Simplify And Solve: -25z + 1 + 35z = 10z + 1

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Simplify and Solve: -25z + 1 + 35z = 10z + 1

Hey math whizzes! Today, we're diving into a super common type of algebra problem that can sometimes trip people up if they're not careful. We're talking about simplifying and solving linear equations. The equation we've got on the table is βˆ’25z+1+35z=10z+1-25z + 1 + 35z = 10z + 1. Now, don't let the negative numbers or the multiple terms scare you off. We're going to break this down step-by-step, making sure we cover all the bases so you can conquer similar problems with confidence. Think of this as your ultimate guide to untangling algebraic expressions and finding that elusive variable, 'z'. We'll cover combining like terms, isolating the variable, and checking our work. Ready to get your math on? Let's go!

Understanding the Equation: Breaking It Down

Alright guys, the first thing we need to do when we look at an equation like βˆ’25z+1+35z=10z+1-25z + 1 + 35z = 10z + 1 is to understand what we're dealing with. This is a linear equation because the highest power of our variable, 'z', is just 1. Our goal here is to find the value of 'z' that makes this equation true. To do that, we need to simplify both sides of the equation first. On the left side, we have βˆ’25z+1+35z-25z + 1 + 35z. Notice that we have two terms with 'z' in them (βˆ’25z-25z and +35z+35z) and a constant term (+1). These are called like terms. Combining like terms is a fundamental skill in algebra. To combine βˆ’25z-25z and +35z+35z, we simply add their coefficients (the numbers in front of the 'z'). So, βˆ’25+35-25 + 35 equals 1010. Therefore, βˆ’25z+35z-25z + 35z simplifies to 10z10z. Now, the entire left side of the equation becomes 10z+110z + 1. Easy peasy, right? On the right side of the equation, we have 10z+110z + 1. This side is already simplified. It has one term with 'z' and one constant term. So, our original equation βˆ’25z+1+35z=10z+1-25z + 1 + 35z = 10z + 1 simplifies to 10z+1=10z+110z + 1 = 10z + 1. See how much cleaner that looks? This simplification step is crucial because it helps us see the structure of the equation more clearly and prepares us for the next steps in solving for 'z'. Remember, the key is to always simplify each side of the equation as much as possible before you start trying to isolate the variable. This involves identifying and combining like terms – any terms that have the same variable raised to the same power, or constant terms (numbers without variables).

Simplifying Both Sides: The First Crucial Step

So, as we just saw, the first big step in solving our equation, βˆ’25z+1+35z=10z+1-25z + 1 + 35z = 10z + 1, is simplification. We need to make both sides of the equation as neat and tidy as possible before we start moving things around. Let's focus on the left side first: βˆ’25z+1+35z-25z + 1 + 35z. Here, the terms βˆ’25z-25z and +35z+35z are like terms because they both contain the variable 'z' raised to the power of 1. The term +1+1 is a constant term. To combine the like terms involving 'z', we add their coefficients: βˆ’25+35=10-25 + 35 = 10. So, βˆ’25z+35z-25z + 35z simplifies to 10z10z. Now, we bring back the constant term, +1+1. So, the entire left side simplifies to 10z+110z + 1. Now, let's look at the right side of the equation: 10z+110z + 1. This side is already in its simplest form. There are no like terms to combine. So, after simplifying the left side, our equation becomes: 10z+1=10z+110z + 1 = 10z + 1. This is a fantastic outcome! It means we've successfully reduced the complexity of the equation. This simplification process is key. It's like clearing away the clutter before you can see the real problem. Always remember to combine all your 'z' terms together, all your 'x' terms together, and all your constant numbers together on each side of the equals sign separately before you start moving terms across the equals sign. This prevents errors and makes the rest of the solving process much more straightforward. You're essentially rewriting the equation in its most basic, understandable form.

Isolating the Variable: Finding the Value of 'z'

Now that we've simplified our equation to 10z+1=10z+110z + 1 = 10z + 1, the next big step is to isolate the variable, 'z'. This means we want to get 'z' all by itself on one side of the equation. To do this, we use inverse operations. Remember, whatever you do to one side of the equation, you must do to the other side to keep it balanced. Let's start by trying to get all the 'z' terms on one side. We have 10z10z on the left and 10z10z on the right. Let's subtract 10z10z from both sides:

(10z+1)βˆ’10z=(10z+1)βˆ’10z(10z + 1) - 10z = (10z + 1) - 10z

On the left side, 10zβˆ’10z=010z - 10z = 0. So, the left side becomes 0+10 + 1, which is just 11.

On the right side, 10zβˆ’10z=010z - 10z = 0. So, the right side becomes 0+10 + 1, which is also just 11.

So, our equation now looks like this: 1=11 = 1.

What does this mean? When you perform operations to isolate the variable and end up with a true statement (like 1=11 = 1, or 5=55 = 5), it means the original equation is true for all possible values of 'z'. This is called an identity. It means no matter what number you plug in for 'z', the equation will always hold true. This is a special case in algebra, and it's important to recognize it. If we had ended up with a false statement, like 1=51 = 5, that would mean there is no solution to the equation. But in this case, 1=11 = 1 tells us that any value of 'z' works. So, the solution set is all real numbers. Pretty neat, huh? The process of isolating the variable involves using subtraction, addition, multiplication, or division to undo operations and get the variable alone. Always work systematically, dealing with addition/subtraction first, then multiplication/division.

What Does '1=11=1' Mean? Understanding Identities

So, we've gone through the steps of simplifying and trying to isolate 'z' from our equation βˆ’25z+1+35z=10z+1-25z + 1 + 35z = 10z + 1, and we've ended up with 1=11 = 1. This isn't a result that gives us a specific numerical value for 'z', like z=5z = 5 or z=βˆ’2z = -2. Instead, it's a statement that is always true. In mathematics, when solving an equation leads to a statement that is always true, regardless of the value of the variable, we call that equation an identity. Think about it: if you plug in any number for 'z' into the original equation, it will work out. Let's try a couple of examples just to prove it.

  • If z=0z = 0: Left side: βˆ’25(0)+1+35(0)=0+1+0=1-25(0) + 1 + 35(0) = 0 + 1 + 0 = 1 Right side: 10(0)+1=0+1=110(0) + 1 = 0 + 1 = 1 So, 1=11 = 1. It works!
  • If z=10z = 10: Left side: βˆ’25(10)+1+35(10)=βˆ’250+1+350=101-25(10) + 1 + 35(10) = -250 + 1 + 350 = 101 Right side: 10(10)+1=100+1=10110(10) + 1 = 100 + 1 = 101 So, 101=101101 = 101. It works!
  • If z=βˆ’5z = -5: Left side: βˆ’25(βˆ’5)+1+35(βˆ’5)=125+1βˆ’175=βˆ’49-25(-5) + 1 + 35(-5) = 125 + 1 - 175 = -49 Right side: 10(βˆ’5)+1=βˆ’50+1=βˆ’4910(-5) + 1 = -50 + 1 = -49 So, βˆ’49=βˆ’49-49 = -49. It works!

See? No matter what value we choose for 'z', the equation holds true. This is because the simplified form of the equation, 10z+1=10z+110z + 1 = 10z + 1, shows that both sides are fundamentally the same expression. When you subtract 10z10z from both sides, you're left with 1=11 = 1. This means the original equation is true for all real numbers. So, the solution isn't a single number, but rather the entire set of real numbers. It's a powerful concept in algebra, showing that sometimes the