transform the following Rational Algebraic Equation to quadratic equation an solve for the value of x show your factoring and checking.
3/X + 5/X+4 = 1

1/X  + 1/X+3 = 1/2

Let's solve each of these Rational Algebraic Equations step by step, transforming them into quadratic equations, then solving for the value of \( x \), and finally checking the solutions.

### Equation 1: \( \frac{3}{x} + \frac{5}{x+4} = 1 \)

#### Step 1: Combine the fractions
To combine the fractions, we'll get a common denominator, which is \( x(x + 4) \).

\[
\frac{3(x + 4) + 5x}{x(x + 4)} = 1
\]

Simplifying the numerator:

\[
\frac{3x + 12 + 5x}{x(x + 4)} = 1
\]

This simplifies to:

\[
\frac{8x + 12}{x(x + 4)} = 1
\]

#### Step 2: Eliminate the denominator
Multiply both sides by the denominator \( x(x + 4) \):

\[
8x + 12 = x(x + 4)
\]

#### Step 3: Expand and rearrange
Expand the right side:

\[
8x + 12 = x^2 + 4x
\]

Move all terms to one side of the equation:

\[
x^2 - 4x - 12 = 0
\]

#### Step 4: Factor the quadratic equation
Now, we factor the quadratic equation:

\[
(x - 6)(x + 2) = 0
\]

#### Step 5: Solve for \( x \)
Set each factor equal to zero:

\[
x - 6 = 0 \quad \text{or} \quad x + 2 = 0
\]

So, \( x = 6 \) or \( x = -2 \).

#### Step 6: Check the solutions
Substitute \( x = 6 \) and \( x = -2 \) back into the original equation to ensure they are valid solutions.

For \( x = 6 \):

\[
\frac{3}{6} + \frac{5}{10} = \frac{1}{2} + \frac{1}{2} = 1 \quad \text{(valid solution)}
\]

For \( x = -2 \):

\[
\frac{3}{-2} + \frac{5}{2} = -\frac{3}{2} + \frac{5}{2} = 1 \quad \text{(valid solution)}
\]

Both solutions \( x = 6 \) and \( x = -2 \) are correct.

### Equation 2: \( \frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2} \)

#### Step 1: Combine the fractions
Let's find a common denominator:

\[
\frac{(x + 3) + x}{x(x + 3)} = \frac{1}{2}
\]

This simplifies to:

\[
\frac{2x + 3}{x(x + 3)} = \frac{1}{2}
\]

#### Step 2: Eliminate the denominator
Cross multiply to eliminate the denominator:

\[
2(2x + 3) = x(x + 3)
\]

#### Step 3: Expand and rearrange
Expand both sides:

\[
4x + 6 = x^2 + 3x
\]

Move all terms to one side:

\[
x^2 - x - 6 = 0
\]

#### Step 4: Factor the quadratic equation
Now, factor the quadratic equation:

\[
(x - 3)(x + 2) = 0
\]

#### Step 5: Solve for \( x \)
Set each factor equal to zero:

\[
x - 3 = 0 \quad \text{or} \quad x + 2 = 0
\]

So, \( x = 3 \) or \( x = -2 \).

#### Step 6: Check the solutions
Substitute \( x = 3 \) and \( x = -2 \) back into the original equation to ensure they are valid solutions.

For \( x = 3 \):

\[
\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{1}{2} \quad \text{(valid solution)}
\]

For \( x = -2 \):

\[
\frac{1}{-2} + \frac{1}{1} = -\frac{1}{2} + 1 = \frac{1}{2} \quad \text{(valid solution)}
\]

Both solutions \( x = 3 \) and \( x = -2 \) are correct.

### Summary of Solutions:
1. For \( \frac{3}{x} + \frac{5}{x+4} = 1 \), the solutions are \( x = 6 \) and \( x = -2 \).
2. For \( \frac{1}{x} + \frac{1}{x+3} = \frac{1}{2} \), the solutions are \( x = 3 \) and \( x = -2 \).

Would you like any more details or have any questions? 

Here are five related questions to consider:
1. How would you solve if the equations were more complex?
2. What are other methods for solving rational equations?
3. How can you apply the quadratic formula to similar problems?
4. What happens if the quadratic equation cannot be factored?
5. Can these methods be used to solve equations with three or more terms?

**Tip:** Always check your solutions in the original equation, as some solutions may not be valid for the rational equation due to restrictions in the domain (e.g., division by zero).

Transform the following Rational Algebraic Equation to a quadratic equation and solve for the value of x. Show your factoring and checking. 
3/X + 5/X+4 = 1

1/X  + 1/X+3 = 1/2

Learn how to transform rational algebraic equations into quadratic equations and solve for the value of x. This problem covers key algebraic concepts such as finding a common denominator, factoring quadratic equations, and checking solutions. Suitable for students in Grades 9-12.

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