Question

Solve the equation \(2-\frac{1}{x+2}=\frac{x}{x+2}+1\) algebraically and graphically, and explain how the results of both methods verify each other.

Short answer

No solution exists for the equation.

Step-by-step solution

- Simplify the equation

Start by rewriting the given equation: \[ 2 - \frac{1}{x+2} = \frac{x}{x+2} + 1 \] Move all terms to one side to simplify it to a single rational expression: \[ 2 - \frac{1}{x+2} - 1 = \frac{x}{x+2} \] Combine constants on the left side: \[ 1 - \frac{1}{x+2} = \frac{x}{x+2} \].

- Combine fractions

To combine fractions, get a common denominator on the left side: \[ 1 - \frac{1}{x+2} = \frac{x+2 - 1}{x+2} = \frac{x+1}{x+2} \]. So the equation becomes: \[ \frac{x+1}{x+2} = \frac{x}{x+2} \].

- Simplify the equation

Because the denominators are the same, equate the numerators: \[ x+1 = x \].

- Solve for x

Subtract x from both sides to solve for x: \[ 1 = 0 \]. This contradiction indicates no solution exists for the equation algebraically.

- Verify graphically

To verify graphically, graph the functions: \( y_1 = 2 - \frac{1}{x+2} \) and \( y_2 = \frac{x}{x+2} + 1 \). Plot both graphs and observe the intersection points. If they do not intersect, it confirms there is no solution.

Key concepts

graphical verification

Graphical verification is an essential way to confirm the solutions obtained algebraically. Let's plot the functions given in the equation to see how they behave on a graph.
To verify graphically, we consider two separate functions.
- Let \( y_1 = 2 - \frac{1}{x+2} \)
- Let \( y_2 = \frac{x}{x+2} + 1 \)
Using a graphing calculator or software, plot both these functions on the same set of axes. Observe where the graphs intersect. If there are no points of intersection, it indicates that there is no solution, matching the algebraic conclusion.
This visual confirmation can help solidify your understanding and give a different perspective on why the equation has no solution.

algebraic manipulation

Algebraic manipulation involves rewriting and simplifying the given equation step-by-step until we reach a form that allows us to solve for the variable.
In the given problem, we start with:
\[ 2 - \frac{1}{x+2} = \frac{x}{x+2} + 1 \]
We simplify by moving all terms to one side:
\[ 1 - \frac{1}{x+2} = \frac{x}{x+2} \]
To combine the fractions, we find a common denominator:
\[ \frac{x+1}{x+2} = \frac{x}{x+2} \]
With the same denominators, we can equate the numerators:
\[ x + 1 = x \]
Subtracting \( x \) from both sides gives us the contradictory statement:
\[ 1 = 0 \]
This contradiction indicates no solution exists for the equation.

rational expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials.
In solving the given problem, we encountered rational expressions several times.
Initially, we had terms like:
\[ \frac{x}{x+2} \] and \[ - \frac{1}{x+2} \]
Combining these required finding a common denominator and simplifying further. Recognizing and manipulating rational expressions is key to solving such equations effectively. Having a strong grasp of handling these expressions helps in simplifying and finding solutions efficiently.

no solution in equations

Sometimes, equations yield contradictions that indicate no solution exists. In our problem, the final step was:
\[ x + 1 = x \]
Subtracting \( x \) from both sides led to:
\[ 1 = 0 \]
Such contradictions signal that the original equation has no possible solution.
This can be an important concept as it tells us that not all mathematical equations have solutions. Graphical verification can further confirm this by showing no points of intersection between the plotted functions, hence corroborating our algebraic findings. Understanding this principle helps in recognizing unsolvable equations early in the process.