Math Equation: Solve For Z In Fractional Equation

Hey guys! Ever stared at an equation that looks like a tangled mess of numbers and variables and thought, "What in the world am I supposed to do here?" Well, you're not alone! Today, we're diving deep into a pretty common type of math problem: solving for a variable in an equation that's loaded with fractions. Specifically, we're going to break down how to tackle this beast: $\frac{2(5 z-3)}{7}-\frac{3(5 z-2)}{5}=\frac{8}{13}$. Don't let those fractions scare you off; we'll walk through it step-by-step, making sure you understand each move. We're talking about linear equations here, which means the highest power of our variable, 'z' in this case, is just one. The trickiest part is dealing with those denominators – the 7, 5, and 13. To make things simpler, we'll use a super handy technique: finding a common denominator. This allows us to combine terms and eventually isolate 'z' so we can find its value. Think of it like tidying up a messy room; you need to put similar things together before you can see what you've got. We'll also cover the importance of distributing those numbers outside the parentheses, like the 2 and the 3, because that's another common spot where mistakes can happen. By the end of this, you'll feel way more confident tackling similar equations. We're going to emphasize clarity and accuracy, showing you precisely how to simplify and solve, so you can impress your math teacher or just feel good about conquering this problem. This isn't just about getting an answer; it's about understanding the process – the logic behind each step that leads us to the solution. So, grab your calculators, maybe a trusty pencil, and let's get this equation solved!

Understanding the Equation: The Anatomy of a Fractional Equation

Alright, let's really get to know the equation we're working with: $\frac{2(5 z-3)}{7}-\frac{3(5 z-2)}{5}=\frac{8}{13}$. First off, notice we've got a variable, 'z', which is what we're trying to find the value of. It's wrapped up inside parentheses, and those parentheses are part of fractions. The core of this problem lies in managing these fractions. We have three denominators: 7, 5, and 13. To solve this equation effectively, our primary goal is to eliminate these denominators. Why? Because working with whole numbers is way easier than constantly juggling fractions. The strategy we'll employ is to multiply both sides of the entire equation by the least common multiple (LCM) of the denominators. This is a golden rule in algebra: whatever you do to one side of an equation, you must do to the other side to maintain the balance. If we multiply each term by the LCM, each denominator will cancel out. Let's figure out the LCM of 7, 5, and 13. Since 7, 5, and 13 are all prime numbers (meaning they are only divisible by 1 and themselves), their LCM is simply their product. So, LCM(7, 5, 13) = 7 * 5 * 13. Calculating that: 7 * 5 = 35, and 35 * 13 = 455. So, 455 is our magic number! We're going to multiply every single term in the equation by 455. Before we jump into the multiplication, it's also crucial to address the parts inside the parentheses. We have $2(5z-3)$ and $3(5z-2)$. This means we need to distribute the numbers outside the parentheses to the terms inside. For $2(5z-3)$, this becomes $(2 \times 5z) - (2 \times 3)$, which simplifies to $10z - 6$. Similarly, for $3(5z-2)$, it becomes $(3 \times 5z) - (3 \times 2)$, simplifying to $15z - 6$. Performing this distribution before clearing the fractions can sometimes make the subsequent steps clearer, though you can also distribute after clearing the fractions. We'll stick with distributing first for this walkthrough. So, our equation, after initial distribution, conceptually looks like $\frac{10z - 6}{7}-\frac{15z - 6}{5}=\frac{8}{13}$. Now, with the distribution done and our LCM (455) identified, we're perfectly poised to clear those pesky denominators and move towards finding the value of 'z'. This initial phase is all about setting up the equation for simplification, and understanding why we're doing each step is key to mastering algebra.

Step-by-Step Solution: Clearing Fractions and Isolating 'z'

Okay guys, we've got our equation $\frac{2(5 z-3)}{7}-\frac{3(5 z-2)}{5}=\frac{8}{13}$, and we've identified that the LCM of our denominators (7, 5, and 13) is 455. We've also mentally prepared to distribute the constants. Now, let's put it all together and actually solve for 'z'. The first major move is to multiply every term in the equation by our LCM, 455. This is the most critical step for eliminating the fractions.

Step 1: Multiply every term by the LCM (455).

$$455 \times \left(\frac{2(5 z-3)}{7}\right) - 455 \times \left(\frac{3(5 z-2)}{5}\right) = 455 \times \left(\frac{8}{13}\right)$$

Now, let's simplify each part. We can simplify before multiplying the numerators fully, which often makes the numbers more manageable.

