Solving Proportions A Step-by-Step Guide To (6x-2)/4 = (2x-10)/5

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Hey guys! Today, we're diving deep into the world of proportions, specifically tackling the equation 6xβˆ’24=2xβˆ’105{ \frac{6x-2}{4} = \frac{2x-10}{5} }. Proportions might seem intimidating at first, but don't worry! We'll break it down step-by-step, making sure you not only understand how to solve this particular problem but also grasp the underlying concepts. So, grab your pencils, and let's get started!

Understanding Proportions

Before we jump into the nitty-gritty of solving this equation, let's quickly recap what proportions are all about. Proportions are essentially statements that two ratios are equal. A ratio, in turn, is a comparison of two quantities. Think of it like this: if you're baking a cake, the ratio of flour to sugar is crucial. If you change that ratio, you change the whole cake!

In mathematical terms, a proportion looks like this: ab=cd{ \frac{a}{b} = \frac{c}{d} }, where a, b, c, and d are numbers (and b and d are not zero, because we can't divide by zero, right?). This equation is saying that the ratio of a to b is the same as the ratio of c to d. When we're faced with a proportion where one of the terms is unknown (like our x in the equation), we need to solve the proportion to find that unknown value. The beauty of proportions lies in their ability to help us solve a wide range of problems, from scaling recipes to calculating distances on a map. This foundational understanding is critical because proportions are used everywhere in math and real-life situations. To really solidify this, imagine you're scaling up a recipe. If the original recipe calls for 2 cups of flour and 1 cup of sugar, you're dealing with a 2:1 ratio. If you want to double the recipe, you'll need to maintain that same ratio, meaning you'll use 4 cups of flour and 2 cups of sugar. That’s a proportion in action! And the most common method we use for solving these proportions is cross-multiplication, which we’ll explore in detail in the next section.

The Cross-Multiplication Method

The most common and effective way to solve proportions is by using a technique called cross-multiplication. This method is a lifesaver because it transforms a proportion, which can look a bit scary with its fractions, into a simple linear equation that's much easier to handle.

So, how does cross-multiplication work? Well, if we have a proportion like ab=cd{ \frac{a}{b} = \frac{c}{d} }, cross-multiplication tells us that aΓ—d=bΓ—c{ a \times d = b \times c }. In other words, we multiply the numerator of the first fraction by the denominator of the second fraction, and we set that equal to the product of the denominator of the first fraction and the numerator of the second fraction. It's like drawing an 'X' across the equals sign, multiplying along each diagonal. The real power of cross-multiplication is that it gets rid of the fractions, making the equation much simpler to manipulate. This is a key step in solving for the unknown variable. Think about it: fractions can be a pain to work with, but once they're gone, we're left with a straightforward equation that we can solve using basic algebraic techniques. For instance, consider the proportion 23=x6{ \frac{2}{3} = \frac{x}{6} }. By cross-multiplying, we get 2Γ—6=3Γ—x{ 2 \times 6 = 3 \times x }, which simplifies to 12=3x{ 12 = 3x }. Now, we just need to divide both sides by 3 to find that x=4{ x = 4 }. See how much easier that was without the fractions? This same principle applies to more complex proportions, like the one we’re tackling today. We'll apply cross-multiplication to 6xβˆ’24=2xβˆ’105{ \frac{6x-2}{4} = \frac{2x-10}{5} }, and you'll see how it neatly transforms the equation into something much more manageable. Remember, the goal is always to isolate the variable, and cross-multiplication is a powerful tool in our arsenal for achieving that.

Solving 6xβˆ’24=2xβˆ’105{ \frac{6x-2}{4} = \frac{2x-10}{5} } Step-by-Step

Okay, let's get our hands dirty and solve this proportion! We're going to take it one step at a time, so you can see exactly how it's done. Remember, the equation we're working with is 6xβˆ’24=2xβˆ’105{ \frac{6x-2}{4} = \frac{2x-10}{5} }.

Step 1: Cross-multiply. This is where the magic happens. We multiply the numerator of the first fraction by the denominator of the second fraction, and vice versa. This gives us:

(6xβˆ’2)Γ—5=(2xβˆ’10)Γ—4{ (6x - 2) \times 5 = (2x - 10) \times 4 }

Step 2: Distribute. Now, we need to get rid of those parentheses. We do this by distributing the numbers on the outside of the parentheses to each term inside. So, we have:

30xβˆ’10=8xβˆ’40{ 30x - 10 = 8x - 40 }

Notice how the fractions have disappeared? We're now dealing with a linear equation, which is much more familiar territory. This distribution step is crucial because it allows us to separate the terms and start isolating x. If we skipped this step, we'd be stuck with those parentheses, and the equation would remain difficult to solve. Think of distribution as unlocking the equation, freeing up the terms so we can rearrange them. For example, on the left side, we multiplied 5 by both 6x{ 6x } and -2, resulting in 30x{ 30x } and -10, respectively. Similarly, on the right side, we multiplied 4 by both 2x{ 2x } and -10, giving us 8x{ 8x } and -40. This careful distribution ensures that we maintain the equality of the equation while simplifying its structure. Next, we’ll want to group like terms, and that’s where the next step comes in handy.

