Problem 81 - Algebra for College Students

Question

Solve each equation. Identify each equation as a conditional equation, an inconsistent equation, or an identity. State the solution sets to the identities using interval notation. $$\frac{1}{x}+\frac{1}{x-1}=\frac{2 x-1}{x^{2}-x}$$

Step-by-Step Solution

Answer

The equation is an identity. The solution set is \ (-\infty, 0) \cup (0, 1) \cup (1, \infty) \.

Step 1 - Simplify Both Sides

Rewrite the equation to make it easier to manage. Notice that the denominator on the right side of the equation, \ (x^2 - x), can be factored as \ ((x)(x-1)). Rewrite the left side of the equation to have this common denominator.\[\frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)}\]

Step 2 - Combine the Fractions on the Left Side

Find a common denominator on the left side so that the two fractions can be combined. The common denominator is \ (x(x-1))):\[\frac{1 \cdot (x-1) + 1 \cdot x}{x(x-1)} = \frac{2x - 1}{x(x-1)}\]Simplify the numerator:\[\frac{x-1 + x}{x(x-1)} = \frac{2x - 1}{x(x-1)}\]

Step 3 - Equate the Numerators

Since the denominators are now the same, we can set the numerators equal to each other:\[2x - 1 = 2x - 1\]

Step 4 - Analyze the Simplified Equation

Observe that the equation \ 2x - 1 = 2x - 1 \ is always true for any value of x except \(x=0\) and \(x=1\) where the original denominators would be zero and thus undefined. This tells us that the equation is an identity.

Step 5 - State the Solution Set

Since the equation is true for all x except \(x=0\) and \(x=1\), write the solution set in interval notation:\[(-\infty, 0) \cup (0, 1) \cup (1, \infty)\]

Step 6 - Identify the Type of Equation

Since the equation is true for all values of x (except where it is undefined), it is an identity.

Key Concepts

conditional equation

A conditional equation is an equation that is true only for specific values of the variable. It isn’t always true. For example, the equation \(2x = 4\) is only true when \(x = 2\). If you plug any other number into the equation for \(x\), it won't be true.
To solve a conditional equation:

• Isolate the variable
• Solve for the variable

Then, check that your solution satisfies the original equation. If it does, your solution is correct! Keep in mind that the solution must work for each variable assigned.

inconsistent equation

An inconsistent equation is an equation that has no solutions. No matter what value you plug into the variable, the equation will never be true.
For example, consider \(x + 5 = x + 6\). If you subtract \(x\) from both sides, you get \(5 = 6\), which is a contradiction.
In aligning with our given exercise, if we had simplified our equation and found a contradictory statement instead of a true one, it would indicate an inconsistent equation.

identity

An identity is an equation that is true for all values of the variable, except where undefined.
In other words, no matter what number you choose to plug in for the variable, the equation will always be true.
For instance, in our exercise, once simplified, we arrived at \(2x - 1 = 2x - 1\). This statement is true for any \(x\). Therefore, the equation is classified as an identity.
However, we must exclude any values that make the original equation undefined; in this case, those values are \(x = 0\) and \(x = 1\). So, our equation is an identity except at those points.

solution sets

The solution set of an equation is the set of all possible values that satisfy the equation.
In the case of a conditional equation, the solution set is the specific values that make the equation true.
For an identity, the solution set is all real numbers, except where the equation could be undefined.
To find the solution set:

• Solve the equation for the variable
• Determine any restrictions or points of discontinuity (such as undefined points)
• Express the solution set in the appropriate notation

In our exercise, we determined that the equation is an identity, and hence the solution set for the given rational equation is all \(x\) except \(x = 0\) and \(x = 1\).

interval notation

Interval notation is a way of writing the set of all solutions to an inequality or an equation.
In interval notation, we use parentheses \(( ) \) and brackets \( [ ] \). Parentheses indicate that an endpoint is not included, while brackets indicate that an endpoint is included.
For example:

• \((-∞, 0)\) indicates all real numbers less than 0, not including 0.
• \([0, 1]\) indicates all real numbers from 0 to 1, including 0 and 1.
• The union symbol \( \cup \) is used to combine intervals.

In the given exercise, our solution set in interval notation is \( (-∞, 0) \cup (0, 1) \cup (1, ∞) \). This tells us that \(x\) can be any number except 0 and 1.