 16\) and \(x = -1\).
1Step 1: Identify Restriction
The rational expression is undefined for certain values of x, namely when the denominator equals zero. For the first rational expression, the denominator is \(2x+3\), so \(2x+3=0\) gives \(x=-\frac{3}{2}\). For the second rational expression, the denominator is \(x-2\), so \(x-2=0\) gives \(x=2\). These two values are restrictions that we need to avoid.
2Step 2: Cross-multiplication
Now, using cross-multiplication to simplify the equation, we get \((x-1)(x-2) = 2(x+3)6\). After cross multiplication, the denominators are gone, and we have a simpler, more straightforward equation to solve.
3Step 3: Expand and Simplify
Expanding both sides gives \(x^2-2x-x+2 = 12x+18\), which simplifies to \(x^2-3x+2=12x+18\). Reordering the terms gives \(x^2-15x-16 = 0\).
4Step 4: Solve the Quadratic Equation
Now, we can solve the quadratic equation \(x^2-15x-16 = 0\). We factorize to obtain \((x-16)(x+1)= 0 \). This gives us two potential solutions: \(x = 16\) and \(x = -1\)
5Step 5: Eliminate Restrictions
When we check these potential solutions against our original restrictions, we see that neither solution is among the restrictions. So, both \(x = 16\) and \(x = -1\) are valid solutions.
Key Concepts
Cross-MultiplicationQuadratic EquationsFactoring Polynomials
Cross-Multiplication
Cross-multiplication is a vital technique used in solving equations involving two rational expressions. In essence, it's a method to eliminate the denominators of the fractions by multiplying each side of the equation by the denominators of the opposite side. For the equation \(\frac{x-1}{2x+3} = \frac{6}{x-2}\), we multiply each side by the denominators of the fractions from the other side, which gives us \( (x-1)(x-2) = 6(2x+3) \). When executed correctly, cross-multiplication simplifies the problem to a polynomial equation, making it easier to solve.
To improve understanding while using this technique, it's recommended to:
To improve understanding while using this technique, it's recommended to:
- Clearly display each step of the cross-multiplication process.
- Ensure the multiplication is done correctly, as a mistake here can lead to incorrect solutions.
- Check your results by plugging them back into the original equation to verify if both sides are equal, confirming the correctness of your solution.
Quadratic Equations
Quadratic equations take the form \(ax^2 + bx + c = 0\), where 'a', 'b', and 'c' are constants, and 'x' represents the unknown variable we're trying to find. After cross-multiplication, we often end up with a quadratic equation to solve, as seen with \(x^2 - 15x - 16 = 0\). There are several methods to solve quadratic equations, including factoring, using the quadratic formula, completing the square, or graphing.
The simplest way to solve a quadratic equation is by factoring, if possible. That involves breaking down the quadratic into two binomial expressions set to zero, like \( (x - p)(x - q) = 0 \), where 'p' and 'q' would give us our solutions when set equal to zero. The advice to facilitate comprehension here might be to:
The simplest way to solve a quadratic equation is by factoring, if possible. That involves breaking down the quadratic into two binomial expressions set to zero, like \( (x - p)(x - q) = 0 \), where 'p' and 'q' would give us our solutions when set equal to zero. The advice to facilitate comprehension here might be to:
- Always start by attempting to factor if the quadratic appears factorable.
- Lay out clear steps in checking for possible factors of 'c' that sum up to 'b' in the quadratic equation.
- Verify the solutions by plugging them back into the original equation to ascertain their validity.
Factoring Polynomials
Factoring polynomials involves expressing the polynomial as a product of its factors, much like factoring a number into its prime factors. Factoring is a key step in solving polynomials, including quadratic equations. When we factor the quadratic equation \(x^2 - 15x - 16 = 0\), we find the factors \(x - 16\) and \(x + 1\), leading to the solutions \(x = 16\) and \(x = -1\).
To enhance the learning experience around factoring, it's important to:
To enhance the learning experience around factoring, it's important to:
- Highlight the importance of looking for a greatest common factor (GCF) before attempting other factoring techniques.
- Use the 'FOIL' method (First, Outer, Inner, Last) as a check to ensure that the factored form is correct when multiplied out.
- Encourage practice with different types of polynomials to build confidence and skill in factoring.
Other exercises in this chapter
Problem 26
Simplify complex rational expression by the method of your choice. \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}\)
View solution Problem 26
Multiply as indicated. $$\frac{8 x+2}{x^{2}-9} \cdot \frac{3-x}{4 x^{2}+x}$$
View solution Problem 27
Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-1}{6}+\frac{x+2}{3}$$
View solution Problem 27
What does it mean if two quantities vary inversely?
View solution