AllRounder.ai

Enrol to start learning

Reading is open to everyone. Enrolling is free, and it is what unlocks the audio lessons, practice tests and progress tracking.

Enrol free

2.3. Reducing Equations to Simpler Form

Interactive Audio Lesson

Session 1: Introduction to Reducing Equations

Unlock the classroom podcast

The transcript is above and free to read. A free account plays the conversation back.

Create a free account
Sarah
SarahInstructor

Today, we will start with the concept of reducing equations to simpler forms. Can anyone tell me why it’s important to simplify equations?

Noah
Noah

It makes them easier to solve!

Sarah
SarahInstructor

Exactly! Simplifying reduces complexity. For instance, consider the equation 6x + 1 = x - 3. Wouldn't it be easier if there were no fractions?

Isabella
Isabella

Yes! How do we remove them?

Sarah
SarahInstructor

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.

Session 2: Finding Common Denominators

Unlock the classroom podcast

The transcript is above and free to read. A free account plays the conversation back.

Create a free account
Robert
RobertInstructor

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?

Akash
Akash

We would do 6(6x + 1/3) = 6(x - 3/6)?

Robert
RobertInstructor

Correct! And what do we get on the left?

Ananya
Ananya

It becomes 12x + 2!

Robert
RobertInstructor

Exactly! Now it’s simplified. Now, we can solve for x easily.

Session 3: Combining Like Terms

Unlock the classroom podcast

The transcript is above and free to read. A free account plays the conversation back.

Create a free account
Sarah
SarahInstructor

After eliminating fractions, what’s the next step?

Noah
Noah

We should combine like terms!

Sarah
SarahInstructor

Correct! For example, let's simplify the equation 12x + 8 = x - 3. What do we get when we move x terms to one side?

Isabella
Isabella

11x + 8 = -3!

Sarah
SarahInstructor

Exactly! Further simplifications lead us to our solution.

Session 4: Example Problems

Unlock the classroom podcast

The transcript is above and free to read. A free account plays the conversation back.

Create a free account
Robert
RobertInstructor

Let's solve an example together. Reduce the equation 5x - 2(2x - 7) = 2(3x - 1). What’s our first step?

Akash
Akash

First, we open the brackets!

Robert
RobertInstructor

Right! And we’ll also combine like terms. What do we end up with?

Ananya
Ananya

We should end with a linear equation!

Robert
RobertInstructor

Exactly! This practice is vital in solving real-world problems.

Session 5: Review of Topics

Unlock the classroom podcast

The transcript is above and free to read. A free account plays the conversation back.

Create a free account
Sarah
SarahInstructor

To wrap up, can someone list the main steps we took today?

Noah
Noah

Find the LCM and multiply, then combine like terms.

Isabella
Isabella

Check our solutions, and make sure LHS equals RHS.

Sarah
SarahInstructor

Perfect! Remember, simplifying helps us grasp harder problems too.

Overview

Short Summary

This section explains how to reduce equations, particularly those involving fractions and variables, to simpler forms for easier solution finding.

Medium Summary

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 Summary

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

4x+73=6x+56\frac{4x + 7}{3} = \frac{6x + 5}{6}

Solution: Multiply both sides of the equation by 6, 6(4x+73)=6(6x+56)6 \cdot \left(\frac{4x + 7}{3}\right) = 6 \cdot \left(\frac{6x + 5}{6}\right)

or 2(4x+7)=6x+52(4x + 7) = 6x + 5

or 8x+14=6x+58x + 14 = 6x + 5

or 2x=92x = -9

or x=92x = -\frac{9}{2}

(required solution)

Example 2: Solve

7x+24=5x13\frac{7x + 2}{4} = \frac{5x - 1}{3}

Solution: Multiply both sides of the equation by 12, 12(7x+24)=12(5x13)12 \cdot \left(\frac{7x + 2}{4}\right) = 12 \cdot \left(\frac{5x - 1}{3}\right)

or 3(7x+2)=4(5x1)3(7x + 2) = 4(5x - 1)

or 21x+6=20x421x + 6 = 20x - 4

or x=10x = -10

(required solution)

Check:

LHS = 7(10)+24=70+24=684=17\frac{7(-10) + 2}{4} = \frac{-70 + 2}{4} = \frac{-68}{4} = -17

RHS = 5(10)13=5013=513=17\frac{5(-10) - 1}{3} = \frac{-50 - 1}{3} = \frac{-51}{3} = -17

Thus, LHS = RHS = (as required)

Example : Solve 4x3(2x+5)=3(4x)+144x - 3(2x + 5) = 3(4 - x) + 14
Solution:
Let’s open the brackets.

LHS=4x6x15(expanding the left side)\text{LHS} = 4x - 6x - 15 \quad \text{(expanding the left side)} RHS=123x+14(expanding the right side)\text{RHS} = 12 - 3x + 14 \quad \text{(expanding the right side)}

The equation is 2x15=263x -2x - 15 = 26 - 3x

(transposing terms)\text{(transposing terms)}

Check: LHS=2x15\text{LHS} = -2x - 15

RHS=263x(as required)\text{RHS} = 26 - 3x \quad \text{(as required)}

Therefore, required solution is: x=1x = 1

Reference YouTube Videos

Audio Book

Voice:
Introduction to Simplifying Equations

Unlock the audio lesson

The script is above and free to read. A free account plays it back, in the voice you pick.

Create a free account

Example 16: Solve +1=6x+1 x−3 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 the audio lesson

The script is above and free to read. A free account plays it back, in the voice you pick.

Create a free account

Solution: Multiplying both sides of the equation by 6, we get 6(6x+1) = 6(x−3).

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 the audio lesson

The script is above and free to read. A free account plays it back, in the voice you pick.

Create a free account

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 – 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 the audio lesson

The script is above and free to read. A free account plays it back, in the voice you pick.

Create a free account

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 the audio lesson

The script is above and free to read. A free account plays it back, in the voice you pick.

Create a free account

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 the audio lesson

The script is above and free to read. A free account plays it back, in the voice you pick.

Create a free account

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. It’s like a final proof that keeps you confident about your findings.

--

Key Concepts

Core takeaways and short definitions to help you quickly recall the key ideas from this section.

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

Step-by-step examples to apply the section's ideas and test your understanding.

1

Example 1: Solve 6x + 1/3 = x - 3/6 by multiplying through by 6.

2

Example 2: Reduce and solve 5x - 2(2x - 7) = 2(3x - 1) by combining like terms.

Memory Aids

Interactive tools to help you remember key concepts

🎵

Rhymes

To find the LCM, don't delay, multiply the smallest, that’s the way!
📖

Stories

Once in math land, LCM was a helpful knight, clearing fractions left and right!
🧠

Memory Tools

Use ‘Reduce, Multiply, Combine’ to help you remember the steps!
🎯

Acronyms

UCMC

Use

Combine

Multiply

Check – for solving equations!

Flash Cards

Glossary

Algebraic Equation

An equation composed of variables and constants, linked by an equality sign.

Linear Equation

An equation in which the highest power of the variable is one.

Least Common Multiple (LCM)

The smallest multiple that is evenly divisible by each of the denominators.

Combine Like Terms

The process of simplifying expressions by adding or subtracting terms with the same variable raised to the same power.