Before we dive into solving equations, let's review some basic algebraic concepts and vocabulary.

First, what is an equation? An equation is a mathematical statement that shows that two expressions are equal. For example, 2x + 3 = 7 is an equation because the expression 2x + 3 is equal to 7.

The expressions on either side of the equal sign are called the left-hand side (LHS) and the right-hand side (RHS) of the equation. In the example above, 2x + 3 is the LHS and 7 is the RHS.

The goal of solving an equation is to find the value of the variable(s) that make the equation true. For example, in the equation 2x + 3 = 7, the value of x that makes the equation true is 2.

To solve an equation, we use algebraic operations to isolate the variable on one side of the equation. The four basic algebraic operations are addition, subtraction, multiplication, and division.

When performing algebraic operations on an equation, it is important to remember that whatever we do to one side of the equation, we must also do to the other side to keep the equation balanced. For example, if we add 2 to the LHS of the equation 2x + 3 = 7, we must also add 2 to the RHS to get 2x + 5 = 9.

Now that we have reviewed some basic algebraic concepts and vocabulary, we are ready to dive into solving equations.

To solve a linear equation with one variable, we need to isolate the variable on one side of the equation. This means that we need to perform the same operation on both sides of the equation to simplify it and bring the variable to one side.Let's take an example equation: 2x + 5 = 11 We want to isolate x on one side of the equation. To do this, we need to subtract 5 from both sides of the equation: 2x + 5 - 5 = 11 - 5 Simplifying, we get: 2x = 6 Now, we need to isolate x completely, so we divide both sides of the equation by 2: 2x/2 = 6/2 Simplifying: x = 3 Therefore, the solution to the equation 2x + 5 = 11 is x = 3. It's important to remember that we always want to simplify the equation as much as possible before we start solving. This means that we should combine like terms and use the distributive property if necessary. Let's look at another example: 3(x - 4) - 2x = 5 First, we simplify by using the distributive property: 3x - 12 - 2x = 5 Combining like terms: x - 12 = 5 Next, we want to isolate x, so we add 12 to both sides of the equation: x - 12 + 12 = 5 + 12 Simplifying: x = 17 Therefore, the solution to the equation 3(x - 4) - 2x = 5 is x = 17. Remember to always check your solution by plugging it back into the original equation. If the equation is true, then your solution is correct.

Inverse operations are mathematical operations that undo each other. For example, addition and subtraction are inverse operations because adding a number and then subtracting the same number will result in the original number. Similarly, multiplication and division are inverse operations because multiplying a number and then dividing the same number will also result in the original number.

To solve an equation using inverse operations, we need to perform the same operation on both sides of the equation to isolate the variable. The goal is to get the variable by itself on one side of the equation and the constant on the other side.

Let's look at an example:

3x + 4 = 13

To isolate x, we need to get rid of the constant on the left side of the equation. We can do this by subtracting 4 from both sides of the equation:

3x = 9

Now, we need to isolate x completely. We can do this by dividing both sides of the equation by 3:

x = 3

Therefore, the solution to the equation 3x + 4 = 13 is x = 3.

It's important to remember to perform the same operation on both sides of the equation to maintain the balance of the equation.

Now, it's your turn to practice solving equations using inverse operations. Try the following problems:

1. 4x - 8 = 12

2. 2y + 3 = 9

3. 5z - 7 = 18

4. 6a + 2 = 26

Remember to use inverse operations to isolate the variable and solve for its value.

One of the most important uses of algebra is in solving real-world problems. By using equations, we can represent and solve problems that involve unknown values. Let's take a look at an example:

Alice and Bob are running a lemonade stand. They have 50 cups of lemonade to sell at $1.50 each. However, they want to earn $100 in total. How many cups of lemonade do they need to sell?

To solve this problem, we need to use an equation. Let x be the number of cups of lemonade they need to sell. The total amount they will earn is $1.50 times the number of cups, which can be expressed as 1.5x. We can set up an equation:

1.5x = 100

To solve for x, we can divide both sides by 1.5:

x = 100 / 1.5 = 66.67

Since they cannot sell a fraction of a cup, they need to sell 67 cups of lemonade to earn $100.

This is just one example of how we can use equations to solve real-world problems. By using variables and setting up equations, we can represent and solve a wide range of problems, from calculating distances and speeds to determining costs and profits. Remember to always carefully read the problem and identify what the unknown value is before setting up an equation.

Now that you have learned how to solve equations, it's time to assess your understanding with a quiz or worksheet. This will not only help you gauge your understanding of the topic but also assist you in identifying areas where you might need to improve.

Below is a sample quiz for you to try out:

1) Solve for x: 5x + 2 = 17

a) x = 5

b) x = 3

c) x = 3.4

d) x = 6

2) Solve for y: 3y - 7 = 8

a) y = 5

b) y = 3

c) y = 15

d) y = -5

3) Solve for z: 2z + 5 = 11

a) z = 3

b) z = 2

c) z = 8

d) z = 3.5

4) Solve for w: 4w - 3 = 13

a) w = 4

b) w = 4.5

c) w = 4.25

d) w = 4.75

5) Solve for x: 2(x + 5) = x - 1

a) x = -11

b) x = -9

c) x = 1

d) x = -3

Once you have completed the quiz, check your answers and see how you did. If you got any questions wrong, review the corresponding lesson section and try again until you fully understand the concept. Remember, practice makes perfect!