Solving Equations: A Step-by-Step Guide

Hey guys! Are you struggling with solving equations? Don't worry, you're not alone! Many students find equations tricky, but with the right approach, you can conquer them. In this guide, we'll break down how to solve a few different types of equations. So, let's jump right in and make math a little less daunting.

Equation a) 9x = 27

Let's start with the first equation: 9x = 27. Our main keyword here is solving linear equations, and this one is a classic example. The goal is to isolate x on one side of the equation. We want to get x all by itself so we know what its value is. To do that, we need to undo the operation that's being done to x. In this case, x is being multiplied by 9. So, what's the opposite of multiplication? Division! That's right, we're going to divide both sides of the equation by 9. Remember, whatever you do to one side of the equation, you must do to the other side to keep things balanced.

So, let’s dive deeper into the process. When we divide both sides by 9, we get:

(9x) / 9 = 27 / 9

On the left side, the 9s cancel each other out, leaving us with just x. On the right side, 27 divided by 9 is 3. So, our equation simplifies to:

x = 3

And that's it! We've solved for x. The solution to the equation 9x = 27 is x = 3. See, that wasn't so bad, was it? This basic principle of isolating the variable by using inverse operations is the foundation for solving many types of equations. Understanding this concept is key to mastering algebra. Keep practicing, and you'll become a pro at solving for x in no time! And remember, if you ever get stuck, don't hesitate to review the steps or ask for help. Everyone learns at their own pace, and the important thing is to keep trying and building your skills.

Equation b) 5x + 2 - 5x + 5x - 1 = 121

Now, let's tackle the second equation: 5x + 2 - 5x + 5x - 1 = 121. This one looks a little more complicated, but don't let it intimidate you! Our main focus here is still on solving linear equations, but this time, we have a few more terms to deal with. The first thing we want to do is simplify the equation by combining like terms. Like terms are those that have the same variable raised to the same power. In this case, we have terms with x and constant terms (numbers without any variables).

Let's gather our x terms: 5x - 5x + 5x. Notice that the first two terms, 5x and -5x, cancel each other out. This is because they are additive inverses – they have the same magnitude but opposite signs. So, 5x - 5x = 0. That leaves us with just 5x.

Next, let's combine our constant terms: +2 - 1. This is a simple subtraction problem: 2 - 1 = 1. So, now we've simplified the left side of the equation to:

5x + 1 = 121

See? It's already looking much simpler! Now we're back to a more familiar form. To isolate the x term, we need to get rid of the +1. The inverse operation of addition is subtraction, so we'll subtract 1 from both sides of the equation:

5x + 1 - 1 = 121 - 1

This simplifies to:

5x = 120

Great! Now we're almost there. We have 5 times x equals 120. To isolate x, we need to undo the multiplication by dividing both sides by 5:

(5x) / 5 = 120 / 5

On the left side, the 5s cancel out, leaving us with x. On the right side, 120 divided by 5 is 24. So, we have:

x = 24

And we've done it! The solution to the equation 5x + 2 - 5x + 5x - 1 = 121 is x = 24. Remember, simplifying the equation first by combining like terms is a crucial step in solving more complex equations. It makes the process much easier and less prone to errors. So, always look for opportunities to simplify before you start isolating the variable. You've got this!

Equation c) 4x - 10 · 2x + 16 = 0

Alright, let's move on to the third equation: 4x - 10 · 2x + 16 = 0. This equation is a bit different from the previous ones because it involves an exponential term. Our focus now shifts to solving exponential equations. The key to solving these types of equations is to recognize the pattern and use a substitution to make it look like a quadratic equation, which we already know how to solve.

First, let's rewrite the equation to make the pattern clearer. Notice that 4x can be written as (2^2)x which is the same as (2x)2. This is a handy trick when dealing with exponential equations. So, we can rewrite our equation as:

(2x)2 - 10 · 2x + 16 = 0

Now, do you see it? It's starting to look like a quadratic equation! To make it even clearer, let's use a substitution. Let's say y = 2x. If we substitute y for 2x in our equation, we get:

y2 - 10y + 16 = 0

Ah, much better! This is a quadratic equation in the form ay2 + by + c = 0, where a = 1, b = -10, and c = 16. We can solve this quadratic equation using several methods, such as factoring, completing the square, or the quadratic formula. Let's use factoring, as it's often the quickest method if the equation factors easily.

We're looking for two numbers that multiply to 16 and add up to -10. Those numbers are -2 and -8. So, we can factor the quadratic equation as:

(y - 2)(y - 8) = 0

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for y:

y - 2 = 0 => y = 2 y - 8 = 0 => y = 8

So, we have two solutions for y: y = 2 and y = 8. But remember, we're not trying to solve for y; we're trying to solve for x. We need to substitute back 2x for y to find the values of x.

Let's start with y = 2:

2x = 2

To solve for x, we need to ask ourselves,