Factoring: How To Factor -3/4 From (3/2)x - (9/4)y

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Hey guys! Today, we're diving into a super useful algebra skill: factoring. Factoring might seem tricky at first, but once you get the hang of it, it's like unlocking a secret code to simplifying expressions. We're going to break down how to factor out a fraction, specifically βˆ’34-\frac{3}{4}, from the expression 32xβˆ’94y\frac{3}{2}x - \frac{9}{4}y. So, grab your pencils, and let's get started!

Understanding Factoring

Before we jump into the problem, let's quickly recap what factoring actually means. Think of it like the reverse of expanding. When you expand, you multiply something into a set of parentheses. Factoring is pulling something out of an expression and writing it outside the parentheses. It's all about finding the common elements in each term of your expression.

In this case, our goal is to factor out βˆ’34-\frac{3}{4} from 32xβˆ’94y\frac{3}{2}x - \frac{9}{4}y. This means we want to rewrite the expression in the form βˆ’34(...)-\frac{3}{4}(...), where the parentheses will contain what's left after we've divided each term by βˆ’34-\frac{3}{4}.

Why is Factoring Important?

You might be wondering, "Why bother with factoring?" Well, factoring is a fundamental skill in algebra and calculus. It helps us:

  • Simplify expressions: Factoring can make complex expressions easier to work with.
  • Solve equations: Factoring is a key step in solving many types of algebraic equations, especially quadratic equations.
  • Identify key features of functions: In higher-level math, factoring helps us find the roots (or zeros) of polynomial functions.

So, mastering factoring now will definitely pay off down the road!

Step-by-Step Guide to Factoring βˆ’34-\frac{3}{4}

Okay, let's get down to business. Here’s how to factor βˆ’34-\frac{3}{4} from the expression 32xβˆ’94y\frac{3}{2}x - \frac{9}{4}y. We'll take it one step at a time to make sure everything's crystal clear.

Step 1: Identify the Term to Factor Out

The first step is recognizing what we need to factor out. In this problem, it's βˆ’34-\frac{3}{4}. We want to rewrite our expression so that βˆ’34-\frac{3}{4} is outside a set of parentheses, like this:

βˆ’34(...)-\frac{3}{4}(...)

Step 2: Divide Each Term by βˆ’34-\frac{3}{4}

This is the heart of the factoring process. We need to divide each term in the original expression, 32x\frac{3}{2}x and βˆ’94y-\frac{9}{4}y, by βˆ’34-\frac{3}{4}. Remember, dividing by a fraction is the same as multiplying by its reciprocal (flipping the fraction). Let's do each term separately:

Dividing 32x\frac{3}{2}x by βˆ’34-\frac{3}{4}

32xΓ·(βˆ’34)=32xβ‹…(βˆ’43)\frac{3}{2}x \div \left(-\frac{3}{4}\right) = \frac{3}{2}x \cdot \left(-\frac{4}{3}\right)

Now, we multiply the fractions. Remember, when multiplying fractions, you multiply the numerators (top numbers) and the denominators (bottom numbers):

32β‹…(βˆ’43)=3β‹…(βˆ’4)2β‹…3=βˆ’126=βˆ’2\frac{3}{2} \cdot \left(-\frac{4}{3}\right) = \frac{3 \cdot (-4)}{2 \cdot 3} = \frac{-12}{6} = -2

So, when we divide 32x\frac{3}{2}x by βˆ’34-\frac{3}{4}, we get βˆ’2x-2x.

Dividing βˆ’94y-\frac{9}{4}y by βˆ’34-\frac{3}{4}

Now let's divide the second term, βˆ’94y-\frac{9}{4}y, by βˆ’34-\frac{3}{4}:

βˆ’94yΓ·(βˆ’34)=βˆ’94yβ‹…(βˆ’43)-\frac{9}{4}y \div \left(-\frac{3}{4}\right) = -\frac{9}{4}y \cdot \left(-\frac{4}{3}\right)

Again, multiply the fractions:

βˆ’94β‹…(βˆ’43)=(βˆ’9)β‹…(βˆ’4)4β‹…3=3612=3-\frac{9}{4} \cdot \left(-\frac{4}{3}\right) = \frac{(-9) \cdot (-4)}{4 \cdot 3} = \frac{36}{12} = 3

So, when we divide βˆ’94y-\frac{9}{4}y by βˆ’34-\frac{3}{4}, we get 3y3y.

Step 3: Write the Factored Expression

Now that we've divided each term by βˆ’34-\frac{3}{4}, we can write the factored expression. We put the results we got in Step 2 inside the parentheses:

βˆ’34(βˆ’2x+3y)-\frac{3}{4}(-2x + 3y)

And that's it! We've successfully factored βˆ’34-\frac{3}{4} from 32xβˆ’94y\frac{3}{2}x - \frac{9}{4}y.

Let's Double-Check Our Work

It's always a good idea to check your work, especially when you're learning something new. We can check our factoring by distributing the βˆ’34-\frac{3}{4} back into the parentheses. If we did it correctly, we should get back our original expression:

βˆ’34(βˆ’2x+3y)=(βˆ’34)(βˆ’2x)+(βˆ’34)(3y)-\frac{3}{4}(-2x + 3y) = \left(-\frac{3}{4}\right)(-2x) + \left(-\frac{3}{4}\right)(3y)

Multiply the fractions:

(βˆ’34)(βˆ’2x)=32x\left(-\frac{3}{4}\right)(-2x) = \frac{3}{2}x

(βˆ’34)(3y)=βˆ’94y\left(-\frac{3}{4}\right)(3y) = -\frac{9}{4}y

Combine the results:

32xβˆ’94y\frac{3}{2}x - \frac{9}{4}y

Look at that! We got back our original expression, which means our factoring was correct. High five!

Common Mistakes to Avoid

Factoring can be a bit tricky, so it's helpful to know some common pitfalls to watch out for:

  • Forgetting to divide every term: Make sure you divide each term in the expression by the factor you're pulling out. It's easy to miss one, especially if you're working with a long expression.
  • Sign errors: Pay close attention to the signs (positive and negative) when you're dividing. A mistake with the signs can throw off your entire answer.
  • Not checking your work: Always, always check your factoring by distributing back into the parentheses. This is the best way to catch errors.

Practice Makes Perfect

The best way to get comfortable with factoring is to practice! Try factoring different expressions with fractions and negative numbers. The more you practice, the more natural the process will become.

Here are a few practice problems you can try:

  1. Factor βˆ’12-\frac{1}{2} from 14aβˆ’32b\frac{1}{4}a - \frac{3}{2}b
  2. Factor βˆ’23-\frac{2}{3} from 43x+89y\frac{4}{3}x + \frac{8}{9}y
  3. Factor βˆ’56-\frac{5}{6} from 52mβˆ’103n\frac{5}{2}m - \frac{10}{3}n

Conclusion

Great job, guys! You've learned how to factor a fraction from an algebraic expression. Factoring βˆ’34-\frac{3}{4} from 32xβˆ’94y\frac{3}{2}x - \frac{9}{4}y might have seemed daunting at first, but by breaking it down step by step, you've conquered it. Remember, factoring is a powerful tool in algebra, so keep practicing, and you'll be a factoring pro in no time! Keep up the awesome work, and I'll catch you in the next math adventure!