Problem 23 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Solve the equation by using the LCD. Check your solution(s). $$\frac{6 x}{x+4}+4=\frac{2 x+2}{x-1}$$

Step-by-Step Solution

Answer

The solution for the given equation is \(x = \frac{2}{3},-3\). However, \(x = -4\) and \(x = 1\) are exclusion values for the original equation, since they make the denominator zero. Therefore, both \(x = \frac{2}{3},-3\) are valid solutions.

Step 1: Identify the LCD

Given equation is \( \frac{6x}{x+4} + 4=\frac{2x+2}{x-1} \) . The denominators are \( (x+4) \) and \( (x-1) \). So the lowest common denominator (LCD) is \( (x+4)(x-1) \).

Step 2: Multiply Each Term by the LCD

Multiply every term in the equation by this LCD to remove the fraction:\[ (x+4)(x-1) \cdot \frac{6x}{x+4} + (x+4)(x-1) \cdot 4 = (x+4)(x-1) \cdot \frac{2x+2 }{x-1} \] Which simplifies to:\[ 6x(x-1) + 4(x+4)(x-1) = (x+4)(2x+2) \]

Step 3: Simplify the Equation

Simplify the equation further:\[ 6x^2-6x + 4x^2 -4 = 2x^2+8x+8 \] \nCombine like terms:\[ 8x^2 -8x -4 = 2x^2+8x+8 \] \nSubtract \(2x^2 + 8x + 8\) from both sides to make the equation equal to zero:\[ 6x^2 -16x -12 = 0 \]

Step 4: Solve for x

Now, we have a quadratic equation which needs to be solved. Dividing all terms by 2 gives:\[ 3x^2 -8x -6 = 0 \] The solutions can be found using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting a=3, b=-8, and c=-6 gives solutions for x.

Step 5: Checking the Solutions

The derived solutions should be validated by substituting them back into the original equation. We need to take care about the values which could make the denominator zero, because these are not valid solutions for the rational equation.

Key Concepts

The Least Common Denominator (LCD)

When solving rational equations, it's crucial to eliminate the fractions for easier handling. The key to doing this is finding the Least Common Denominator (LCD). The LCD is the smallest expression that both denominators can divide into without a remainder. In our original equation, the denominators are \( x+4 \) and \( x-1 \). Hence, the LCD is \((x+4)(x-1)\).

Once the LCD is determined, we multiply every term in the equation by this LCD. This step helps in getting rid of the fractions. So, the equation becomes:

• \((x+4)(x-1) \cdot \frac{6x}{x+4}\)
• \((x+4)(x-1) \cdot 4\)
• \((x+4)(x-1) \cdot \frac{2x+2}{x-1}\)

This simplification allows us to easily move on to solving the equation by focusing only on the numerators.

Applying the Quadratic Formula

After simplification steps, the rational equation often translates into a quadratic equation form, which can be represented generally as \( ax^2 + bx + c = 0 \). In our example, this becomes simplified to \( 6x^2 - 16x - 12 = 0 \).

To solve this quadratic equation, we use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows finding the solutions for \( x \) by substituting the values of \( a \), \( b \), and \( c \). In our case, we have \( a = 3 \), \( b = -8 \), and \( c = -6 \). Substituting these into the formula, simplifies the process of finding the numerical solutions for \( x \).

This formula is invaluable in algebra, especially when factoring is not feasible or when the roots are complex.

Checking Your Solutions

Checking your solutions is a critical step when dealing with rational equations. The reason is some solutions might not be valid due to making a denominator zero, which is generally not allowed in mathematics.

For our problem, after obtaining solutions from the quadratic formula, it is essential to reintroduce them into the original equation to ensure neither \( x + 4 \) nor \( x - 1 \) equals zero. If any solution leads to a zero in these terms, such a solution would be termed extraneous and considered invalid for the given rational equation.

This is crucial because it helps confirm that our solutions are correct and applicable, preventing errors in real-world applications or further mathematical problems.

Combining Like Terms

When simplifying equations, particularly after multiplying each term by the LCD, combining like terms is vital. In mathematics, like terms refer to terms that have the same variable raised to the same power. They can be summed up to simplify expressions.

In our scenario, after expanding, the equation becomes \( 6x^2 - 6x + 4x^2 - 4 = 2x^2 + 8x + 8 \). By grouping the \( x^2 \), \( x \), and constant terms together, we combine like terms:

• The \( x^2 \) terms: \(6x^2 + 4x^2 = 10x^2\)
• The \( x \) terms: \(-6x\)
• Constant terms are simply moved together: \(- 4\)

This step forwards the balancing of equations, leading to a more straightforward path towards a solution, ultimately enabling us to use tools like the quadratic formula effectively.