Solving Rational Equations

Rational equations are invaluable tools that have a significant impact on various aspects of our daily lives, from personal finances to complex engineering tasks. They enable us to make informed decisions, optimize processes, and solve real-world problems. By understanding and mastering rational equations, we gain access to a versatile mathematical resource that can be employed in a multitude of applications, such as calculating speeds, determining average costs, and modeling various phenomena.

A rational equation is an equation that involves one or more rational expressions. It is important to remember that a rational expression is an algebraic expression in which a rational number, or a fraction, is the denominator. These expressions can represent relationships or proportions between quantities, and when properly manipulated, they can provide valuable insights into the situations they represent.

Consider the following example of a rational expression:

$\frac{7}{x + 2} = \frac{x}{2} + 2$

This expression may appear simple at first glance, but it holds the key to solving a wide range of problems, from basic arithmetic to intricate calculations in various fields such as economics, physics, and engineering. By learning the methods for solving rational expressions, we unlock the potential to comprehend and analyze complex relationships in the world around us, thereby enhancing our capacity to make informed decisions and solve practical problems.

Solving rational expressions using cross-multiplication

Cross-multiplying is a simple way to solve rational expressions when there is a single rational expression on each side of the equation. Some textbooks call this the means/extremes property.

$\frac{a}{b} = \frac{c}{d}$

When we cross-multiply this, we see that $a \times d = b \times c$ . Let's look at this in action.

Example 1

Solve using cross-multiplication: $\frac{4}{x + 2} = \frac{8}{4 x - 2}$

The first step is to cross-multiply. We multiply $4 \times (4 x - 2)$ and $8 \times (x + 2)$ .

$4 \times (4 x - 2) = 8 \times (x + 2)$

We will use the distributive property to perform these multiplication problems.

$16 x - 8 = 8 x + 16$

Then subtract 8x from each side.

$(16 x - 8 x - 8) = (8 x + 16 - 8 x)$

Simplify by performing the subtraction.

$8 x - 8 = 16$

Then add 8 to each side.

$8 x - 8 + 8 = 16 + 8$

Simplify by performing the addition.

$8 x = 24$

Divide each side by 8.

$\frac{8 x}{8} = \frac{24}{8}$

Simplify by performing the division.

$x = 3$

Example 2

Solve using cross-multiplication: $\frac{8}{3 x - 2} = \frac{2}{x - 1}$

To cross-multiply, we should multiply 8 by $(x - 1)$ and 2 by $(3 x - 2)$ .

$8 (x - 1) = 2 (3 x - 2)$

We will use the distributive property to perform the calculations.

$8 x - 8 = 6 x - 4$

To simplify, subtract 6x from each side:

$8 x - 8 - 6 x = 6 x - 4 - 6 x$

Simplify by performing the subtraction:

$2 x - 8 = - 4$

Next, add 8 to each side:

$2 x - 8 + 8 = - 4 + 8$

Simplify by performing the addition:

$2 x = 4$

Finally, divide each side by 2:

$\frac{2 x}{2} = \frac{4}{2}$

When we perform the division, we find the solution:

$x = 2$

Solving rational equations using least common denominators (LCD)

If we have a rational equation that isn't in the form that supports cross-multiplication, we can still solve it using the least common denominator method. The least common denominator is the easiest common denominator to use in solving these types of problems. To find the LCD, we must factor each expression and multiply all of the unique factors to reveal the LCD of two or more rational expressions.

Then each side of the equation is multiplied by the least common denominator to solve the rational equation. Let's see it in action to understand this method better.

Example 3

Solve $(\frac{3}{x}) + (\frac{5}{4}) = (\frac{8}{x})$ using the least common denominator method.

The denominators here are x, 4, and x again. We will find the least common denominator by multiplying 4 and x, which is 4x. Next, we will multiply the LCD by the equation on each side to find the solution.

$4 x \times (\frac{3}{x} + \frac{5}{4}) = 4 x * (\frac{8}{x})$

$4 x \times \frac{3}{x} = 4 \times 3$

$4 x \times \frac{5}{4} = 5 x$

$4 x \times \frac{8}{x} = 4 \times 8$

So now we have

$12 + 5 x = 32$

Next, we will simplify further by subtracting 12 from each side.

$12 + 5 x - 12 = 32 - 12$

Perform the subtraction.

$5 x = 20$

Finally, divide each side by 5.

$x = 4$

We can check our solution by plugging 4 into the variables in the original equation.

$(\frac{3}{4}) + (\frac{5}{4}) = (\frac{8}{4})$

This is a true statement, so we have found the correct solution.

Example 4:

Find all real solutions to the rational equation: $\frac{3 x + 2}{x^{2} - x} = \frac{x + 4}{x^{2} + x}$

Step 1: Find a common denominator.

In this case, the common denominator is $(x^{2} - x) (x^{2} + x)$ . We'll multiply both sides of the equation by the common denominator:

$(\frac{3 x + 2}{x^{2} - x}) \times (x^{2} - x) \times (x^{2} + x) = (\frac{x + 4}{x^{2} + x}) \times (x^{2} - x) \times (x^{2} + x)$

Step 2: Simplify.

The denominators will cancel out on both sides:

$(3 x + 2) (x^{2} + x) = (x + 4) (x^{2} - x)$

Step 3: Apply the distributive property.

$3 x^{3} + 3 x^{2} + 2 x^{2} + 2 x = x^{3} - x^{2} + 4 x^{2} - 4 x$

Step 4: Combine like terms.

$3 x^{3} + 5 x^{2} + 2 x = x^{3} + 3 x^{2} - 4 x$

Step 5: Subtract the right side of the equation from the left side.

$2 x^{3} + 2 x^{2} + 6 x = 0$

Step 6: Factor out the greatest common divisor (GCD) from the left side of the equation.

$2 x (x^{2} + x + 3) = 0$

Step 7: Set each factor to zero and solve for x.

$2 x = 0 \geq x = 0$

$x^{2} + x + 3 = 0$

The quadratic equation $x^{2} + x + 3 = 0$ doesn't have real solutions since its discriminant is negative:

$b^{2} - 4 a c = 1 - 4 (3) = - 11$

Topics related to the Solving Rational Equations

Simplifying Rational Expressions

Dividing Rational Expressions

Adding and Subtracting Rational Expressions with Like Denominators

Flashcards covering the Solving Rational Equations

Precalculus Flashcards

CLEP Precalculus Flashcards

Practice tests covering the Solving Rational Equations

Precalculus Diagnostic Tests

Get help learning about solving rational equations

Tutoring is an excellent way for your student to learn how to solve rational equations outside of the classroom. Sometimes there's not enough time in the classroom to completely comprehend the processes involved in solving rational equations, and if this is the case for your student, you can't do better than having them study with a private tutor.

A private tutor will work at your student's pace through problems until your student demonstrates a thorough understanding of the concepts and processes. A tutor can also discover your student's learning style and customize their tutoring methods to complement it. Whether your student learns best visually, audibly, or kinesthetically, a tutor can accommodate them.

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