How to Solve for z in 4z + 1 = bz + c

Hey guys! Ever stared at an equation and felt like it was speaking a foreign language? We’ve all been there, especially when letters start popping up everywhere. Today, we’re diving deep into how to solve for z in the equation 4z + 1 = bz + c . Don’t sweat it; we’re going to break this down step-by-step, making it super clear and easy to follow. This kind of problem is super common in algebra, and once you get the hang of it, you’ll be zipping through similar equations in no time. We’re assuming, of course, that there is a solution for z, which is key!

Our main mission here is to isolate the variable ‘z’ on one side of the equation. Think of it like trying to get your favorite gadget all to yourself – you gotta move everything else out of the way! The equation we’re working with is 4z + 1 = bz + c . Notice we’ve got ‘z’ terms on both sides (4z and bz) and some constant terms on both sides (+1 and +c). To solve for z, we need to gather all the ‘z’ terms on one side and all the other terms (the constants) on the other side. This is the golden rule of equation solving: whatever you do to one side, you must do to the other to keep things balanced. It’s all about maintaining that equality.

So, first things first, let’s get those ‘z’ terms together. We have 4z on the left and bz on the right. To move bz from the right side to the left, we need to perform the opposite operation. Since bz is being added on the right (or, more accurately, it’s a positive term), we’ll subtract bz from both sides of the equation. This is crucial! Doing this gives us:

4z - bz + 1 = bz - bz + c

On the right side, bz - bz cancels out, leaving us with just c . On the left side, we now have 4z - bz + 1 . So, the equation transforms into:

4z - bz + 1 = c

See? We’re one step closer to getting ‘z’ by itself. Now, we need to move the constant term +1 from the left side to the right side. Just like before, we do the opposite operation. Since +1 is being added, we’ll subtract 1 from both sides.

4z - bz + 1 - 1 = c - 1

This simplifies to:

4z - bz = c - 1

Awesome! All our ‘z’ terms are now on the left, and all our constants are on the right. The next big move is to factor out ‘z’ from the terms on the left side. Both 4z and bz have ‘z’ as a common factor. So, we can rewrite the left side as z(4 - b) . This step is like packaging all the ‘z’s together so we can see how many we actually have. The equation now looks like this:

z(4 - b) = c - 1

We’re in the home stretch, guys! Our goal is to get ‘z’ completely alone. Right now, ‘z’ is being multiplied by the term (4 - b) . To undo this multiplication, we need to divide. And just like always, we divide both sides of the equation by (4 - b) . This is where we need to make a quick but important note: this step is only possible if (4 - b) is not equal to zero. If (4 - b) = 0 , it means b = 4 . In that specific case, we’d have a different situation, possibly no unique solution or infinite solutions, but the prompt assumes a solution exists. So, assuming 4 - b ≠ 0 :

z(4 - b) / (4 - b) = (c - 1) / (4 - b)

The (4 - b) on the left side cancels out, leaving ‘z’ all by its lonesome. And there you have it! The solution for ‘z’ is:

z = (c - 1) / (4 - b)

Boom! You’ve successfully solved for ‘z’. It looks a little complex with all those letters, but if you follow the steps – gather variables, isolate terms, factor, and divide – you can tackle any similar equation. Keep practicing, and you’ll be an algebra whiz in no time!

Understanding the Steps in Detail

Let’s really dig into why each step works and what it means. When we’re solving for z , we’re essentially trying to find the value of ‘z’ that makes the equation true. The equation 4z + 1 = bz + c is like a balanced scale. Whatever we do to one side, we must mirror on the other to keep it balanced. Our objective is to get ‘z’ by itself on one side of the scale.

First, we addressed the ‘z’ terms. We had 4z on the left and bz on the right. To bring them together, we subtracted bz from both sides. Why subtract? Because subtraction is the inverse operation of addition. If bz is on the right, adding it to the equation (which is what +bz implies) means we need to take it away to move it. So, 4z + 1 - bz = bz + c - bz . This simplifies to 4z - bz + 1 = c . This step is fundamental: grouping like terms . We want all terms with the variable we’re solving for on one side and all constant terms on the other.

Next, we tackled the constants. We had +1 on the left and +c on the right. We wanted to move +1 over to the side with c . Again, we use the inverse operation. Since 1 is added, we subtract 1 from both sides: 4z - bz + 1 - 1 = c - 1 . This leaves us with 4z - bz = c - 1 . This stage is about isolating the variable term . We’ve successfully put all the ‘z’ components on one side and all the non-‘z’ components on the other.

