Math Problem Solved: Slope-Intercept Form Explained

Hey math enthusiasts! Ever felt like deciphering algebraic equations is like trying to solve a mystery? Well, you're in the right place. Today, we're going to break down a common problem using the slope-intercept form, a fundamental concept in algebra that helps us understand linear relationships. We'll walk through each step, making sure everything is crystal clear, just like a perfectly graphed line. So, grab your notebooks and let's dive into this step-by-step solution, transforming confusion into clarity. By the end of this article, you'll feel more confident tackling similar problems, proving that mathematics can be both accessible and rewarding. We'll explore the 'why' behind each move, not just the 'how,' making this a truly valuable learning experience. Get ready to boost your math skills!

Understanding the Slope-Intercept Form: The Key to Linear Equations

The slope-intercept form of a linear equation is your best friend when it comes to graphing and understanding lines. It's typically written as y = mx + b, where 'm' represents the slope of the line and 'b' represents the y-intercept. Think of 'm' as the steepness of the line – a positive 'm' means the line goes uphill from left to right, while a negative 'm' means it goes downhill. The 'b' value is simply where the line crosses the y-axis. This form is incredibly useful because it directly gives you two crucial pieces of information about the line, making it easy to plot or analyze. For instance, if you have the equation y = 2x + 3, you immediately know the slope is 2 and the y-intercept is 3. This means the line rises 2 units for every 1 unit it moves to the right, and it crosses the y-axis at the point (0, 3). Mastery of this form is essential for many areas of mathematics, from basic algebra to calculus and beyond. It provides a standardized way to represent and compare different linear functions, allowing for deeper insights into their behavior and relationships. The elegance of the slope-intercept form lies in its simplicity and directness, making complex linear relationships easily interpretable. It’s the foundation upon which much of our understanding of linear systems is built.

Step-by-Step Solution: Decoding the Equation

Let's tackle the problem step-by-step, just as presented. Our starting point is the equation: Step 1: 5=-2 imesrac{-3}{10}+b. Here, we're given a specific point (likely an x and y value, though not explicitly stated as coordinates) and a slope, and we need to find the y-intercept, 'b'. The goal is to isolate 'b' to fully define the line using the slope-intercept form. In this first step, we see the value '5' on the left side, which is likely our 'y' value. On the right side, we have $-2$ multiplied by rac{-3}{10}, which represents the $m imes x$ part of the equation, and '+b' is what we want to find. So, the equation is set up to solve for 'b'. This is a typical scenario in algebra where you're given partial information about a line and need to find the missing piece. The numbers themselves might look a bit intimidating at first glance, especially with the fraction, but remember that each operation follows standard arithmetic rules. We're essentially plugging in known values into the general $y = mx + b$ formula and solving for the unknown. This process is designed to be systematic, ensuring accuracy and understanding.

Simplifying the Multiplication: Step 2

Moving on to Step 2: 5=rac{6}{10}+b. The crucial operation here is the multiplication in Step 1. We multiplied $-2$ by rac{-3}{10}. Remember that a negative number multiplied by a negative number results in a positive number. So, -2 imes rac{-3}{10} = rac{-2}{1} imes rac{-3}{10}. When multiplying fractions, you multiply the numerators together and the denominators together: rac{(-2) imes (-3)}{1 imes 10} = rac{6}{10}. This simplification is key. We've successfully combined the two numbers on the right side of the equation that were multiplied, leaving us with a simpler expression. This step demonstrates the importance of carefully applying the rules of multiplication, especially with negative numbers and fractions. Each step builds upon the previous one, moving us closer to our goal of finding 'b'. The reduction of the terms makes the equation much more manageable for the subsequent calculations. It's a common practice in algebra to simplify expressions as you go to avoid errors and make the overall process smoother. The fraction rac{6}{10} can also be simplified to rac{3}{5}, but for the purpose of direct subtraction in the next step, keeping it as rac{6}{10} might be more convenient to match the whole number 5 when converted to tenths. This flexibility in handling fractions is a useful skill to develop.

Isolating 'b': The Subtraction Step (Step 3)

Now we arrive at Step 3: 5-rac{6}{10}=rac{6}{10}-rac{6}{10}+b. Our objective is to get 'b' all by itself on one side of the equation. To do this, we need to eliminate the rac{6}{10} that is being added to 'b'. The mathematical principle we use here is the subtraction property of equality. Whatever we do to one side of the equation, we must do to the other side to maintain the balance. Therefore, we subtract rac{6}{10} from both sides. On the right side, rac{6}{10} - rac{6}{10} equals zero, effectively canceling out the term and leaving just 'b'. On the left side, we perform the subtraction: 5 - rac{6}{10}. This step is fundamental in solving algebraic equations. It shows how we systematically manipulate the equation to isolate the variable we're interested in. The concept of maintaining equality is paramount; without it, the equation would no longer be true. This methodical approach ensures that our solution for 'b' will be accurate.

