Interactive Audio Lesson
Listen to a student-teacher conversation explaining the topic in a relatable way.
Introduction to Reducing Equations
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Today, we will start with the concept of reducing equations to simpler forms. Can anyone tell me why itβs important to simplify equations?
It makes them easier to solve!
Exactly! Simplifying reduces complexity. For instance, consider the equation 6x + 1 = x - 3. Wouldn't it be easier if there were no fractions?
Yes! How do we remove them?
Great question! One method is to find the least common multiple, or LCM, of the denominators. In our case, we would multiply both sides by 6.
Finding Common Denominators
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Now let's look at how we can apply the LCM. For 6x + 1/3 = x - 3/6, the LCM is 6. So, what happens when we multiply through by 6?
We would do 6(6x + 1/3) = 6(x - 3/6)?
Correct! And what do we get on the left?
It becomes 12x + 2!
Exactly! Now itβs simplified. Now, we can solve for x easily.
Combining Like Terms
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
After eliminating fractions, whatβs the next step?
We should combine like terms!
Correct! For example, let's simplify the equation 12x + 8 = x - 3. What do we get when we move x terms to one side?
11x + 8 = -3!
Exactly! Further simplifications lead us to our solution.
Example Problems
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Let's solve an example together. Reduce the equation 5x - 2(2x - 7) = 2(3x - 1). Whatβs our first step?
First, we open the brackets!
Right! And weβll also combine like terms. What do we end up with?
We should end with a linear equation!
Exactly! This practice is vital in solving real-world problems.
Review of Topics
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
To wrap up, can someone list the main steps we took today?
Find the LCM and multiply, then combine like terms.
Check our solutions, and make sure LHS equals RHS.
Perfect! Remember, simplifying helps us grasp harder problems too.
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
In this section, we will learn about techniques for simplifying equations, especially those that contain fractions. Strategies like finding the least common multiple (LCM) and rearranging terms will be discussed. Several examples illustrate how to approach these equations systematically.
Detailed
Detailed Summary
This section focuses on methods to simplify and solve equations, particularly those containing fractions and multiple terms. Simplifying equations can often involve opening brackets, rearranging terms, and finding a common denominator. The examples provided guide students through these processes, showing how to multiply both sides of an equation by the least common multiple (LCM) of denominators to eliminate fractions, and how to effectively combine like terms for clarity. Through systematic steps, students learn to transform complicated expressions into a linear form that is easier to handle.
Example 1: Solve
$$\frac{4x + 7}{3} = \frac{6x + 5}{6}$$
Solution: Multiply both sides of the equation by 6,
$$6 \cdot \left(\frac{4x + 7}{3}\right) = 6 \cdot \left(\frac{6x + 5}{6}\right)$$
or
$$2(4x + 7) = 6x + 5$$
or
$$8x + 14 = 6x + 5$$
or
$$2x = -9$$
or
$$x = -\frac{9}{2}$$
(required solution)
Example 2: Solve
$$\frac{7x + 2}{4} = \frac{5x - 1}{3}$$
Solution: Multiply both sides of the equation by 12,
$$12 \cdot \left(\frac{7x + 2}{4}\right) = 12 \cdot \left(\frac{5x - 1}{3}\right)$$
or
$$3(7x + 2) = 4(5x - 1)$$
or
$$21x + 6 = 20x - 4$$
or
$$x = -10$$
(required solution)
Check:
LHS = $$\frac{7(-10) + 2}{4} = \frac{-70 + 2}{4} = \frac{-68}{4} = -17$$
RHS = $$\frac{5(-10) - 1}{3} = \frac{-50 - 1}{3} = \frac{-51}{3} = -17$$
Thus, LHS = RHS = (as required)
Example : Solve \( 4x - 3(2x + 5) = 3(4 - x) + 14 \)
Solution:
Letβs open the brackets.
\[
\text{LHS} = 4x - 6x - 15 \quad \text{(expanding the left side)}
\]
\[
\text{RHS} = 12 - 3x + 14 \quad \text{(expanding the right side)}
\]
The equation is \[
-2x - 15 = 26 - 3x \]
\[
\text{(transposing terms)}
\]
Check: \( \text{LHS} = -2x - 15 \)
\[
\text{RHS} = 26 - 3x \quad \text{(as required)}
\]
Therefore, required solution is: \( x = 1 \)
Youtube Videos
Audio Book
Dive deep into the subject with an immersive audiobook experience.
Introduction to Simplifying Equations
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Example 16: Solve +1=6x+1 xβ3 3 6 Why 6? Because it is the smallest multiple (or LCM) of the given denominators.
