Solve 6a-9=15, 30-5a=20, 2(2a+1)=a+10: Step-by-Step
Hey guys! Today, we're diving deep into the world of algebraic equations. Don't worry if that sounds intimidating – we're going to break it down step-by-step, making it super easy to understand. We'll be tackling three equations: 6a-9=15, 30-5a=20, and 2(2a+1)=a+10. By the end of this guide, you'll not only know how to solve these specific problems but also grasp the fundamental principles behind solving a wide range of algebraic equations. So, grab your pencils, and let's get started!
Unlocking the Secrets of Algebraic Equations
Before we jump into the equations themselves, let's quickly recap what algebraic equations are all about. At their core, they are mathematical statements that show the equality between two expressions. These expressions often contain variables, which are just letters that represent unknown numbers. Our goal in solving an equation is to find the value of the variable that makes the equation true. This involves using various operations to isolate the variable on one side of the equation. Think of it like solving a puzzle – we need to carefully manipulate the pieces (numbers and operations) to reveal the hidden answer (the value of the variable).
In the context of our equations, the variable we're dealing with is 'a.' So, when we solve 6a-9=15, 30-5a=20, and 2(2a+1)=a+10, we're essentially finding the specific value of 'a' that satisfies each equation. This might involve adding or subtracting numbers from both sides, multiplying or dividing both sides by a constant, or even using the distributive property. The key is to perform the same operation on both sides to maintain the balance of the equation. Remember, algebra is all about maintaining equality. Whatever you do to one side, you must do to the other.
Now, let's talk about why understanding algebraic equations is so important. It's not just about getting good grades in math class! Algebraic equations are the foundation for many real-world applications, from calculating finances to designing buildings. They're used in physics, engineering, computer science, and countless other fields. By mastering the art of solving equations, you're equipping yourself with a powerful tool that can be applied to a wide range of problems. So, take your time, practice consistently, and don't be afraid to ask questions. The journey to mastering algebra is a rewarding one!
Solving 6a - 9 = 15: A Step-by-Step Approach
Let's kick things off with our first equation: 6a - 9 = 15. This equation involves a variable ('a') multiplied by a constant (6), and then a constant (-9) subtracted from the result. Our goal is to isolate 'a' on one side of the equation. To do this, we'll use the concept of inverse operations. Inverse operations are operations that undo each other. For example, addition and subtraction are inverse operations, and multiplication and division are inverse operations. We'll use these operations to strategically eliminate the numbers around 'a' until it stands alone.
The first step is to get rid of the -9. To do this, we'll add 9 to both sides of the equation. Remember, whatever we do to one side, we must do to the other! This maintains the balance of the equation and ensures that the equality remains true. So, we have:
6a - 9 + 9 = 15 + 9
Simplifying this, we get:
6a = 24
Now, 'a' is being multiplied by 6. To undo this multiplication, we'll divide both sides of the equation by 6. This is the inverse operation of multiplication. So, we have:
6a / 6 = 24 / 6
Simplifying this, we finally get:
a = 4
And there you have it! We've solved our first equation. The value of 'a' that satisfies the equation 6a - 9 = 15 is 4. But how can we be sure our answer is correct? The best way to check is to substitute the value of 'a' back into the original equation. If the equation holds true, then we know we've found the correct solution. Let's do that now:
6(4) - 9 = 15
24 - 9 = 15
15 = 15
As you can see, the equation holds true! This confirms that our solution, a = 4, is indeed correct. So, we've successfully navigated the first equation. You're doing great! Now, let's move on to the next challenge.
Tackling 30 - 5a = 20: A Different Perspective
Our second equation, 30 - 5a = 20, presents a slightly different scenario. Here, the variable 'a' is being multiplied by a negative number (-5) and then subtracted from a constant (30). While the principles remain the same, we need to be extra careful with the signs. The same rules apply, we want to isolate 'a' on one side of the equation. So, we'll use inverse operations to peel away the numbers surrounding 'a' step by step.
