Solving Rational Equations A Step-by-Step Guide

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Hey everyone! Today, we're diving deep into the world of rational equations and tackling a specific problem. Rational equations, at first glance, might seem intimidating with their fractions and variables in the denominator. But don't worry, we'll break it down step by step, making it super easy to understand. We'll solve the equation 10xβˆ’4=xβˆ’7\frac{10}{x-4} = x - 7, finding all its solutions, and making sure we identify the smallest one first. So, grab your calculators, and let's get started!

Understanding Rational Equations

Before we jump into solving our equation, let's quickly recap what rational equations are. Rational equations are essentially equations that contain at least one fraction where the numerator and/or the denominator are polynomials. Our equation, 10xβˆ’4=xβˆ’7\frac{10}{x-4} = x - 7, fits this definition perfectly. The key thing to remember with these equations is that we need to be mindful of values that would make the denominator zero. Why? Because division by zero is a big no-no in mathematics! These values are called undefined points or restrictions and we'll need to watch out for them when we solve.

When dealing with rational equations, the primary goal is to eliminate the fractions. Think of it like this: fractions often complicate things, so getting rid of them simplifies the equation and makes it easier to solve. We do this by finding the least common denominator (LCD) of all the fractions in the equation and then multiplying both sides of the equation by the LCD. This clever trick cancels out the denominators, leaving us with a more manageable equation, usually a polynomial equation.

However, as we solve rational equations, there's a crucial step we must never skip: checking for extraneous solutions. Extraneous solutions are those values we get as solutions during the solving process but don't actually satisfy the original equation. They often arise when we square both sides of an equation or, as in this case, when we multiply by an expression containing a variable. These extraneous solutions might make the denominator of the original rational equation equal to zero, which, as we discussed, is undefined. So, always, always, always check your solutions in the original equation!

Step-by-Step Solution of the Equation

Okay, let's tackle our equation: 10xβˆ’4=xβˆ’7\frac{10}{x-4} = x - 7. We'll break it down into manageable steps, so you can follow along easily.

1. Identify the LCD

The first step is to identify the least common denominator (LCD). In our equation, we have only one fraction, 10xβˆ’4\frac{10}{x-4}. The denominator is (xβˆ’4)(x - 4). We can think of the right side of the equation, (xβˆ’7)(x - 7), as having a denominator of 1. So, the LCD is simply (xβˆ’4)(x - 4). Identifying the LCD is crucial because it's the key to eliminating the fraction and simplifying the equation.

2. Multiply Both Sides by the LCD

Next, we multiply both sides of the equation by the LCD, which is (xβˆ’4)(x - 4). This is where the magic happens! Multiplying both sides ensures that we maintain the equality while eliminating the fraction. Here's how it looks:

(xβˆ’4)β‹…10xβˆ’4=(xβˆ’4)β‹…(xβˆ’7)(x - 4) \cdot \frac{10}{x-4} = (x - 4) \cdot (x - 7)

On the left side, the (xβˆ’4)(x - 4) in the numerator and denominator cancel out, leaving us with just 10. On the right side, we need to distribute (xβˆ’4)(x - 4) across (xβˆ’7)(x - 7).

This simplifies to:

10=(xβˆ’4)(xβˆ’7)10 = (x - 4)(x - 7)

3. Expand and Simplify

Now, we need to expand the right side of the equation by multiplying the two binomials: (xβˆ’4)(xβˆ’7)(x - 4)(x - 7). We can use the FOIL method (First, Outer, Inner, Last) to ensure we multiply each term correctly.

(xβˆ’4)(xβˆ’7)=x2βˆ’7xβˆ’4x+28(x - 4)(x - 7) = x^2 - 7x - 4x + 28

Combining the like terms, βˆ’7x-7x and βˆ’4x-4x, we get:

x2βˆ’11x+28x^2 - 11x + 28

So, our equation now looks like this:

10=x2βˆ’11x+2810 = x^2 - 11x + 28

To solve for x, we need to set the equation to zero. Subtract 10 from both sides:

0=x2βˆ’11x+180 = x^2 - 11x + 18

4. Solve the Quadratic Equation

We now have a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0. There are several ways to solve quadratic equations, including factoring, using the quadratic formula, or completing the square. In this case, factoring is the easiest approach.

We need to find two numbers that multiply to 18 and add up to -11. Those numbers are -2 and -9. So, we can factor the quadratic equation as follows:

0=(xβˆ’2)(xβˆ’9)0 = (x - 2)(x - 9)

Setting each factor equal to zero gives us our potential solutions:

xβˆ’2=0x - 2 = 0 or xβˆ’9=0x - 9 = 0

Solving for x in each equation, we get:

x=2x = 2 or x=9x = 9

5. Check for Extraneous Solutions

This is a critical step! We need to check if our potential solutions, x=2x = 2 and x=9x = 9, actually work in the original equation, 10xβˆ’4=xβˆ’7\frac{10}{x-4} = x - 7. We need to make sure that these values don't make the denominator zero, and that they satisfy the equation.

Check x = 2:

Plug x=2x = 2 into the original equation:

102βˆ’4=2βˆ’7\frac{10}{2 - 4} = 2 - 7

10βˆ’2=βˆ’5\frac{10}{-2} = -5

βˆ’5=βˆ’5-5 = -5

This is true! So, x=2x = 2 is a valid solution.

Check x = 9:

Plug x=9x = 9 into the original equation:

109βˆ’4=9βˆ’7\frac{10}{9 - 4} = 9 - 7

105=2\frac{10}{5} = 2

2=22 = 2

This is also true! So, x=9x = 9 is also a valid solution.

6. Identify the Smallest Solution

We have two solutions: x=2x = 2 and x=9x = 9. The question asks for the smallest solution. Comparing the two values, it's clear that 2 is smaller than 9.

Therefore, the smallest solution is x=2x = 2.

Common Mistakes to Avoid

Solving rational equations can be tricky, and there are a few common pitfalls to watch out for. Let's discuss some of these, so you can avoid them.

Forgetting to Check for Extraneous Solutions

This is the biggest mistake people make! As we've emphasized, always check your solutions in the original equation. If you skip this step, you might end up including extraneous solutions in your answer, which are incorrect. Remember, these solutions arise because of the algebraic manipulations we perform, but they don't actually satisfy the original equation.

Incorrectly Identifying the LCD

The least common denominator (LCD) is the key to eliminating fractions. If you misidentify the LCD, you'll likely end up with a more complicated equation that's harder to solve. Make sure you correctly identify all the denominators in the equation and find their least common multiple.

Making Sign Errors

Sign errors are easy to make, especially when dealing with negative numbers and distributing terms. Pay close attention to the signs when you're expanding expressions and combining like terms. A small sign error can lead to a completely wrong answer.

Dividing Instead of Multiplying by the LCD

Remember, we eliminate the fractions by multiplying both sides of the equation by the LCD, not dividing. Dividing would just make the fractions even more complex. Multiplying cancels out the denominators, which is what we want.

Incorrectly Factoring Quadratic Equations

If you end up with a quadratic equation, make sure you factor it correctly. If you're not comfortable with factoring, you can always use the quadratic formula. But if you choose to factor, double-check your work to ensure you've found the correct factors.

Real-World Applications of Rational Equations

You might be wondering,