Solve (x+3)(2x+5)+30=(2x+3)(x+5): Step-by-Step

by Felix Dubois 47 views

Hey guys! Today, we're diving into a super interesting math problem that involves solving an algebraic equation. Equations like these might seem daunting at first, but trust me, with a step-by-step approach, they become much more manageable. We're going to break down the equation (x+3)(2x+5)+30=(2x+3)(x+5), find the value of 'x', and make sure you understand every little detail along the way. So, grab your pencils, and let's get started!

Understanding the Equation

Before we jump into solving, let's take a moment to really understand what this equation is telling us. The equation (x+3)(2x+5)+30=(2x+3)(x+5) is a polynomial equation, specifically a quadratic equation in disguise. Our mission is to find the value(s) of 'x' that make the left-hand side (LHS) equal to the right-hand side (RHS). To do this, we need to simplify both sides, get all the terms on one side, and then solve for 'x'.

When we look at the equation, we see products of binomials (expressions with two terms) and a constant. This means we'll be using the distributive property (also known as FOIL – First, Outer, Inner, Last) to expand these products. We also have a constant term (+30) on the left side that we need to consider. By systematically expanding and simplifying, we'll transform the equation into a more recognizable form, which will allow us to use standard methods for solving quadratic equations. Remember, the key here is to take it one step at a time and ensure accuracy in each step. Math is like building blocks; each step relies on the previous one, so getting the basics right is super important!

Step-by-Step Solution

Let's get our hands dirty and solve this equation step-by-step. We'll break it down into manageable chunks so you can follow along easily.

Step 1: Expand Both Sides of the Equation

The first thing we need to do is expand the products on both sides of the equation. This means applying the distributive property (FOIL method) to get rid of the parentheses.

On the left-hand side (LHS), we have (x+3)(2x+5)+30. Let's expand (x+3)(2x+5):

  • First: x * 2x = 2x²
  • Outer: x * 5 = 5x
  • Inner: 3 * 2x = 6x
  • Last: 3 * 5 = 15

So, (x+3)(2x+5) = 2x² + 5x + 6x + 15 = 2x² + 11x + 15. Don't forget to add the +30 from the original equation. Thus, the LHS becomes 2x² + 11x + 15 + 30 = 2x² + 11x + 45.

Now, let's expand the right-hand side (RHS), which is (2x+3)(x+5):

  • First: 2x * x = 2x²
  • Outer: 2x * 5 = 10x
  • Inner: 3 * x = 3x
  • Last: 3 * 5 = 15

So, (2x+3)(x+5) = 2x² + 10x + 3x + 15 = 2x² + 13x + 15.

After expanding both sides, our equation looks like this: 2x² + 11x + 45 = 2x² + 13x + 15.

Step 2: Simplify the Equation

The next step is to simplify the equation by moving all the terms to one side. This will help us set the equation to zero, which is a standard form for solving quadratic equations.

Subtract 2x² from both sides: 2x² + 11x + 45 - 2x² = 2x² + 13x + 15 - 2x², which simplifies to 11x + 45 = 13x + 15.

Now, subtract 11x from both sides: 11x + 45 - 11x = 13x + 15 - 11x, which simplifies to 45 = 2x + 15.

Next, subtract 15 from both sides: 45 - 15 = 2x + 15 - 15, which simplifies to 30 = 2x.

Step 3: Solve for x

Now we're in the home stretch! We have a simple equation: 30 = 2x. To solve for 'x', we need to isolate 'x' by dividing both sides by 2.

Divide both sides by 2: 30 / 2 = 2x / 2, which gives us 15 = x. So, x = 15.

Step 4: Verify the Solution

It's always a good idea to verify our solution by plugging the value of 'x' back into the original equation to make sure it holds true. This helps prevent errors and gives us confidence in our answer.

Original equation: (x+3)(2x+5)+30=(2x+3)(x+5)

Plug in x = 15: (15+3)(2(15)+5)+30=(2(15)+3)(15+5)

Simplify: (18)(30+5)+30=(30+3)(20)

Further simplify: (18)(35)+30=(33)(20)

Calculate: 630+30=660

And: 660=660

Since both sides of the equation are equal, our solution x = 15 is correct! Yay us!

Common Mistakes to Avoid

When solving equations like this, there are a few common pitfalls that students often stumble into. Being aware of these can help you avoid making the same mistakes.

One common mistake is incorrect application of the distributive property. Remember, you need to multiply each term inside the parentheses by the term outside. For example, in (x+3)(2x+5), you need to make sure you multiply x by both 2x and 5, and 3 by both 2x and 5. A small oversight here can throw off the entire solution.

Another mistake is forgetting to combine like terms properly. After expanding, you'll often have terms that can be combined, such as 5x and 6x. Make sure you add or subtract these terms correctly. A simple arithmetic error can lead to the wrong answer.

Sign errors are also common, especially when dealing with negative numbers. Pay close attention to the signs of the terms as you move them from one side of the equation to the other. Remember, when you move a term across the equals sign, you need to change its sign.

Finally, don't forget to verify your solution! Plugging your answer back into the original equation is a crucial step to ensure you haven't made any mistakes along the way. It's like a final check that can save you from submitting an incorrect answer.

Alternative Methods for Solving

While we solved this equation by expanding and simplifying, there are other methods we could have used. Let's briefly touch on a couple of alternative approaches.

One alternative is to look for patterns or shortcuts that might simplify the equation more quickly. However, in this particular case, expanding and simplifying is probably the most straightforward method. Sometimes, trying to find a shortcut can actually make the problem more complicated, especially if you're not entirely confident in your pattern-recognition skills.

Another approach, which is more applicable to quadratic equations in general, is to use the quadratic formula. However, in this case, after simplifying, we ended up with a linear equation (an equation where the highest power of 'x' is 1), so the quadratic formula wouldn't be necessary or particularly helpful. The quadratic formula is most useful when you have a quadratic equation in the form ax² + bx + c = 0 that can't be easily factored.

In summary, while there are often multiple ways to solve a math problem, choosing the most efficient and straightforward method is key. For this equation, expanding and simplifying was the way to go.

Real-World Applications

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