• For the first term: $455 \div 7 = 65$. So, we have $65 \times 2(5z-3)$.
• For the second term: $455 \div 5 = 91$. So, we have $91 \times 3(5z-2)$.
• For the third term (the right side): $455 \div 13 = 35$. So, we have $35 \times 8$.

Our equation now looks like this:

$$65 \times 2(5z-3) - 91 \times 3(5z-2) = 35 \times 8$$

Step 2: Perform the multiplications and distributions.

Let's simplify further by multiplying the constants with the terms in the parentheses. Remember the distributive property!

• $65 \times 2 = 130$. So, the first term becomes $130(5z-3)$.
• $91 \times 3 = 273$. So, the second term becomes $273(5z-2)$.
• $35 \times 8 = 280$. This is the value of the right side.

Now, distribute these new coefficients:

• $130(5z-3) = (130 \times 5z) - (130 \times 3) = 650z - 390$
• $273(5z-2) = (273 \times 5z) - (273 \times 2) = 1365z - 546$

Substituting these back into our equation:

$$(650z - 390) - (1365z - 546) = 280$$

Step 3: Combine like terms.

This is where we group all the 'z' terms together and all the constant terms together. Pay close attention to the minus sign in front of the second set of parentheses. It means we subtract both terms inside.

$$650z - 390 - 1365z + 546 = 280$$

Combine the 'z' terms: $650z - 1365z = -715z$

Combine the constant terms: $-390 + 546 = 156$

So, our simplified equation is:

$$-715z + 156 = 280$$

Step 4: Isolate the variable term.

To get the '-715z' term by itself, we need to subtract 156 from both sides of the equation:

$$-715z + 156 - 156 = 280 - 156$$

$$-715z = 124$$

Step 5: Solve for 'z'.

Finally, to get 'z' by itself, we divide both sides by the coefficient of 'z', which is -715:

$$\frac{-715z}{-715} = \frac{124}{-715}$$

$$z = -\frac{124}{715}$$

And there you have it! The value of 'z' is $-\frac{124}{715}$. We managed to clear the fractions, simplify, and isolate 'z' step-by-step. It might look complex at first, but by breaking it down and systematically applying the rules of algebra, we get to the answer. Remember to double-check your calculations, especially with the signs and distribution!

Verification: Plugging Our Solution Back In

So, we've landed on our answer: $z = -\frac{124}{715}$. Now, in the world of math, especially when dealing with equations that have this many moving parts, it's always a seriously good idea to verify your solution. This means plugging our calculated value of 'z' back into the original equation and making sure that both sides are, in fact, equal. If they are, we know for sure we've nailed it. If not, well, it's time to go back and find that sneaky little mistake. It's like proofreading your work – essential for accuracy!

Our original equation is: $\frac{2(5 z-3)}{7}-\frac{3(5 z-2)}{5}=\frac{8}{13}$.

Let's substitute $z = -\frac{124}{715}$ into the expression $5z-3$ and $5z-2$ first.

• For $5z-3$: $5\left(-\frac{124}{715}\right) - 3 = -\frac{5 \times 124}{715} - 3$. Since $715 = 5 \times 143$, we can simplify this: $-\frac{124}{143} - 3$. To subtract 3, we need a common denominator, which is 143. So, $-3 = -\frac{3 \times 143}{143} = -\frac{429}{143}$. Therefore, $5z-3 = -\frac{124}{143} - \frac{429}{143} = -\frac{553}{143}$.

• For $5z-2$: $5\left(-\frac{124}{715}\right) - 2 = -\frac{124}{143} - 2$. Again, we need a common denominator of 143. So, $-2 = -\frac{2 \times 143}{143} = -\frac{286}{143}$. Therefore, $5z-2 = -\frac{124}{143} - \frac{286}{143} = -\frac{410}{143}$.

Now, let's plug these results back into the left side of the original equation:

Left Side = $\frac{2(\frac{-553}{143})}{7}-\frac{3(\frac{-410}{143})}{5}$

Left Side = $\frac{\frac{-1106}{143}}{7}-\frac{\frac{-1230}{143}}{5}$

To divide by 7, we multiply by $\frac{1}{7}$: $\frac{-1106}{143 \times 7} = \frac{-1106}{1001}$.

