Solving Linear Equations Step-by-Step Guide With Examples

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Hey guys! πŸ‘‹ Are you ready to dive into the exciting world of linear equations? If you've ever felt a little puzzled by algebraic expressions, you're in the right place! In this article, we're going to break down five different linear equations, step-by-step, so you can master the art of solving them. We'll use simple language, clear explanations, and a dash of fun to make sure you not only understand the solutions but also the why behind each step. Get your pencils ready, and let's jump in!

Okay, let's start with our first equation: 198=154+7xβˆ’68198 = 154 + 7x - 68. The goal here is to isolate x on one side of the equation. This means we need to get x all by itself, with no other numbers hanging around. To do this, we'll follow a few simple steps. First, let's simplify the right side of the equation by combining the constant terms. We have 154 and -68, so let's add those together. 154 minus 68 is 86. So our equation now looks like this: 198=86+7x198 = 86 + 7x. Much cleaner, right? Now, we need to get rid of that 86 on the right side. To do this, we'll subtract 86 from both sides of the equation. Remember, whatever you do to one side, you have to do to the other to keep the equation balanced. So, 198 minus 86 is 112. On the right side, 86 minus 86 cancels out, leaving us with just 7x. Our equation is now 112=7x112 = 7x. We're almost there! Now, we just need to get x by itself. It's currently being multiplied by 7, so to undo that, we'll divide both sides by 7. 112 divided by 7 is 16. On the right side, 7x divided by 7 is just x. So our final answer is x=16x = 16. πŸŽ‰ Easy peasy, right? Let's recap: we simplified, subtracted, and divided to isolate x. These are the core moves in solving linear equations, so keep them in your toolkit! Now, let's move on to the next problem and keep building our equation-solving skills.

Alright, let's tackle our next equation: βˆ’131=βˆ’5(3xβˆ’8)+6x-131 = -5(3x - 8) + 6x. This one looks a little more complex, but don't worry, we'll break it down! The first thing we need to do is get rid of those parentheses. We'll use the distributive property, which means we'll multiply the -5 by both terms inside the parentheses. So, -5 times 3x is -15x, and -5 times -8 is +40. Remember, a negative times a negative is a positive! Our equation now looks like this: βˆ’131=βˆ’15x+40+6x-131 = -15x + 40 + 6x. Now, let's simplify by combining like terms. We have -15x and +6x on the right side, so let's add those together. -15x plus 6x is -9x. Our equation is now βˆ’131=βˆ’9x+40-131 = -9x + 40. We're making progress! Next, we want to isolate the term with x, so we need to get rid of that +40. We'll subtract 40 from both sides of the equation. -131 minus 40 is -171. On the right side, 40 minus 40 cancels out, leaving us with -9x. Our equation is now βˆ’171=βˆ’9x-171 = -9x. We're almost there! Now, we just need to get x by itself. It's currently being multiplied by -9, so we'll divide both sides by -9. Remember, a negative divided by a negative is a positive! -171 divided by -9 is 19. On the right side, -9x divided by -9 is just x. So our final answer is x=19x = 19. πŸ₯³ You nailed it! We used the distributive property, combined like terms, subtracted, and divided to solve this equation. Keep practicing these steps, and you'll become an equation-solving pro in no time!

Let's move on to problem number eight: βˆ’7xβˆ’10=18+3x-7x - 10 = 18 + 3x. In this equation, we have x terms on both sides, so our first goal is to gather all the x terms on one side and the constant terms on the other. This will make it easier to isolate x. Let's start by getting rid of the -3x on the right side. To do this, we'll subtract 3x from both sides of the equation. -7x minus 3x is -10x. On the right side, 3x minus 3x cancels out. Our equation now looks like this: βˆ’10xβˆ’10=18-10x - 10 = 18. Great! Now, we need to get rid of the -10 on the left side. We'll add 10 to both sides of the equation. -10 plus 10 cancels out on the left side. On the right side, 18 plus 10 is 28. Our equation is now βˆ’10x=28-10x = 28. We're almost there! Now, we just need to get x by itself. It's currently being multiplied by -10, so we'll divide both sides by -10. 28 divided by -10 is -2.8. On the left side, -10x divided by -10 is just x. So our final answer is x=βˆ’2.8x = -2.8. 🌟 Excellent work! We gathered the x terms, moved the constants, and then divided to solve for x. This is a common strategy when you have variables on both sides of the equation, so remember this technique!

