Week 14 in Precalc 11 – Solving Rational Equations

This in Precalc we learned how to solve rational equation I choose this subject because it was an interesting application of simplyfiying rational expressions using addition , subtractions, multiplication and division. Solving rational equations is important because they are a useful tools for representing real-life situations and for finding answers to real problems and finding unkowns.

Steps to Solve Rational Equations (note: this is one method to solve Rational Equations using the LCD )

1.Identify the Denominators: Identify all the denominators in the equation if needed find the value for x on the deminator that would make the denomiator equal 0 that is your non-permissable value.

2. Find the Least Common Denominator (LCD): Determine the least common denominator of all the rational expressions in the equation.

3. Clear the Denominators: Multiply both sides of the equation by the LCD to eliminate the denominators. This step turns the rational equation into a polynomial equation.

4. Simplify the Equation: Distribute and combine like terms to simplify the equation.

5. Solve the Resulting Equation: Solve the simplified polynomial equation using appropriate methods (factoring, quadratic formula, etc.).

6. Check for Extraneous Solutions: Substitute the solutions back into the original rational equation to ensure they do not make any denominator zero. Discard any solutions that do.

Example:

Solve the rational equation: $\frac{2}{x+1}+\frac{3}{x-2}=1$ .

Step 1: Identify the Denominators
The denominators are $x+1$ and $x-2$ so are non permissible values for x are -1, 2.

Step 2: Find the Least Common Denominator (LCD)
The LCD of $x+1$ and $x-2$ is $\(x+1)(x-2)\$ .

Step 3: Clear the Denominators
Multiply both sides of the equation by the LCD $\(x+1)(x-2)\$ :

$(x+1)(x-2)\left(\frac{2}{x+1}+\frac{3}{x-2}\right)=(x+1)(x-2)\cdot 1$

Distribute the LCD to each term:

$2(x-2)+3(x+1)=(x+1)(x-2)$

Step 4: Simplify the Equation
Distribute and combine like terms:

$2x-4+3x+3=x^2-x-2$

Combine like terms on the left side:

$5x-1=x^2-x-2$

Move all terms to one side to set the equation to zero:

$0=x^2-6x-1$

Step 5: Solve the Resulting Equation
This is a quadratic equation, so we can solve it using the quadratic formula \ $(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)$ , where $a=1$ , $b=-6$ , and $c=-1$ :

$x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1\cdot(-1)}}{2\cdot 1}$
$x=\frac{6\pm\sqrt{36+4}}{2}$
$x=\frac{6\pm\sqrt{40}}{2}$
$x=\frac{6\pm 2\sqrt{10}}{2}$
$x=3\pm\sqrt{10}$

The solutions are $x=3+\sqrt{10}$ and $x=3-\sqrt{10}$ .

Step 6: Check for Extraneous Solutions
Substitute $x=3+\sqrt{10}$ and $x=3-\sqrt{10}$ back into the original equation to ensure they do not make any denominator zero. Since neither value makes $x+1$ or $x-2$ equal to zero, both are valid solutions.

Final Answer

The solutions to the rational equation $\frac{2}{x+1}+\frac{3}{x-2}=1$ are $x=3+\sqrt{10}$ and $x=3-\sqrt{10}$ .

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Week 14 in Precalc 11 – Solving Rational Equations

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