Finding a Common Denominator: Preparing for Subtraction (Step 4)

Step 4 is rac{50}{10}-rac{6}{10}=b. Before we can accurately subtract rac{6}{10} from 5, we need to express the whole number 5 as a fraction with the same denominator, which is 10. This is known as finding a common denominator. To convert 5 into a fraction with a denominator of 10, we can multiply it by rac{10}{10} (which is equivalent to 1, so it doesn't change the value of 5). So, 5 imes rac{10}{10} = rac{5 imes 10}{1 imes 10} = rac{50}{10}. Now that both terms have the same denominator, we can subtract the numerators while keeping the denominator the same. This step highlights the importance of understanding fraction operations. Working with fractions often requires adjustments like finding common denominators to perform addition or subtraction accurately. This process ensures that we are comparing and combining quantities of the same size. It’s a fundamental skill that underpins much of arithmetic and algebra.

The Final Value of 'b': Step 5

We've reached Step 5: rac{44}{10}=b. This is the result of the subtraction performed in the previous steps. We successfully calculated 5 - rac{6}{10} by converting 5 to rac{50}{10} and then subtracting: rac{50}{10} - rac{6}{10} = rac{50 - 6}{10} = rac{44}{10}. So, the value of our y-intercept, 'b', is rac{44}{10}. This fraction can be simplified further. Both 44 and 10 are divisible by 2, so rac{44}{10} simplifies to rac{22}{5}. Alternatively, as a decimal, rac{44}{10} is equal to 4.4. This is the crucial value that completes our slope-intercept equation. Finding 'b' is often the final step when you're given a slope and a point, as it allows you to write the complete equation of the line. This numerical result is the culmination of all the preceding algebraic manipulations. It's a tangible outcome that allows us to fully describe the linear relationship.

Constructing the Final Equation: Step 6

Finally, we have Step 6: $y=-2x-4.4$. This is the culmination of our entire process. We started with the general slope-intercept form, $y = mx + b$. We were given, or deduced from the problem context, that the slope ($m$) is $-2$. Through our step-by-step calculations, we found the y-intercept ($b$) to be rac{44}{10}, which we converted to the decimal 4.4. Now, we substitute these values back into the $y = mx + b$ formula. So, $y = (-2)x + (4.4)$. However, the provided Step 6 shows $y=-2x-4.4$. Let's re-examine the initial setup. If the initial equation in Step 1 was indeed 5=-2 imesrac{-3}{10}+b, then $b$ calculates to $+4.4$. This would result in the equation $y = -2x + 4.4$. The provided Step 6, $y=-2x-4.4$, suggests that the 'b' value should have been $-4.4$. This discrepancy indicates a potential error in the original problem statement or the provided steps leading to Step 6. Assuming the calculations in Steps 1-5 are correct, the final equation should be $y = -2x + 4.4$. If the intention was for the final equation to be $y=-2x-4.4$, then the initial setup in Step 1 would need to be different to yield a negative 'b' value. For instance, if the point used had a different y-value or if the slope calculation resulted in a negative term that was then added to 'b'. It's important to always double-check the substitution and ensure the final equation accurately reflects the calculated slope and intercept. This final step is where all the pieces come together, providing a complete description of the linear function.

Conclusion: Mastering Linear Equations

We've successfully navigated through the process of finding the equation of a line in slope-intercept form. Each step, from simplifying the multiplication to isolating the variable and finally substituting the values, is crucial. Understanding the slope-intercept form ($y=mx+b$) is a foundational skill in mathematics that unlocks the ability to analyze and visualize linear relationships. Whether you're solving for an unknown intercept, determining the equation of a line given points, or interpreting graphs, this form is indispensable. Remember, math problems are like puzzles; breaking them down into smaller, manageable steps makes them solvable. Practice is key to building confidence and fluency. Keep working through problems, and don't be afraid to ask questions or revisit concepts. The journey of learning mathematics is continuous, and each solved problem is a step forward. If you're looking to deepen your understanding of linear equations and their applications, exploring resources like Khan Academy can provide further explanations, practice exercises, and video tutorials that reinforce these concepts.