Step 3: Group like terms. Our goal is to get all the x terms on one side of the equation and all the constant terms on the other side. Let's subtract 8x{ 8x } from both sides:

30xβˆ’8xβˆ’10=8xβˆ’8xβˆ’40{ 30x - 8x - 10 = 8x - 8x - 40 }

22xβˆ’10=βˆ’40{ 22x - 10 = -40 }

Then, add 10 to both sides:

22xβˆ’10+10=βˆ’40+10{ 22x - 10 + 10 = -40 + 10 }

22x=βˆ’30{ 22x = -30 }

This step of grouping like terms is essential for isolating the variable. It’s like sorting your laundry – you want to put all the socks together, all the shirts together, and so on. In our equation, we're grouping the x terms and the constant terms. Subtracting 8x{ 8x } from both sides allows us to move the x term from the right side to the left side. Adding 10 to both sides does the same for the constant term, moving it from the left side to the right side. Remember, whatever we do to one side of the equation, we must do to the other side to maintain balance. This principle of equality is fundamental to solving equations. Each operation we perform is carefully chosen to bring us closer to isolating x. So, after this step, we've simplified the equation to a point where we have the x term on one side and a constant on the other, making the final step of solving for x much clearer. This organized approach makes complex equations less daunting and more manageable. Let's finalize this and get x all by itself.

Step 4: Solve for x. Finally, to isolate x, we divide both sides by 22:

22x22=βˆ’3022{ \frac{22x}{22} = \frac{-30}{22} }

x=βˆ’3022{ x = \frac{-30}{22} }

Step 5: Simplify the fraction. We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

x=βˆ’1511{ x = \frac{-15}{11} }

So, the solution to the proportion is x=βˆ’1511{ x = \frac{-15}{11} }. This final step of simplifying the fraction is crucial because it presents the solution in its most basic form. Fractions, like any mathematical expression, should always be simplified to their lowest terms. This makes them easier to understand and compare. In our case, dividing both -30 and 22 by 2 gives us -15 and 11, respectively. The fraction βˆ’1511{ \frac{-15}{11} } cannot be simplified further because 15 and 11 have no common factors other than 1. Presenting the solution in this simplified form demonstrates a thorough understanding of fraction manipulation and a commitment to accuracy. It also makes the solution easier to interpret and use in further calculations if needed. Remember, simplifying fractions is not just about getting the right answer; it's about presenting that answer in the clearest and most elegant way possible.

Checking Your Solution

Before we celebrate too much, it's always a fantastic idea to check our solution. This way, we can be absolutely sure we haven't made any mistakes along the way. To check our solution, we'll substitute x=βˆ’1511{ x = \frac{-15}{11} } back into the original equation:

6xβˆ’24=2xβˆ’105{ \frac{6x-2}{4} = \frac{2x-10}{5} }

Substitute x=βˆ’1511{ x = \frac{-15}{11} }:

6(βˆ’1511)βˆ’24=2(βˆ’1511)βˆ’105{ \frac{6(\frac{-15}{11})-2}{4} = \frac{2(\frac{-15}{11})-10}{5} }

Let's simplify each side separately.

Left side:

6(βˆ’1511)βˆ’24=βˆ’9011βˆ’24{ \frac{6(\frac{-15}{11})-2}{4} = \frac{\frac{-90}{11}-2}{4} }

To combine the terms in the numerator, we need a common denominator. So, we rewrite 2 as 2211{ \frac{22}{11} }:

βˆ’9011βˆ’22114=βˆ’112114{ \frac{\frac{-90}{11}-\frac{22}{11}}{4} = \frac{\frac{-112}{11}}{4} }

Now, we divide by 4, which is the same as multiplying by 14{ \frac{1}{4} }:

βˆ’11211Γ—14=βˆ’11244{ \frac{-112}{11} \times \frac{1}{4} = \frac{-112}{44} }

Simplify the fraction by dividing both numerator and denominator by 4:

βˆ’2811{ \frac{-28}{11} }

Right side:

2(βˆ’1511)βˆ’105=βˆ’3011βˆ’105{ \frac{2(\frac{-15}{11})-10}{5} = \frac{\frac{-30}{11}-10}{5} }

Rewrite 10 as 11011{ \frac{110}{11} }:

βˆ’3011βˆ’110115=βˆ’140115{ \frac{\frac{-30}{11}-\frac{110}{11}}{5} = \frac{\frac{-140}{11}}{5} }

Divide by 5 (multiply by 15{ \frac{1}{5} }):

βˆ’14011Γ—15=βˆ’14055{ \frac{-140}{11} \times \frac{1}{5} = \frac{-140}{55} }

Simplify the fraction by dividing both numerator and denominator by 5:

βˆ’2811{ \frac{-28}{11} }

Comparison:

Left side: βˆ’2811{ \frac{-28}{11} }

Right side: βˆ’2811{ \frac{-28}{11} }

Since both sides are equal, our solution x=βˆ’1511{ x = \frac{-15}{11} } is correct! This verification process is invaluable. It not only confirms the accuracy of our solution but also reinforces our understanding of the steps involved. By substituting the solution back into the original equation and simplifying, we ensure that the equality holds true. If the two sides didn't match, we'd know there was an error somewhere in our calculations, and we could go back and review each step. Checking our work is a critical habit to develop in mathematics, as it builds confidence and prevents careless mistakes from going unnoticed. So, always remember to check your solutions – it's the final piece of the puzzle!

Conclusion

Awesome! We've successfully solved the proportion 6xβˆ’24=2xβˆ’105{ \frac{6x-2}{4} = \frac{2x-10}{5} }, and we've done it with a clear understanding of each step. Remember, the key to solving proportions is cross-multiplication, followed by careful algebraic manipulation to isolate the variable. And don't forget to check your answer! Proportions are fundamental in mathematics, and mastering them will open doors to more advanced concepts. Keep practicing, and you'll become a proportion pro in no time! This comprehensive approach, from understanding the basics of proportions to the final check, equips you with the skills and confidence to tackle any proportion problem that comes your way. And the best part is, these skills are transferable to many other areas of math and even real-life situations where ratios and proportions are used. So, keep honing your skills, and you'll find math becoming less daunting and more engaging. You've got this!