Problem 44 - Elementary Algebra

Question

For the following problems, solve each conditional equation. If the equation is not conditional, identify it as an identity or a contradiction. $$ 2(x-7)=2 x+5 $$

Step-by-Step Solution

Answer

Answer: The given equation is a conditional equation. After simplifying and attempting to solve for x, we discovered a contradiction (-14 = 5). Therefore, there are no solutions for the variable x.

Step 1: Distribute the number on the left side of the equation

To start, we need to distribute the number 2 to both terms inside the parentheses on the left side of the equation. $$ 2(x-7) = 2 \cdot x - 2 \cdot 7 $$ $$ 2x - 14 = 2x + 5 $$

Step 2: Simplify the equation

Now our equation is simplified, we will try to isolate x: $$ 2x - 14 = 2x + 5 $$

Step 3: Subtract 2x from both sides of the equation

We want to get rid of the x term on one side of the equation to solve for x: $$ 2x - 2x - 14 = 2x - 2x + 5 $$ $$ -14 = 5 $$

Step 4: Determine the type of equation

We find that we have a contradiction (-14 = 5) that makes no sense. Since there are no solutions that make this equation true, we can conclude that this is a contradiction.

Key Concepts

Conditional Equations

Conditional equations are equations that hold true only for specific values of the variable involved. These are very common in algebra, where we solve for the value of a variable.
For example, if an equation like \(2x + 3 = 7\) is given, it will only be true for \(x = 2\). This specific solution makes the equation a conditional equation, since it does not apply for any other value of \(x\).
Solving such equations involves:

• Isolating the variable on one side
• Performing basic arithmetic operations
• Checking if the obtained solution satisfies the original equation

Remember, conditional equations are only valid when the variable equals a particular value, unlike identities or contradictions.

Identity Equations

Identity equations are always true, no matter what value the variable takes. Imagine an equation like \(x + x = 2x\). This identity holds for any real number.
In identity equations:

• Both sides of the equation simplify to the same expression
• They are true for all permissible values of the variable
• For example, \(3(x + 4) = 3x + 12\)

Even if you plug in any number for \(x\), the equation remains true. These types of equations help show deep algebraic properties and are useful for proving mathematical theorems and identities.

Contradiction in Algebra

Contradictory equations are equations for which there are no possible solutions. In this type, you will find an inconsistency as you solve the equation.
According to our exercise, simplifying the equation ended in \(-14 = 5\), which is a contradiction. There's no value of \(x\) that will make this true, as it's simply incorrect mathematically.
Important aspects of contradiction in algebra include:

• End result shows an impossible statement
• No value of the variable satisfies the equation
• May emerge from errors in the equation set-up or simplifying process

Being able to identify contradictions quickly is key to understanding the problem and any potential mistakes.

Solving Linear Equations

Solving linear equations involves finding the value of a variable that makes the equation true. Linear equations have the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants.
Here are steps to solve them:

• Simplify each side of the equation separately if necessary
• Use inverse operations to isolate the variable on one side (like adding, subtracting, multiplying, dividing)
• Check your solution by substituting back into the original equation

The process teaches valuable skills for algebra and forms the basis for solving more complex equations.