Problem 65 - College Algebra Essentials

Question

Solve equation. Then determine whether the equation is an identity, a conditional equation, or an inconsistent equation. \(10 x+3=8 x+3\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = 0\). It is a conditional equation as it is true only for \(x = 0\), but not for other values of x.

Step 1: Simplify the Equation

In the given equation \(10x + 3 = 8x + 3\), both sides have a '+3', which can be cancelled out for simplification of the equation. The equation, after simplification, is \(10x = 8x\).

Step 2: Solve the Equation

To isolate x, subtract 8x from both sides of the equation which results in \(2x = 0\). When further divided by 2 on both sides to solve for x, the result is \(x =0\).

Step 3: Identify the Type of Equation

Testing the original equation with \(x = 0\), we get \(3 = 3\), which is true. Therefore, we conclude that it's a conditional equation as it is true only for \(x = 0\). It's not an identity equation as it doesn't hold true for all x, and it's not an inconsistent equation as it has a solution (x = 0).

Key Concepts

Types of Equations

There are different types of equations in mathematics that you can encounter. These types include identity equations, conditional equations, and inconsistent equations:

• **Identity Equations:** These equations are true no matter what value you plug in for the variable. For example, the equation \(2x + 3x = 5x\) is an identity since it holds true for any value of \(x\).
• **Conditional Equations:** These are only true for certain values of the variables involved. In the exercise example \(10x + 3 = 8x + 3\), it simplifies to \(2x = 0\) which means it is satisfied only when \(x = 0\).
• **Inconsistent Equations:** These have no solution. They are never true for any value of the variable. An example is \(x + 1 = x + 2\), which is a contradiction, as no value of \(x\) can satisfy the equation.

Understanding these types helps in identifying how to solve them and what results to expect.

Solving Linear Equations

Linear equations like the one provided, \(10x + 3 = 8x + 3\), can be solved using simple algebraic techniques. The goal is to isolate the variable, usually \(x\), to find its value. Here is a method to solve linear equations step-by-step:

• **Simplify the Equation:** Look for like terms on each side of the equation. In our example, \(10x + 3 = 8x + 3\), both sides had a \(+3\), which we can subtract to simplify to \(10x = 8x\).
• **Eliminate Variable Terms on One Side:** Subtract \(8x\) from both sides to begin isolating \(x\), making it \(2x = 0\).
• **Divide to Solve for \(x\):** Finally, divide each side by \(2\) to find \(x = 0\).

This straightforward method of solving linear equations is essential for revealing the variable's value.

Equations and Solutions

Understanding the solution of an equation involves verifying whether your answer makes the original equation true. Let's revisit our example:

• **Testing the Solution:** When you find \(x = 0\), substitute it back into the original equation to ensure it holds: \[10(0) + 3 = 8(0) + 3 \] or simply \[3 = 3\], confirming our solution.
• **Identifying Equation Types with Solutions:** After confirming \(x = 0\) as a solution, we identified the equation as a conditional equation since it holds true only for this particular value.

It’s critical to cross-check the solution with the original equation to ascertain correctness and understand the classification. This practice not only verifies your work but helps solidify your grasp of solution-based equation types.