Detailed Explanation
In this example, we are given an equation with fractions. The equation says that some expression related to 'x' equals another. The fractions have denominators of 3 and 6. To solve the equation more easily, we can find a number called the Least Common Multiple (LCM) for the denominators. Here, the LCM of 3 and 6 is 6. This step will help eliminate the fractions when we multiply through by this number.
Examples & Analogies
Imagine you are trying to find a common time to meet with friends who have different schedules. You need to find a time that works for everyone. Similarly, finding the LCM is like finding that common time, so we can work with the equation more easily.
Multiplying Both Sides of the Equation
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Solution: Multiplying both sides of the equation by 6, we get 6(6x+1) = 6(xβ3).
Detailed Explanation
To eliminate the fractions, we multiplied every term in the equation by 6. This operation keeps the equation balanced because whatever we do to one side, we must do to the other. After this multiplication, the equation will be easier to work with since there won't be any fractions left. The left side becomes 6 times '6x + 1' while the right side becomes 6 times '(x - 3)'.
Examples & Analogies
Think of a balance scale. If you have weights on both sides and you want to add more weight to one side, you must add the same amount to the other side to keep it balanced. In the same way, we multiply both sides by the same number (6 here) to keep the equations balanced.
Opening the Brackets
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
or 2 (6x + 1) + 6 = x - 3 or 12x + 2 + 6 =x - 3 (opening the brackets).
Detailed Explanation
After multiplying through, we opened the parentheses and simplified the equation further. Opening the brackets means distributing the multiplied factor (in this case, 6) to each term inside the parenthesis. When we did this, we collected like terms to get a simpler representation of our equation.
Examples & Analogies
Imagine unwrapping a present that is enclosed in several boxes. Opening the outer boxes allows you to see the smaller boxes inside β it makes the whole thing simpler to understand. Similarly, opening the brackets helps us simplify the equation and see the terms clearly.
Combining Like Terms
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
or 12x + 8 = x - 3 or 12x - x + 8 = -3.
Detailed Explanation
After combining like terms on both sides of the equation, we brought all 'x' terms to one side and constant terms to the other side. This process involves arithmetic where we add or subtract terms that are similar. This gives us a clearer view of our equation and moves us closer to finding the solution for 'x'.
Examples & Analogies
Imagine sorting your toys into two piles: one for cars and another for action figures. Combining like terms is like gathering all similar items into one place, making it easier to count or organize them just as we do with 'x' terms and constants in an equation.
Isolating the Variable
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
or 11x = -3 - 8 or 11x = -11 or x = -1 (required solution).
Detailed Explanation
Finally, we isolate 'x' by performing operations that allow us to get 'x' by itself on one side. This includes dividing both sides by the coefficient of 'x'. Here, we divided by 11 to find that 'x' equals -1. This is the solution to our equation.
Examples & Analogies
Think of trying to find out how many cookies someone has when you already know the total number of cookies in a jar. To find that specific count, you separate the unknown amount from the total. This process of isolating is similar to how we isolate 'x' to find its value.
Verification of the Solution
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Check: LHS = +1= +1 = + = = 3 3 3 3 3 3 LHS = RHS. (as required).
Detailed Explanation
After finding the solution, it's essential to verify if it is correct by substituting 'x' back into the original equation. We do this by replacing 'x' with -1 and checking if both sides of the equation are equal. If they are, we confirm that our solution is valid.
Examples & Analogies
Just like checking your work on a math test, verifying your solution ensures that the answer is correct. Itβs like a final proof that keeps you confident about your findings.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
-
Reducing Equations: Simplifying complex equations for easier solving.
-
Using LCM: Finding the least common multiple to eliminate fractions.
-
Combining Like Terms: Rearranging and simplifying equations.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
-
Example 1: Solve 6x + 1/3 = x - 3/6 by multiplying through by 6.
-
Example 2: Reduce and solve 5x - 2(2x - 7) = 2(3x - 1) by combining like terms.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
-
To find the LCM, don't delay, multiply the smallest, thatβs the way!
π Fascinating Stories
-
Once in math land, LCM was a helpful knight, clearing fractions left and right!
π§ Other Memory Gems
-
Use βReduce, Multiply, Combineβ to help you remember the steps!
π― Super Acronyms
UCMC
- Use
- Combine
- Multiply
- Check β for solving equations!
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
-
Term: Algebraic Equation
Definition:
An equation composed of variables and constants, linked by an equality sign.
-
Term: Linear Equation
Definition:
An equation in which the highest power of the variable is one.
-
Term: Least Common Multiple (LCM)
Definition:
The smallest multiple that is evenly divisible by each of the denominators.
-
Term: Combine Like Terms
Definition:
The process of simplifying expressions by adding or subtracting terms with the same variable raised to the same power.