How To Determine The Values Of A, B, And C In Algebraic Equations
Hey guys! 👋 Ever stumbled upon a math problem that looks like a cryptic puzzle? You're not alone! Today, we're diving deep into determining the values of A, B, and C in algebraic expressions. This might seem daunting at first, but trust me, with a step-by-step approach, it's totally conquerable. We'll break down three different scenarios and equip you with the skills to tackle similar problems with confidence. So, grab your pencils, and let's get started! 🚀
a. (2x + 1)(3x - 7) = Ax² + Bx - 7
In this first scenario, we're presented with an equation that involves expanding binomials and then comparing coefficients. Our mission? To find the values of A and B that make the equation true. Let's break it down step by step:
Step 1: Expand the Left Side
The first thing we need to do is expand the left side of the equation, which is (2x + 1)(3x - 7). Remember the FOIL method? (First, Outer, Inner, Last) It's our best friend here.
- First: (2x)(3x) = 6x²
- Outer: (2x)(-7) = -14x
- Inner: (1)(3x) = 3x
- Last: (1)(-7) = -7
Now, let's add these terms together: 6x² - 14x + 3x - 7. Combining the like terms (-14x and 3x), we get:
6x² - 11x - 7
Step 2: Compare Coefficients
Now we have the expanded form of the left side: 6x² - 11x - 7. Our original equation is (2x + 1)(3x - 7) = Ax² + Bx - 7. We've simplified the left side to 6x² - 11x - 7, so we can rewrite the equation as:
6x² - 11x - 7 = Ax² + Bx - 7
The key here is to compare the coefficients of the corresponding terms on both sides of the equation. Coefficients are the numbers that multiply the variables (like x² and x).
- The coefficient of x² on the left side is 6, and on the right side, it's A. So, A = 6. 🎉
- The coefficient of x on the left side is -11, and on the right side, it's B. So, B = -11. 🎉
- The constant term (the number without any x) is -7 on both sides, which confirms that our expansion and comparison are correct.
Step 3: The Solution
And there you have it! We've successfully determined the values of A and B:
- A = 6
- B = -11
This problem demonstrates the power of expanding expressions and comparing coefficients. It's a fundamental technique in algebra, and you'll find it super useful in many different scenarios. Remember, the key is to break down the problem into smaller, manageable steps. Expanding, simplifying, and comparing – that's the name of the game!
Why is this important, you ask? Well, these skills aren't just for textbooks. They pop up in various fields, from engineering to computer science, where manipulating algebraic expressions is crucial for solving real-world problems. So, mastering this technique now will set you up for success down the road. 👍
b. A(x - 2)(x + 1) + B(x - 1)(x + 3) + C(x + 2)(x - 3) = 9x + 17
Okay, team, let's tackle our second challenge! This one looks a bit more complex, but don't sweat it. We'll use a similar strategy: expand, simplify, and compare. This time, we have three unknowns – A, B, and C – which means we'll need a slightly different approach to solve for them. The core idea remains the same, though: we need to manipulate the equation to a point where we can isolate and determine these values.
Step 1: Expand the Terms
The first order of business is to expand each term on the left side of the equation. Remember our trusty FOIL method? It's time to put it back into action!
- A(x - 2)(x + 1):
- (x - 2)(x + 1) = x² + x - 2x - 2 = x² - x - 2
- So, A(x - 2)(x + 1) = Ax² - Ax - 2A
- B(x - 1)(x + 3):
- (x - 1)(x + 3) = x² + 3x - x - 3 = x² + 2x - 3
- So, B(x - 1)(x + 3) = Bx² + 2Bx - 3B
- C(x + 2)(x - 3):
- (x + 2)(x - 3) = x² - 3x + 2x - 6 = x² - x - 6
- So, C(x + 2)(x - 3) = Cx² - Cx - 6C
Step 2: Combine Like Terms
Now that we've expanded each term, let's substitute these expressions back into our original equation:
Ax² - Ax - 2A + Bx² + 2Bx - 3B + Cx² - Cx - 6C = 9x + 17
Next, we need to group the terms with the same powers of x together:
(Ax² + Bx² + Cx²) + (-Ax + 2Bx - Cx) + (-2A - 3B - 6C) = 9x + 17
Now, factor out the x² and x:
(A + B + C)x² + (-A + 2B - C)x + (-2A - 3B - 6C) = 9x + 17
Step 3: Equate Coefficients and Solve the System of Equations
This is where things get interesting! We now have a polynomial on the left side that we can directly compare to the polynomial on the right side (which, in this case, is 9x + 17). Remember, for two polynomials to be equal, their corresponding coefficients must be equal.
