Solving Algebraic Equations Step By Step R + 10 + 13 = 15

Hey guys! Let's dive into the fascinating world of algebra and tackle some equations together. In this guide, we'll break down how to solve for 'r' in a series of equations. We'll start with simpler equations and gradually move to more complex ones, ensuring you grasp each step along the way. Whether you're a student brushing up on your algebra skills or just curious about problem-solving, this article is for you. So, grab your pencil and paper, and let's get started!

Unpacking the Basics of Algebraic Equations

Before we jump into solving specific equations, it's crucial to understand the fundamental principles of algebra. Algebra, at its core, is about finding the unknown. We use symbols, often letters like 'r', to represent these unknowns. An equation is a statement that two expressions are equal. Our goal is to isolate the variable (in this case, 'r') on one side of the equation to determine its value. Think of an equation like a balance scale; whatever you do to one side, you must do to the other to maintain the balance. This principle is the cornerstone of solving algebraic equations.

When dealing with equations, we use inverse operations to undo operations that are attached to the variable. For example, if 'r' is being added to a number, we subtract that number from both sides of the equation. If 'r' is being multiplied by a number, we divide both sides by that number. These inverse operations help us peel away the layers surrounding 'r' until it stands alone, revealing its value. Remember, consistency is key! Always perform the same operation on both sides of the equation to keep it balanced and accurate. So, with these basics in mind, let's begin our journey of solving equations, starting with the first one on our list.

Solving the Equation: r + 10 + 13 = 15

Let's kick things off with our first equation: r + 10 + 13 = 15. The aim here is to isolate 'r' on one side of the equation. To do this, we need to simplify the equation step by step. First, let’s combine the constants on the left side. We have 10 and 13, which add up to 23. So, the equation now looks like this: r + 23 = 15.

Now, to get 'r' by itself, we need to undo the addition of 23. The inverse operation of addition is subtraction. Therefore, we will subtract 23 from both sides of the equation. This gives us: r + 23 - 23 = 15 - 23. On the left side, +23 and -23 cancel each other out, leaving just 'r'. On the right side, 15 - 23 equals -8. So, the solution to our equation is r = -8.

To ensure our solution is correct, it’s always a good idea to check our work. We can do this by substituting -8 back into the original equation in place of 'r'. So, the equation becomes: -8 + 10 + 13 = 15. Simplifying the left side, we get -8 + 23 = 15, which is indeed true. This confirms that our solution, r = -8, is correct. So, we've successfully solved our first equation! Let's move on to the next one, where we'll continue to apply these fundamental principles of algebra.

Tackling the Equation: r + 3 = 28

Moving on to our second equation, we have r + 3 = 28. This one is quite straightforward and gives us a great opportunity to reinforce the principles we discussed earlier. Our mission, as always, is to isolate 'r' on one side of the equation. Looking at the equation, we can see that 'r' is being added to 3. To undo this addition, we need to perform the inverse operation, which is subtraction.

Therefore, we will subtract 3 from both sides of the equation. This ensures that we maintain the balance of the equation, keeping both sides equal. Subtracting 3 from both sides gives us: r + 3 - 3 = 28 - 3. On the left side, the +3 and -3 cancel each other out, leaving us with just 'r'. On the right side, 28 - 3 equals 25. So, the solution to this equation is r = 25.

Just like before, it's crucial to verify our solution to ensure its accuracy. We can do this by substituting 25 back into the original equation in place of 'r'. This gives us: 25 + 3 = 28. Simplifying the left side, we get 28, which is indeed equal to the right side. This confirms that our solution, r = 25, is correct. We're making great progress in our equation-solving journey! Now, let's step up the complexity a bit and tackle our third equation, which involves a slight twist with multiplication.

Deciphering the Equation: 8r - 1 = 7

Now let's tackle our third equation: 8r - 1 = 7. This equation introduces a new element: a coefficient (the number 8) multiplied by our variable 'r'. Don't worry, though! We'll break it down step by step. Remember our goal: to isolate 'r' on one side of the equation.

Looking at the equation, we see that 'r' is being multiplied by 8 and then has 1 subtracted from it. To isolate 'r', we need to reverse these operations in the correct order. We start by undoing the subtraction. The inverse operation of subtraction is addition, so we'll add 1 to both sides of the equation. This gives us: 8r - 1 + 1 = 7 + 1. On the left side, -1 and +1 cancel each other out, leaving us with 8r. On the right side, 7 + 1 equals 8. So, the equation now looks like this: 8r = 8.

Next, we need to undo the multiplication. 'r' is being multiplied by 8, so we'll perform the inverse operation, which is division. We'll divide both sides of the equation by 8. This gives us: 8r / 8 = 8 / 8. On the left side, the 8s cancel each other out, leaving us with just 'r'. On the right side, 8 / 8 equals 1. Therefore, the solution to this equation is r = 1.

As always, let's verify our solution. We substitute 1 back into the original equation: 8(1) - 1 = 7. Simplifying the left side, we get 8 - 1 = 7, which is true. This confirms that our solution, r = 1, is correct. We're getting the hang of this! Now, let's move on to our final equation, where we'll apply all the skills we've learned so far.

Conquering the Equation: 3r + 5 = 10

Finally, let's dive into our last equation: 3r + 5 = 10. This equation combines both multiplication and addition, giving us a chance to flex our algebra muscles. Remember, our ultimate goal is to isolate 'r' on one side of the equation, and we'll do this by carefully reversing the operations.

Looking at the equation, we see that 'r' is being multiplied by 3 and then has 5 added to it. To isolate 'r', we need to undo these operations in the reverse order. We start by addressing the addition. The inverse operation of addition is subtraction, so we'll subtract 5 from both sides of the equation. This gives us: 3r + 5 - 5 = 10 - 5. On the left side, the +5 and -5 cancel each other out, leaving us with 3r. On the right side, 10 - 5 equals 5. So, the equation now looks like this: 3r = 5.

Next, we need to tackle the multiplication. 'r' is being multiplied by 3, so we'll perform the inverse operation, which is division. We'll divide both sides of the equation by 3. This gives us: 3r / 3 = 5 / 3. On the left side, the 3s cancel each other out, leaving us with just 'r'. On the right side, 5 / 3 is an improper fraction, which we can leave as it is or convert to a mixed number (1 2/3). Therefore, the solution to this equation is r = 5/3.

Let's verify our solution one last time. We substitute 5/3 back into the original equation: 3(5/3) + 5 = 10. Simplifying the left side, we get 5 + 5 = 10, which is indeed true. This confirms that our solution, r = 5/3, is correct. Fantastic job! We've successfully solved all the equations, and you've gained valuable practice in the art of algebraic problem-solving.

Conclusion: Mastering the Art of Equation Solving

Great job, guys! We've journeyed through a series of equations, from simple addition to those involving multiplication and both. You've learned how to systematically isolate variables by using inverse operations and maintaining the balance of the equation. Remember, the key to success in algebra is practice, so don't hesitate to tackle more equations and challenge yourself.

Solving equations is not just a math skill; it's a valuable tool for problem-solving in many areas of life. It teaches you to think logically, break down complex problems into smaller steps, and find solutions. So, keep practicing, keep exploring, and you'll continue to sharpen your algebraic skills. Until next time, happy equation solving!