Solving Rational Equations: Step-by-Step Guide

Hey guys! Today, we're diving deep into the world of rational equations. Solving these equations can seem tricky, but with a step-by-step approach, you'll master them in no time. Let's break down each equation and solve them together. Ready? Let's get started!

a) Solving $\frac{4}{x-2} + \frac{6}{x+3} = \frac{1}{3}$

Alright, let's tackle this equation. Our goal is to find the value(s) of x that satisfy the equation. Here’s how we can do it:

1. Find the Least Common Denominator (LCD):

   The denominators are (x - 2), (x + 3), and 3. So, the LCD is 3(x - 2)(x + 3).

2. Multiply Each Term by the LCD:

   We're going to multiply every single term in the equation by our LCD. This will eliminate the fractions, making the equation much easier to handle. Trust me, this is a game-changer!

   • $\frac{4}{x-2} * 3(x-2)(x+3) = 12(x+3)$
   • $\frac{6}{x+3} * 3(x-2)(x+3) = 18(x-2)$
   • $\frac{1}{3} * 3(x-2)(x+3) = (x-2)(x+3)$

   So, our equation becomes: 12(x + 3) + 18(x - 2) = (x - 2)(x + 3)

3. Expand and Simplify:

   Now, let's expand those expressions:

   • 12x + 36 + 18x - 36 = x² + 3x - 2x - 6
   • Combine like terms: 30x = x² + x - 6

4. Rearrange into a Quadratic Equation:

   Let’s move everything to one side to form a quadratic equation:

   • x² + x - 30x - 6 = 0
   • x² - 29x - 6 = 0

5. Solve the Quadratic Equation:

   We can use the quadratic formula to solve for x: $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

   Here, a = 1, b = -29, and c = -6.

   • x = $\frac{29 ± \sqrt{(-29)^2 - 4(1)(-6)}}{2(1)}$
   • x = $\frac{29 ± \sqrt{841 + 24}}{2}$
   • x = $\frac{29 ± \sqrt{865}}{2}$

   So, the solutions are: x = $\frac{29 + \sqrt{865}}{2}$ and x = $\frac{29 - \sqrt{865}}{2}$

6. Check for Extraneous Solutions:

   Make sure to plug these values back into the original equation to ensure they don't make any denominator zero. In this case, both solutions are valid.

b) Solving $\frac{x+1}{3x-2} + \frac{2x+1}{x-5} = \frac{3}{2}$

Next up, this equation looks a bit more complex, but don't worry, we'll break it down. Let’s find the values of x that make this equation true.

1. Find the Least Common Denominator (LCD):

   The denominators are (3x - 2), (x - 5), and 2. So, the LCD is 2(3x - 2)(x - 5).

2. Multiply Each Term by the LCD:

   Multiply each term by 2(3x - 2)(x - 5):

   • $\frac{x+1}{3x-2} * 2(3x-2)(x-5) = 2(x+1)(x-5)$
   • $\frac{2x+1}{x-5} * 2(3x-2)(x-5) = 2(2x+1)(3x-2)$
   • $\frac{3}{2} * 2(3x-2)(x-5) = 3(3x-2)(x-5)$

   So, our equation becomes: 2(x + 1)(x - 5) + 2(2x + 1)(3x - 2) = 3(3x - 2)(x - 5)

3. Expand and Simplify:

   Expand those expressions:

   • 2(x² - 5x + x - 5) + 2(6x² - 4x + 3x - 2) = 3(3x² - 15x - 2x + 10)
   • 2(x² - 4x - 5) + 2(6x² - x - 2) = 3(3x² - 17x + 10)
   • 2x² - 8x - 10 + 12x² - 2x - 4 = 9x² - 51x + 30

   Combine like terms: 14x² - 10x - 14 = 9x² - 51x + 30

4. Rearrange into a Quadratic Equation:

   Move everything to one side:

   • 14x² - 9x² - 10x + 51x - 14 - 30 = 0
   • 5x² + 41x - 44 = 0

5. Solve the Quadratic Equation:

   Use the quadratic formula again: $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

   Here, a = 5, b = 41, and c = -44.

   • x = $\frac{-41 ± \sqrt{(41)^2 - 4(5)(-44)}}{2(5)}$
   • x = $\frac{-41 ± \sqrt{1681 + 880}}{10}$
   • x = $\frac{-41 ± \sqrt{2561}}{10}$

   So, the solutions are: x = $\frac{-41 + \sqrt{2561}}{10}$ and x = $\frac{-41 - \sqrt{2561}}{10}$

6. Check for Extraneous Solutions:

   Plug these values back into the original equation to make sure they are valid. Both solutions should work here as well.

c) Solving $\frac{4x-2}{x^2+2x+1} + \frac{3}{2} = \frac{x-5}{x+1}$

Now, let’s solve this equation. It includes a quadratic term in the denominator, but don't let that intimidate you! Let’s find the values of x that satisfy this equation.