Now comes a crucial algebraic maneuver: factoring . Look at the left side: 4z - bz . Both terms share a common factor, which is ‘z’. By factoring out ‘z’, we’re essentially saying, “How many ‘z’s do we have in total?” We have 4 ‘z’s and we’re subtracting b ‘z’s. So, we have (4 - b) groups of ‘z’. This is represented as z(4 - b) . Our equation becomes z(4 - b) = c - 1 . Factoring is powerful because it consolidates the variable, making it easier to isolate.

Finally, we reach the division step . Currently, z is being multiplied by the entire expression (4 - b) . To get z by itself, we must perform the inverse operation of multiplication, which is division. We divide both sides by (4 - b) . So, z(4 - b) / (4 - b) = (c - 1) / (4 - b) . On the left, (4 - b) divided by (4 - b) equals 1, leaving us with z * 1 , which is just z . On the right, we have the expression (c - 1) / (4 - b) . This gives us our final solution: z = (c - 1) / (4 - b) . This step requires the denominator, (4 - b) , to be non-zero. If 4 - b = 0 (meaning b = 4 ), the original equation would simplify differently, and we might not have a unique solution. But since the problem states we assume a solution exists, we proceed with the division.

Why This Matters in Algebra

Understanding how to solve for z in equations like 4z + 1 = bz + c is fundamental to algebra. It’s not just about memorizing steps; it’s about grasping the underlying principles of equation manipulation. This skill is the bedrock upon which more complex mathematical concepts are built. Whether you’re dealing with linear equations, quadratic equations, or even systems of equations, the core strategy often involves isolating the variable you’re interested in.

Think about real-world applications, guys. In physics, you might need to solve for time (’t’) in a motion equation where other variables like velocity (‘v’) and acceleration (‘a’) are involved. In economics, you could be solving for a quantity (‘q’) in a supply or demand function. In computer science, algorithms often rely on solving for variables to determine efficiency or resource allocation. In all these scenarios, the ability to systematically rearrange and solve equations is paramount.

Our equation 4z + 1 = bz + c is a linear equation in one variable, ‘z’, although ‘b’ and ‘c’ are treated as parameters or constants for the purpose of solving for ‘z’. The structure az + d = bz + e is very common. The solution z = (e - d) / (a - b) is the general form, and our specific problem fits this mold perfectly with a=4 , d=1 , b=b , and e=c . Recognizing these patterns helps you solve problems faster and with more confidence.

Furthermore, the process teaches valuable logical reasoning. You learn to think sequentially, anticipate the next step, and understand the consequences of each operation. It’s like solving a puzzle or playing a strategic game. Each move (each operation) must be precise and logical to reach the desired outcome.

When you encounter situations where b = 4 , remember that algebra doesn’t break; it just behaves differently. If b=4 , our equation becomes 4z + 1 = 4z + c . If c is also 1 , then 4z + 1 = 4z + 1 , which is true for any value of ‘z’. This means there are infinite solutions . If c is not 1 (e.g., 4z + 1 = 4z + 5 ), then 1 = 5 , which is impossible. In this case, there is no solution . These special cases are important to recognize and understand, as they reveal the nuances of mathematical relationships.

Mastering the basics of solving for z or any other variable empowers you to tackle more complex mathematical challenges. It builds confidence and analytical skills that extend far beyond the classroom. So, the next time you see an equation, don’t be intimidated. Break it down, apply the rules, and remember you’ve got this!

Putting It All Together: A Quick Recap

Alright, let’s do a lightning-fast recap of how we conquered the equation 4z + 1 = bz + c to solve for z . Remember, the goal is always to get ‘z’ all by its lonesome on one side of the equals sign.

1. Group ‘z’ terms: We moved the bz term from the right side to the left by subtracting it from both sides. This gave us 4z - bz + 1 = c .
2. Group constant terms: We moved the +1 term from the left side to the right by subtracting it from both sides. This resulted in 4z - bz = c - 1 .
3. Factor out ‘z’: We recognized that ‘z’ was a common factor on the left side and factored it out, turning 4z - bz into z(4 - b) . The equation became z(4 - b) = c - 1 .
4. Isolate ‘z’: Finally, to get ‘z’ completely alone, we divided both sides by (4 - b) , assuming 4 - b is not zero. This gave us our clean solution: z = (c - 1) / (4 - b) .

See? It’s a logical flow, like a recipe. Follow the ingredients (the steps), and you get the perfect dish (the solution). This methodical approach works for countless algebraic problems. Keep practicing these core techniques, and you’ll find yourself becoming more and more comfortable with algebra. Happy solving, everyone!