So, we can set up a system of equations by equating the coefficients:
- Coefficient of x²: A + B + C = 0 (Since there is no x² term on the right side, its coefficient is 0)
- Coefficient of x: -A + 2B - C = 9
- Constant term: -2A - 3B - 6C = 17
We now have a system of three linear equations with three unknowns. There are several ways to solve this system (substitution, elimination, matrices), but let's use elimination:
- Add the first and second equations:
- (A + B + C) + (-A + 2B - C) = 0 + 9
- 3B = 9
- B = 3 🎉
- Substitute B = 3 into the first and third equations:
- A + 3 + C = 0 => A + C = -3
- -2A - 3(3) - 6C = 17 => -2A - 6C = 26 => A + 3C = -13
- Subtract the equation A + C = -3 from A + 3C = -13:
- (A + 3C) - (A + C) = -13 - (-3)
- 2C = -10
- C = -5 🎉
- Substitute C = -5 into A + C = -3:
- A + (-5) = -3
- A = 2 🎉
Step 4: The Solution
Woohoo! We've cracked the code! The values of A, B, and C are:
- A = 2
- B = 3
- C = -5
This problem highlighted the importance of systematically expanding, simplifying, and then setting up a system of equations. These skills are essential for solving more complex algebraic problems, and they lay the foundation for advanced mathematical concepts. Remember, don't be intimidated by the complexity – break it down, step by step, and you'll get there!
Why this is super useful in the real world? Think about any situation where you have multiple interacting factors and you need to figure out the influence of each one. This could be anything from designing a bridge to optimizing a business process. The ability to set up and solve systems of equations is a powerful tool in these scenarios. 💪
c. -4x + 13 = A(2x - 5) + B(3x - 6)
Alright, let's move on to our final challenge! This one might look a bit simpler than the last one, but it still requires us to use our algebraic prowess to determine the values of A and B. Our strategy remains the same: expand, simplify, and then compare coefficients. Let's dive in!
Step 1: Expand the Right Side
First, we need to expand the right side of the equation. This involves distributing A and B across the parentheses:
-4x + 13 = A(2x - 5) + B(3x - 6) -4x + 13 = 2Ax - 5A + 3Bx - 6B
Step 2: Group Like Terms
Now, let's group the terms with 'x' and the constant terms together:
-4x + 13 = (2A + 3B)x + (-5A - 6B)
Step 3: Equate Coefficients and Solve the System of Equations
Just like in the previous problem, we'll now equate the coefficients of the corresponding terms on both sides of the equation. This gives us a system of two linear equations with two unknowns:
- Coefficient of x: 2A + 3B = -4
- Constant term: -5A - 6B = 13
Let's solve this system of equations. We can use either substitution or elimination. This time, let's use elimination:
- Multiply the first equation by 2:
- 2(2A + 3B) = 2(-4)
- 4A + 6B = -8
- Add the modified first equation to the second equation:
- (4A + 6B) + (-5A - 6B) = -8 + 13
- -A = 5
- A = -5 🎉
- Substitute A = -5 into the first equation (2A + 3B = -4):
- 2(-5) + 3B = -4
- -10 + 3B = -4
- 3B = 6
- B = 2 🎉
Step 4: The Solution
Fantastic! We've found the values of A and B:
- A = -5
- B = 2
This problem reinforced the importance of careful expansion, grouping like terms, and solving systems of equations. These are fundamental skills that will serve you well in your mathematical journey. The ability to manipulate algebraic expressions and solve for unknowns is a cornerstone of many scientific and technical disciplines. Remember, practice makes perfect, so keep honing these skills!
Why is this important in the real world? Think about situations where you have a limited amount of resources and you need to distribute them optimally. This kind of problem pops up in fields like economics, finance, and operations research. The skills we used here to solve for A and B can be applied to these kinds of optimization problems. 💡
Final Thoughts
So there you have it, guys! We've successfully navigated three different scenarios of determining the values of A, B, and C. We've seen how to expand expressions, compare coefficients, and solve systems of equations. These are powerful tools in your mathematical arsenal. Remember, the key is to break down complex problems into smaller, manageable steps. Keep practicing, and you'll become algebraic masters in no time! You got this! 💪