1. Factor the Denominators:

   Notice that x² + 2x + 1 = (x + 1)². So, we can rewrite the equation as:

   • $\frac{4x-2}{(x+1)^2} + \frac{3}{2} = \frac{x-5}{x+1}$

2. Find the Least Common Denominator (LCD):

   The denominators are (x + 1)², 2, and (x + 1). So, the LCD is 2(x + 1)².

3. Multiply Each Term by the LCD:

   Multiply each term by 2(x + 1)²:

   • $\frac{4x-2}{(x+1)^2} * 2(x+1)^2 = 2(4x-2)$
   • $\frac{3}{2} * 2(x+1)^2 = 3(x+1)^2$
   • $\frac{x-5}{x+1} * 2(x+1)^2 = 2(x-5)(x+1)$

   So, the equation becomes: 2(4x - 2) + 3(x + 1)² = 2(x - 5)(x + 1)

4. Expand and Simplify:

   Expand those expressions:

   • 8x - 4 + 3(x² + 2x + 1) = 2(x² + x - 5x - 5)
   • 8x - 4 + 3x² + 6x + 3 = 2(x² - 4x - 5)
   • 3x² + 14x - 1 = 2x² - 8x - 10

5. Rearrange into a Quadratic Equation:

   Move everything to one side:

   • 3x² - 2x² + 14x + 8x - 1 + 10 = 0
   • x² + 22x + 9 = 0

6. Solve the Quadratic Equation:

   Use the quadratic formula: $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

   Here, a = 1, b = 22, and c = 9.

   • x = $\frac{-22 ± \sqrt{(22)^2 - 4(1)(9)}}{2(1)}$
   • x = $\frac{-22 ± \sqrt{484 - 36}}{2}$
   • x = $\frac{-22 ± \sqrt{448}}{2}$
   • x = $\frac{-22 ± 8\sqrt{7}}{2}$
   • x = -11 ± 4\sqrt{7}$

   So, the solutions are: x = -11 + 4\sqrt{7} and x = -11 - 4\sqrt{7}

7. Check for Extraneous Solutions:

   Make sure to plug these values back into the original equation. We need to ensure that x ≠ -1, which is satisfied by both solutions.

d) Solving $\frac{3}{x-1} - \frac{6}{x+4} = \frac{1}{4}$

Last but not least, let's get through this final equation. By now, you're practically a pro! Let’s find the values of x that make this equation hold true.

1. Find the Least Common Denominator (LCD):

   The denominators are (x - 1), (x + 4), and 4. So, the LCD is 4(x - 1)(x + 4).

2. Multiply Each Term by the LCD:

   Multiply each term by 4(x - 1)(x + 4):

   • $\frac{3}{x-1} * 4(x-1)(x+4) = 12(x+4)$
   • $\frac{6}{x+4} * 4(x-1)(x+4) = 24(x-1)$
   • $\frac{1}{4} * 4(x-1)(x+4) = (x-1)(x+4)$

   So, the equation becomes: 12(x + 4) - 24(x - 1) = (x - 1)(x + 4)

3. Expand and Simplify:

   Expand those expressions:

   • 12x + 48 - 24x + 24 = x² + 4x - x - 4
   • -12x + 72 = x² + 3x - 4

4. Rearrange into a Quadratic Equation:

   Move everything to one side:

   • x² + 3x + 12x - 4 - 72 = 0
   • x² + 15x - 76 = 0

5. Solve the Quadratic Equation:

   Use the quadratic formula: $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

   Here, a = 1, b = 15, and c = -76.

   • x = $\frac{-15 ± \sqrt{(15)^2 - 4(1)(-76)}}{2(1)}$
   • x = $\frac{-15 ± \sqrt{225 + 304}}{2}$
   • x = $\frac{-15 ± \sqrt{529}}{2}$
   • x = $\frac{-15 ± 23}{2}$

   So, the solutions are: x = $\frac{-15 + 23}{2} = 4$ and x = $\frac{-15 - 23}{2} = -19$

6. Check for Extraneous Solutions:

   Plug these values back into the original equation. We need to ensure that x ≠ 1 and x ≠ -4, which is satisfied by both solutions.

Conclusion

Alright, guys! We've walked through solving four rational equations step by step. Remember, the key is to find the LCD, multiply each term by it, simplify, and solve the resulting equation. And don’t forget to check for extraneous solutions! Keep practicing, and you’ll become a pro at solving these equations. You've got this! High five! 🖐️