Problem 20 - Algebra Form and Function

Question

Without solving them, say whether the equations have a positive solution, a negative solution, a zero solution, or no solution. Give a reason for your answer. $$ 3 u-7=5 $$

Step-by-Step Solution

Answer

Answer: The equation has a positive solution.

Step 1: Identify the equation type

We have the equation $$3u - 7 = 5.$$ This is a linear equation in one variable.

Step 2: Check for any contradiction

Look for any contradiction in the equation. There is no term that contradicts the equation, so we can continue to the next step.

Step 3: Test the solution types using the equation

Let's consider the different types of solution without actually solving the equation: - Positive Solution: If \(u\) were positive, \(3u\) would also be positive. \(3u - 7\) would then be a positive number minus \(7\). If the result is greater than \(5\), the equation holds true. There's a possibility for a positive solution. - Negative Solution: If \(u\) were negative, \(3u\) would also be negative. \(3u - 7\) would then be a negative number minus \(7\). Since adding a negative number to a negative number results in a more negative number, \(3u - 7\) would be less than \(5\). There is no negative solution. - Zero solution: If \(u\) were zero, then the equation would be \(-7 = 5\). This is incorrect, so there is no zero solution.

Step 4: Make a conclusion about the solution type

Based on our analysis in Step 3, we can conclude that the equation $$3u - 7 = 5$$ has a positive solution.

Key Concepts

Positive Solution

A positive solution in the context of linear equations refers to any real number solution that is greater than zero. When we have an equation like \(3u - 7 = 5\), we are looking for a value of \(u\) that will make both sides of the equation equal.

If we suspect that \(u\) might be positive, then \(3u\), which means 3 times \(u\), will naturally be a positive value as well. When we subtract 7 from this positive product, we need the result to still be a positive number to affirm the equation. This means that \(3u\) has to be a number larger than 7 in order for the difference, \(3u - 7\), to match the positive value on the other side of the equation, which is 5.

Thus, because \(3u\) can indeed be greater than 7 with appropriate positive values for \(u\), it indicates that there is a positive solution for this equation.

Negative Solution

A negative solution involves a value that is less than zero. Here, the investigation is into whether \(u\) can be negative for the equation \(3u - 7 = 5\) to balance itself.

Let's break it down: if \(u\) is negative, then \(3u\) meaning 3 times negative \(u\) will be negative too. Removing 7 from this negative product results in becoming even more negative, due to subtraction adding to the negative total. This means the left side, \(3u - 7\), will always be much less than 5.

Since no negative \(u\) will balance the equation to equal the positive 5 on the right, there can be no negative solution for this linear equation. This is a crucial part of analyzing equations as it allows you to intuitively understand the nature of the solution set without actual computation.

Linear Equation Analysis

Linear equation analysis is a method to understand the nature of solutions that a given linear equation might provide. In the example of the equation \(3u - 7 = 5\), this involves several logical steps that help to predict possible solutions without direct calculation.

Firstly, recognizing the equation as a linear one-variable equation means knowing it graphs as a straight line. Linear equations can have one solution, no solution, or infinitely many solutions depending on their form.

In analyzing the equation \(3u - 7 = 5\), you consider possible values one by one:

• Positive values for solutions, testing if they satisfy the equation conditions.
• Negative and zero values are similarly tested for feasibility.

Through the logic of balancing both sides of the equation, we understand that positive \(u\) values provide the only potential for a valid solution. This type of analysis helps sharpen logical reasoning and enhance comprehension of algebraic structures.