Algebra Problem: Solving Equations And Formulas

Hey guys! Today, we're diving into an algebra problem that involves calculating values using formulas and solving for variables. We'll be working with the formula P = 4a to find both P and a under different conditions. Plus, we'll tackle another equation, y = 3x - 9, to find values for x and y. Let's break it down step by step!

Part 1: Using the Formula P = 4a

In this first part, we're going to use the formula P = 4a. This formula might look simple, but it's a fundamental concept in algebra and geometry. It essentially tells us that P is equal to 4 times the value of a. Think of it like the perimeter of a square, where P is the perimeter and a is the length of one side. The cool thing about formulas is that they allow us to quickly find a missing value if we know the others.

1) Calculating P when a = 2.5, 15, and 3

Okay, so the first task is to find the value of P when we know the value of a. We have three different values for a: 2.5, 15, and 3. For each value, we simply plug it into the formula P = 4a and do the math. This is where the substitution part of algebra comes in handy – we're substituting the value of a into the equation to find P.

• When a = 2.5: P = 4 * 2.5 = 10 So, when a is 2.5, P is 10.
• When a = 15: P = 4 * 15 = 60 Here, when a is 15, P jumps up to 60.
• When a = 3: P = 4 * 3 = 12 And finally, when a is 3, P is 12.

See how easy that was? By just plugging in the value of a, we quickly found the corresponding value of P. This is a key skill in algebra – being able to manipulate formulas to find what you need.

2) Calculating a when P = 8, 1.6, and 1

Now, let's switch things up! This time, we know the value of P and need to find a. We're still using the same formula, P = 4a, but we're going to rearrange it a little bit. To isolate a, we need to divide both sides of the equation by 4. This gives us a new formula: a = P / 4. This is what we call solving for a variable – we're manipulating the equation to get the variable we want (in this case, a) all by itself on one side.

Now, we've got three values for P: 8, 1.6, and 1. Just like before, we'll substitute each value into our new formula, a = P / 4, to find the corresponding value of a.

• When P = 8: a = 8 / 4 = 2 So, when P is 8, a is 2.
• When P = 1.6: a = 1.6 / 4 = 0.4 When P is 1.6, a is 0.4.
• When P = 1: a = 1 / 4 = 0.25 And when P is 1, a is 0.25.

Again, not too shabby, right? By rearranging the formula and substituting the known values, we've successfully calculated the value of a for each given P. This part demonstrates the power of algebraic manipulation – you can tweak formulas to solve for any variable you need.

Part 2: Working with the Equation y = 3x - 9

Okay, guys, let’s shift gears a bit! We're now going to tackle the equation y = 3x - 9. This is a linear equation, meaning that if you were to graph it, you'd get a straight line. Linear equations are super common in math and real-world applications, so understanding how to work with them is key.

This equation tells us the relationship between two variables, x and y. For any value of x we plug in, we can calculate a corresponding value of y, and vice versa. We're going to explore this relationship by finding specific values of x and y that satisfy the equation. This is like finding points that lie on the line if we were to graph it.

1) Finding the value of x when y = 0

The first challenge is to find the value of x that makes y equal to 0. This is a common type of problem in algebra, and it's often called finding the x-intercept of the line (where the line crosses the x-axis). To solve this, we substitute 0 for y in our equation, giving us: 0 = 3x - 9. Now, we need to solve for x. This involves isolating x on one side of the equation.

Here’s how we do it:

1. Add 9 to both sides: This gets rid of the -9 on the right side, leaving us with 9 = 3x.
2. Divide both sides by 3: This isolates x, giving us x = 3.

So, when y is 0, x is 3. This means the line represented by the equation y = 3x - 9 crosses the x-axis at the point (3, 0). Understanding how to find intercepts like this is super useful in graphing and analyzing linear equations.

2) Finding the value of y when x is a given number (unspecified)

Alright, for the second part, we need to find the value of y when we have a value for x. The problem statement doesn’t give us a specific value for x, but that’s totally okay! We can still talk about the process and how it works. Let's say, just for example, that we wanted to find y when x is 5. The process would be exactly the same as before – we simply substitute the value of x into the equation and do the math.

So, if x = 5, then:

• y = 3 * 5 - 9
• y = 15 - 9
• y = 6

In this case, when x is 5, y is 6. This means the point (5, 6) lies on the line represented by the equation y = 3x - 9. You can do this for any value of x – just plug it into the equation and solve for y! This is the beauty of equations – they give you a way to connect variables and find corresponding values.

Wrapping it Up

So, guys, we've tackled an algebra problem that covered a couple of key concepts: using formulas and solving linear equations. We started with the formula P = 4a, where we calculated P when we knew a, and then flipped it to find a when we knew P. This showed us how to substitute values into formulas and rearrange them to solve for different variables. Then, we moved on to the equation y = 3x - 9, where we found the value of x that makes y zero, and talked about how to find y for any given x. These are the building blocks of algebra, and mastering them will set you up for success in more advanced math topics. Keep practicing, and you'll be solving equations like a pro in no time!