Solving The Equation: 2x-2/x+3 + X+3/x-3 = 5

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Hey guys! Today, we're diving deep into the awesome world of algebra to tackle a pretty common, yet sometimes tricky, equation. You know, the kind that makes you stare at it for a sec and think, "Okay, how do I even start this?" We're going to break down how to solve the equation 2x-2/x+3 + x+3/x-3 =5. This isn't just about getting the right answer; it's about understanding the process and building those crucial algebra muscles. So, grab your thinking caps, maybe a snack, and let's get this done!

Understanding the Equation Structure

Understanding the Equation Structure

First off, let's just look at this equation: 2x-2/x+3 + x+3/x-3 =5. See those fractions? That's usually the first thing that makes people pause. We've got terms like 2x-2 divided by x+3, and then x+3 divided by x-3. The goal here, as with most algebraic equations, is to isolate the variable, which in this case is 'x'. But we can't just add or subtract terms all willy-nilly because of those denominators. We need a strategy. The main strategy when dealing with equations involving fractions is to get rid of the denominators. How do we do that? By finding a common denominator. Think of it like this: if you have 1/2 of a pizza and your friend has 1/3 of a pizza, you can't just add them and say you have 2/5 of a pizza. You need to cut them into smaller, equal slices (find a common denominator) to actually combine them properly. In our equation, the denominators are (x+3) and (x-3). To find a common denominator, we multiply these two together: (x+3)(x-3). This is going to be our magic wand to clear out those pesky fractions.

Finding the Common Denominator and Clearing Fractions

So, we've identified our common denominator as (x+3)(x-3). Now, the cool part: we're going to multiply every single term in the equation by this common denominator. Don't worry, it sounds more complicated than it is! Let's take it step-by-step. Remember, whatever you do to one side of the equation, you must do to the other side to keep it balanced.

Our original equation is:

(2x-2)/(x+3) + (x+3)/(x-3) = 5

Now, let's multiply each term by (x+3)(x-3):

[(x+3)(x-3) * (2x-2)/(x+3)] + [(x+3)(x-3) * (x+3)/(x-3)] = [5 * (x+3)(x-3)]

Look at what happens! In the first term, the (x+3) in the numerator cancels out the (x+3) in the denominator. So, we're left with (x-3) * (2x-2).

In the second term, the (x-3) cancels out. We're left with (x+3) * (x+3).

And on the right side, we have 5 * (x+3)(x-3).

Our equation now looks much simpler:

(x-3)(2x-2) + (x+3)(x+3) = 5(x+3)(x-3)

This is a huge win, guys! We've successfully eliminated the fractions, which were the main hurdle. The next steps involve expanding these products and then simplifying the resulting polynomial equation.

Expanding and Simplifying the Equation

Alright, we've cleared the fractions and have this beast: (x-3)(2x-2) + (x+3)(x+3) = 5(x+3)(x-3). Now, it's time to break out the distributive property (or FOIL, if that's your jam!). Let's expand each part.

First part: (x-3)(2x-2)

  • x * 2x = 2x^2
  • x * -2 = -2x
  • -3 * 2x = -6x
  • -3 * -2 = 6

So, (x-3)(2x-2) = 2x^2 - 2x - 6x + 6, which simplifies to 2x^2 - 8x + 6.

Second part: (x+3)(x+3) (This is a perfect square trinomial, (a+b)^2 = a^2 + 2ab + b^2)

  • x * x = x^2

  • x * 3 = 3x

  • 3 * x = 3x

  • 3 * 3 = 9

So, (x+3)(x+3) = x^2 + 3x + 3x + 9, which simplifies to x^2 + 6x + 9.

Third part: 5(x+3)(x-3)

First, let's handle (x+3)(x-3). This is a difference of squares, (a+b)(a-b) = a^2 - b^2.

  • x * x = x^2

  • x * -3 = -3x

  • 3 * x = 3x

  • 3 * -3 = -9

So, (x+3)(x-3) = x^2 - 3x + 3x - 9, which simplifies to x^2 - 9.

Now, multiply this by 5: 5 * (x^2 - 9) = **5x^2 - 45**.

Now, let's substitute these expanded forms back into our equation:

(2x^2 - 8x + 6) + (x^2 + 6x + 9) = 5x^2 - 45

We're getting closer, guys! The next step is to combine like terms on the left side of the equation.

Combine x^2 terms: 2x^2 + x^2 = 3x^2 Combine x terms: -8x + 6x = -2x Combine constant terms: 6 + 9 = 15

So, the left side simplifies to 3x^2 - 2x + 15.