Now, let's dive into problem nine: 12x+8βˆ’15=βˆ’2(3xβˆ’82)12x + 8 - 15 = -2(3x - 82). This one has a mix of things going on, so we'll take it step by step. First, let's simplify both sides of the equation as much as possible. On the left side, we have +8 and -15. Let's combine those: 8 minus 15 is -7. So the left side becomes 12xβˆ’712x - 7. On the right side, we have -2 multiplied by the parentheses (3xβˆ’82)(3x - 82). We'll use the distributive property again. -2 times 3x is -6x, and -2 times -82 is +164. Remember, a negative times a negative is a positive! So the right side becomes βˆ’6x+164-6x + 164. Our equation now looks like this: 12xβˆ’7=βˆ’6x+16412x - 7 = -6x + 164. Much cleaner, right? Now, let's gather the x terms on one side and the constant terms on the other. Let's add 6x to both sides of the equation to get rid of the -6x on the right. 12x plus 6x is 18x. On the right side, -6x plus 6x cancels out. Our equation is now 18xβˆ’7=16418x - 7 = 164. Next, we need to get rid of the -7 on the left side. We'll add 7 to both sides of the equation. -7 plus 7 cancels out on the left side. On the right side, 164 plus 7 is 171. Our equation is now 18x=17118x = 171. We're almost there! Now, we just need to get x by itself. It's currently being multiplied by 18, so we'll divide both sides by 18. 171 divided by 18 is 9.5. On the left side, 18x divided by 18 is just x. So our final answer is x=9.5x = 9.5. πŸš€ Awesome job! We simplified, distributed, gathered like terms, and then divided to solve for x. You're becoming a true equation-solving wizard!

Finally, let's tackle our last equation: βˆ’(12xβˆ’6)=(12x+6)-(12x - 6) = (12x + 6). This one has parentheses with a negative sign in front, so let's be extra careful with our signs! First, we need to distribute the negative sign on the left side. This is like multiplying everything inside the parentheses by -1. So, -1 times 12x is -12x, and -1 times -6 is +6. Remember, a negative times a negative is a positive! The left side becomes βˆ’12x+6-12x + 6. Our equation now looks like this: βˆ’12x+6=12x+6-12x + 6 = 12x + 6. Now, let's gather the x terms on one side and the constant terms on the other. Let's add 12x to both sides of the equation to get rid of the -12x on the left. -12x plus 12x cancels out on the left side. On the right side, 12x plus 12x is 24x. Our equation is now 6=24x+66 = 24x + 6. Next, we need to get rid of the +6 on the right side. We'll subtract 6 from both sides of the equation. 6 minus 6 is 0 on the left side. On the right side, 6 minus 6 cancels out. Our equation is now 0=24x0 = 24x. We're almost there! Now, we just need to get x by itself. It's currently being multiplied by 24, so we'll divide both sides by 24. 0 divided by 24 is 0. On the right side, 24x divided by 24 is just x. So our final answer is x=0x = 0. πŸŽ‰ You crushed it! We distributed the negative sign, gathered like terms, and then divided to solve for x. This problem shows us that sometimes the answer can be zero, which is perfectly fine!

And there you have it! We've successfully solved five different linear equations, step-by-step. You've learned how to simplify, distribute, combine like terms, and isolate variables. Remember, the key to mastering linear equations is practice, practice, practice! The more you work through problems, the more comfortable and confident you'll become. So, keep those pencils sharp, and keep solving! You've got this! πŸ’ͺ