Solve the Inequality for c 2c+6< 2c+9

Solution:

Step 1: Isolate the variable

Begin by moving all terms with the variable $c$ to one side of the inequality.

Step 1.1: Eliminate $c$ from the right side

Subtract $2c$ from both sides to get $2c + 6 - 2c < 2c + 9 - 2c$.

Step 1.2: Simplify the expression

Now, simplify the left side by combining like terms.

Step 1.2.1: Cancel out $2c$

The $2c$ terms cancel each other out, leaving $0 + 6 < 9$.

Step 1.2.2: Simplify further

Simplify the expression to get $6 < 9$.

Step 2: Evaluate the simplified inequality

Since $6$ is less than $9$, the inequality holds true for all values of $c$.

Step 3: Express the solution

The inequality is true for all real numbers.

Interval Notation: $(-\infty, \infty)$

Knowledge Notes:

The problem at hand is a simple linear inequality in one variable, $c$. The goal is to find the set of all possible values for $c$ that satisfy the inequality $2c + 6 < 2c + 9$.

Relevant knowledge points include:

1. Isolating the Variable: To solve an inequality, it's often necessary to isolate the variable on one side. This is done by performing the same operation on both sides of the inequality, ensuring the balance is maintained.

2. Combining Like Terms: Terms that contain the same variable to the same power can be combined by adding or subtracting them. In this case, $2c - 2c$ equals $0$.

3. Properties of Inequalities: When you subtract the same number from both sides of an inequality, the direction of the inequality remains the same.

4. Simplification: Simplifying the inequality can sometimes lead to a statement that is always true (or always false). In this case, $6 < 9$ is always true, which means the original inequality holds for all values of $c$.

5. Interval Notation: This is a way of writing the set of solutions to an inequality. For an inequality that is always true, the interval notation is $(-\infty, \infty)$, representing all real numbers.

6. No Solution vs. All Real Numbers: It's important to distinguish between an inequality that has no solution (never true) and one that is always true. In this case, since the inequality simplifies to a true statement independent of $c$, the solution includes all real numbers.