One approach is to first get rid of the 30. Since it's being added (or, more accurately, a positive 30 is present), we'll subtract 30 from both sides of the equation. Remember, balance is key! So, we have:
30 - 5a - 30 = 20 - 30
Simplifying, we get:
-5a = -10
Now, 'a' is being multiplied by -5. To undo this multiplication, we'll divide both sides of the equation by -5. Again, pay close attention to the signs! Dividing a negative number by a negative number results in a positive number. So, we have:
-5a / -5 = -10 / -5
Simplifying, we get:
a = 2
Great! We've solved our second equation. The value of 'a' that satisfies the equation 30 - 5a = 20 is 2. As with the previous equation, let's check our answer by substituting it back into the original equation:
30 - 5(2) = 20
30 - 10 = 20
20 = 20
The equation holds true, confirming that our solution, a = 2, is correct. See? You're becoming an equation-solving pro! This equation highlights the importance of paying attention to signs when working with algebraic equations. A small mistake with a sign can lead to a completely incorrect answer. So, always double-check your work and be mindful of the positive and negative numbers. Now, let's move on to our final equation, which involves the distributive property.
Mastering 2(2a + 1) = a + 10: Distributive Property in Action
Our third equation, 2(2a + 1) = a + 10, introduces a new element: the distributive property. The distributive property is a fundamental concept in algebra that allows us to simplify expressions where a number is multiplied by a sum or difference inside parentheses. In this case, we have 2 multiplied by the expression (2a + 1). The distributive property states that we must multiply the 2 by each term inside the parentheses. So, 2(2a + 1) becomes 2 * 2a + 2 * 1, which simplifies to 4a + 2. Let's apply this to our equation:
2(2a + 1) = a + 10
4a + 2 = a + 10
Now that we've applied the distributive property, we have a more manageable equation. Our goal is still to isolate 'a', but this time, 'a' appears on both sides of the equation. To solve this, we need to gather all the 'a' terms on one side and all the constant terms on the other side. Let's start by subtracting 'a' from both sides of the equation:
4a + 2 - a = a + 10 - a
Simplifying, we get:
3a + 2 = 10
Now, we have 'a' only on the left side of the equation. To isolate it further, we'll subtract 2 from both sides:
3a + 2 - 2 = 10 - 2
Simplifying, we get:
3a = 8
Finally, we'll divide both sides by 3 to solve for 'a':
3a / 3 = 8 / 3
Simplifying, we get:
a = 8/3
So, the value of 'a' that satisfies the equation 2(2a + 1) = a + 10 is 8/3. This is a fraction, which is perfectly acceptable as a solution. Let's check our answer by substituting it back into the original equation:
2(2(8/3) + 1) = 8/3 + 10
2(16/3 + 1) = 8/3 + 10
2(16/3 + 3/3) = 8/3 + 30/3
2(19/3) = 38/3
38/3 = 38/3
The equation holds true, confirming that our solution, a = 8/3, is correct. This equation demonstrates the importance of the distributive property and how to handle variables on both sides of the equation. You've successfully navigated a more complex equation, and that's something to be proud of!
Congratulations! You've Conquered Algebraic Equations!
Wow, guys! You've made it through three challenging algebraic equations. You've learned how to isolate variables, use inverse operations, apply the distributive property, and check your answers. You've also seen how algebraic equations are fundamental tools that can be applied to various real-world problems. Remember, the key to mastering algebra is practice, practice, practice! The more you work with equations, the more comfortable and confident you'll become. So, keep practicing, keep exploring, and never stop learning!
If you're feeling a bit overwhelmed, that's perfectly normal. Algebra can be challenging at first, but with consistent effort, you'll gradually build your skills and understanding. Don't be afraid to make mistakes – they're a natural part of the learning process. When you make a mistake, take the time to understand why you made it and how you can avoid it in the future. And most importantly, don't be afraid to ask for help! There are plenty of resources available, from teachers and tutors to online tutorials and study groups. Keep up the great work, and you'll be solving even the most complex algebraic equations in no time!
