Solving Equations: What's The Next Step?

by Dimemap Team 41 views

Hey guys! Ever find yourself staring at an equation and wondering, "Okay, what do I do now?" Let's break down a common algebra problem step-by-step. We're going to look at an equation that's already had a couple of moves made, and figure out the best way to keep solving it. So, grab your pencils (or styluses, if you're cool like that), and let's dive in!

The Problem

Here's the equation we're tackling:

34(20y−8)+5=12y+14(20y+8)\frac{3}{4}(20 y-8)+5=\frac{1}{2} y+\frac{1}{4}(20 y+8)

And here are the first two steps that have already been taken:

  1. 15y−6+5=12y+5y+215 y-6+5=\frac{1}{2} y+5 y+2
  2. 15y−1=112y+215 y-1=\frac{11}{2} y+2

Our mission? Figure out the best next step to solve for 'y'.

Understanding the Steps So Far

Before we jump into what could be next, let's make sure we're all on the same page about what's already happened. This is super important because if we don't get the foundation right, the rest of the solution will be wobbly, you know?

Step 1: Distribution

Step one is all about distributing. Look at the original equation:

34(20y−8)+5=12y+14(20y+8)\frac{3}{4}(20 y-8)+5=\frac{1}{2} y+\frac{1}{4}(20 y+8)

On the left side, we see 34\frac{3}{4} multiplying the entire expression (20y−8)(20y - 8). That means we have to multiply 34\frac{3}{4} by both 20y20y and −8-8. Let's do it:

  • 34∗20y=15y\frac{3}{4} * 20y = 15y
  • 34∗−8=−6\frac{3}{4} * -8 = -6

So, the left side becomes 15y−6+515y - 6 + 5. Don't forget that '+ 5' hanging out at the end!

Now, let's look at the right side. We have 14\frac{1}{4} multiplying (20y+8)(20y + 8). Again, we distribute:

  • 14∗20y=5y\frac{1}{4} * 20y = 5y
  • 14∗8=2\frac{1}{4} * 8 = 2

This makes the right side 12y+5y+2\frac{1}{2}y + 5y + 2. So, after the distribution, we get:

15y−6+5=12y+5y+215 y-6+5=\frac{1}{2} y+5 y+2

Which is exactly what step one shows. Great job, step one!

Step 2: Combining Like Terms

Step two is about simplifying. We are combining like terms. Like terms are terms that have the same variable raised to the same power (or are just constants). In this case, we're mostly looking for constants that we can add or subtract.

On the left side, we have −6+5-6 + 5. That's a no-brainer: −6+5=−1-6 + 5 = -1. So, the left side simplifies to 15y−115y - 1.

On the right side, we have 12y+5y\frac{1}{2}y + 5y. To combine these, we need a common denominator. We can rewrite 5y5y as 102y\frac{10}{2}y. Then we have:

12y+102y=112y\frac{1}{2}y + \frac{10}{2}y = \frac{11}{2}y

So, the right side becomes 112y+2\frac{11}{2}y + 2. Putting it all together, we get:

15y−1=112y+215 y-1=\frac{11}{2} y+2

And that's step two! See? Not so scary when we break it down.

What's the Best Next Step?

Okay, so we're at 15y−1=112y+215 y-1=\frac{11}{2} y+2. What should we do next? The goal is to isolate 'y' on one side of the equation. To do this efficiently, we need to think strategically.

Here's a good approach:

  1. Get all the 'y' terms on one side: We need to decide whether to move the 15y15y to the right side, or the 112y\frac{11}{2}y to the left side. To avoid dealing with negative coefficients (which can be confusing), it's often easiest to move the smaller 'y' term. Since 112\frac{11}{2} (which is 5.5) is smaller than 15, let's subtract 112y\frac{11}{2}y from both sides of the equation.
  2. Get all the constant terms on the other side: After moving the 'y' terms, we'll have a constant term on both sides. We need to move the constant term from the side with the 'y' to the other side. In our case, that means adding 1 to both sides.
  3. Simplify and solve for 'y': Finally, we'll simplify both sides of the equation and then divide to isolate 'y'.

So, the best next step is to subtract 112y\frac{11}{2}y from both sides. This will keep the 'y' term positive and make the following steps easier. It will look like this:

15y−1−112y=112y+2−112y15y - 1 - \frac{11}{2}y = \frac{11}{2}y + 2 - \frac{11}{2}y

Which simplifies to:

15y−112y−1=215y - \frac{11}{2}y - 1 = 2

Now, let's rewrite 15y15y with a denominator of 2: 15y=302y15y = \frac{30}{2}y. So we have:

302y−112y−1=2\frac{30}{2}y - \frac{11}{2}y - 1 = 2

Combining the 'y' terms:

192y−1=2\frac{19}{2}y - 1 = 2

Now we are on the right track. Add 1 to both sides to get:

192y=3\frac{19}{2}y = 3

Multiply both sides by 219\frac{2}{19} to get:

y = 619\frac{6}{19}

Why This Approach?

You might be wondering, "Why not just add 1 to both sides first?" Well, you could, but it's generally more efficient to deal with the variable terms first. This is because you'll eventually have to isolate 'y' anyway, so getting those terms together early simplifies the process.

Also, avoiding negative coefficients is a huge help. Negative signs are notorious for causing errors, so anything we can do to minimize them is a win!

In Conclusion

Solving equations is like following a recipe. Each step builds upon the last, and the order matters. By understanding the underlying principles of distribution, combining like terms, and strategic isolation, you can tackle even the trickiest equations with confidence.

So, next time you're faced with an algebraic challenge, remember to break it down, stay organized, and don't be afraid to ask for help. You got this!