To divide by 5, we multiply by $\frac{1}{5}$: $\frac{-1230}{143 \times 5} = \frac{-1230}{715}$.

So, Left Side = $\frac{-1106}{1001} - \frac{-1230}{715}$.

We need a common denominator for 1001 and 715. Notice that $1001 = 7 \times 11 \times 13$ and $715 = 5 \times 11 \times 13$. The LCM is $5 \times 7 \times 11 \times 13 = 5005$.

• $\frac{-1106}{1001} = \frac{-1106 \times 5}{1001 \times 5} = \frac{-5530}{5005}$
• $\frac{-1230}{715} = \frac{-1230 \times 7}{715 \times 7} = \frac{-8610}{5005}$

So, Left Side = $\frac{-5530}{5005} - \frac{-8610}{5005} = \frac{-5530 + 8610}{5005} = \frac{3080}{5005}$.

Now, let's simplify the fraction $\frac{3080}{5005}$. Both numbers are divisible by 5: $\frac{616}{1001}$. Both are divisible by 7: $\frac{88}{143}$. Both are divisible by 11: $\frac{8}{13}$.

Wow! The left side simplifies to $\frac{8}{13}$.

This matches the right side of our original equation, which was $\frac{8}{13}$.

Since Left Side = Right Side, our solution $z = -\frac{124}{715}$ is correct. This verification process, while sometimes lengthy, is super important for building confidence in your algebraic skills. Nicely done, guys!

Key Takeaways and Tips for Future Equations

Alright team, we've journeyed through a pretty involved fractional equation and emerged victorious! Let's recap what we learned and distill it into some solid takeaways that will help you crush future math problems. The equation we tackled was $\frac{2(5 z-3)}{7}-\frac{3(5 z-2)}{5}=\frac{8}{13}$. The core challenge here was managing those fractions, and we employed a powerful strategy: clearing the denominators.

Key Takeaway 1: Embrace the Least Common Multiple (LCM). The most effective way to deal with equations involving fractions is to eliminate the denominators. We found the LCM of 7, 5, and 13, which was 455. By multiplying every single term on both sides of the equation by this LCM, we transformed the equation into one without fractions. Remember, this principle applies to any equation with fractions. Always find the LCM of the denominators and multiply through. This is your golden ticket to simplification.

Key Takeaway 2: Distribution is Your Friend (and Sometimes Your Foe!). We had coefficients (2 and 3) multiplying expressions within parentheses. It's crucial to distribute these numbers correctly. This means multiplying the coefficient by each term inside the parentheses. For example, $2(5z-3)$ becomes $10z - 6$. Be super careful with negative signs when distributing; a common mistake is forgetting to apply the negative to all terms inside the second set of parentheses, like we saw with $-(1365z - 546)$ becoming $-1365z + 546$. Double-checking your distribution can save you a lot of headache.

Key Takeaway 3: Combine Like Terms Systematically. Once the fractions are gone and distributions are done, you'll be left with an equation that's much simpler, usually with variable terms and constant terms. Group all the variable terms on one side and all the constant terms on the other. In our case, $650z - 1365z$ combined to $-715z$, and $-390 + 546$ combined to $156$. Careful addition and subtraction here are vital. Keep track of your positive and negative signs!

Key Takeaway 4: Isolate and Solve for the Variable. After combining like terms, you'll typically have an equation in the form $Ax = B$ or $Ax + B = C$. The final step is to isolate the variable. This usually involves division. To solve for 'z' in $-715z = 124$, we simply divided both sides by -715, giving us $z = \frac{124}{-715}$, or $z = -\frac{124}{715}$.

Key Takeaway 5: Verification is Non-Negotiable. We spent time plugging our answer back into the original equation. This might seem tedious, but it's the ultimate confirmation that your answer is correct. If you can substitute your solution back into the initial equation and both sides balance out, you've definitely solved it right. It's a powerful self-checking mechanism.

General Tips for Success:

• Read Carefully: Make sure you understand what the question is asking and what operations are involved.
• Show Your Work: Write down every step. This makes it easier to track your logic and to find errors if you make them.
• Stay Organized: Keep your numbers and symbols neat. A messy equation leads to messy thinking.
• Practice, Practice, Practice: The more you practice solving different types of equations, the more comfortable and quicker you'll become.

Dealing with equations like this is a fundamental skill in mathematics. By mastering these steps and tips, you'll be well-equipped to handle even more complex algebraic challenges. Keep up the great work, guys!