Our equation now reads:

3x^2 - 2x + 15 = 5x^2 - 45

Rearranging into a Quadratic Equation

We're almost there! We've simplified the equation to 3x^2 - 2x + 15 = 5x^2 - 45. This looks like a quadratic equation, which is an equation of the form ax^2 + bx + c = 0. To get it into that standard form, we need to move all the terms to one side, making sure the x^2 term is positive (though it's not strictly necessary, it often makes things easier).

Let's move all terms from the left side to the right side. We can do this by subtracting 3x^2, adding 2x, and subtracting 15 from both sides of the equation.

Starting equation: 3x^2 - 2x + 15 = 5x^2 - 45

Subtract 3x^2 from both sides: -2x + 15 = 5x^2 - 3x^2 - 45 -2x + 15 = 2x^2 - 45

Add 2x to both sides: 15 = 2x^2 + 2x - 45

Subtract 15 from both sides: 0 = 2x^2 + 2x - 45 - 15 0 = 2x^2 + 2x - 60

So, our quadratic equation in standard form is 2x^2 + 2x - 60 = 0.

Before we proceed, we can make this equation even simpler! Notice that all the coefficients (2, 2, and -60) are even numbers. This means we can divide the entire equation by 2 to get smaller, easier-to-work-with numbers.

Divide by 2: (2x^2)/2 + (2x)/2 - 60/2 = 0/2

This gives us the simplified quadratic equation: x^2 + x - 30 = 0.

This is fantastic! Working with smaller numbers drastically reduces the chance of silly calculation errors. Now, we just need to find the roots (the values of x that make this equation true) of this simplified quadratic.

Solving the Quadratic Equation: Factoring

We've reached the point where we need to solve the quadratic equation x^2 + x - 30 = 0. There are a few ways to do this: factoring, using the quadratic formula, or completing the square. Factoring is often the quickest and most elegant method if the quadratic is factorable. Let's give it a shot!

We're looking for two numbers that:

  1. Multiply to give us the constant term (-30).
  2. Add up to give us the coefficient of the x term (+1).

Let's think about pairs of numbers that multiply to -30:

  • 1 and -30 (sum = -29)
  • -1 and 30 (sum = 29)
  • 2 and -15 (sum = -13)
  • -2 and 15 (sum = 13)
  • 3 and -10 (sum = -7)
  • -3 and 10 (sum = 7)
  • 5 and -6 (sum = -1)
  • -5 and 6 (sum = 1)

Bingo! The pair -5 and 6 works perfectly. They multiply to -30 and add up to +1.

Now we can factor the quadratic x^2 + x - 30 into (x - 5)(x + 6).

So, our equation becomes: (x - 5)(x + 6) = 0.

For this product of two terms to equal zero, at least one of the terms must be zero. This gives us two possible scenarios:

  1. x - 5 = 0 If we add 5 to both sides, we get x = 5.

  2. x + 6 = 0 If we subtract 6 from both sides, we get x = -6.

So, the potential solutions are x = 5 and x = -6.

Checking for Extraneous Solutions

This is a super important step, especially when you started with fractions in your equation. Remember those denominators? (x+3) and (x-3)? We can't have them equal to zero, because division by zero is undefined. So, we need to check if our solutions make any of the original denominators zero.

Let's check our solutions:

For x = 5:

  • Denominator 1: x + 3 = 5 + 3 = 8 (Not zero, okay!)
  • Denominator 2: x - 3 = 5 - 3 = 2 (Not zero, okay!)

Since neither denominator is zero when x = 5, this is a valid solution.

For x = -6:

  • Denominator 1: x + 3 = -6 + 3 = -3 (Not zero, okay!)
  • Denominator 2: x - 3 = -6 - 3 = -9 (Not zero, okay!)

Since neither denominator is zero when x = -6, this is also a valid solution.

Both x = 5 and x = -6 are valid roots for the equation. You guys did it! We successfully navigated through fractions, expanded expressions, simplified a quadratic, and found the solutions.

Conclusion: Mastering Algebraic Equations

So there you have it, team! We've thoroughly solved the equation 2x-2/x+3 + x+3/x-3 =5. We went from a complex fractional equation to a simple quadratic by finding a common denominator, clearing fractions, expanding terms, and simplifying. Then, we used factoring to find the roots, and finally, we performed a crucial check for extraneous solutions. This process is fundamental in algebra, and practicing these steps will make you a master of solving rational equations and quadratic equations. Remember, the key is to take it step-by-step, stay organized, and not be afraid of the fractions or the algebra involved. Keep practicing, and you'll conquer any equation that comes your way!