Problem 57 - Algebra for College Students

Question

Solve each equation. $$\frac{5}{4 x-2}-\frac{1}{1-2 x}=\frac{7}{3 x+6}$$

Step-by-Step Solution

Answer

Solve the equation by combining fractions, expanding, simplifying terms, and isolating the variable.

Step 1: Identify the common denominators

First, identify and factor the denominators. The denominators are \(4x - 2\), \(1 - 2x\), and \(3x + 6\). Notice that \(1 - 2x\) can be rewritten as \(-2(2x-1)\).

Step 2: Rewrite the equation with a common denominator

Rewrite each term to have a common denominator. We can express the whole equation over the common denominator \((4x-2)(1-2x)(3x+6)\).

Step 3: Combine the fractions

Combine the fractions over the common denominator: \[\frac{5 \times (1-2x)(3x+6) - 1 \times (4x-2)(3x+6) = 7 \times (4x-2)(1-2x)}{(4x-2)(1-2x)(3x+6)}\].

Step 4: Expand the numerators

Expand the numerators in the equation: \[5(1-2x)(3x+6) = 5(3x + 6 - 6x^2 - 12x)\] and \[- (4x-2)(3x+6) = -(12x^2 + 24x - 6x - 12)\] and on the right side \[7(4x-2)(1-2x) = 28x - 14 - 56x^2 + 28\].

Step 5: Simplify the expanded form

Simplify both sides of the equation: \[5(-6x^2 -12x + 3x + 6) - (12x^2 + 24x - 6x - 12) = 7(-56x^2 + 28x - 14)\].Combine like terms to continue simplification.

Step 6: Isolate the variable

Rearrange the equation to isolate the variable \(x\). Solve for \(x\) by factoring, completing the square, or using the quadratic formula if necessary.

Key Concepts

common denominator

When solving rational equations, it's important to find a common denominator. This helps to combine all fractions into a single equation, making it easier to solve. In our example, we start with three denominators: \(4x - 2\), \(1 - 2x\), and \(3x + 6\). Though they seem different, we can factor and rewrite them to find a common ground. Notice that \(1 - 2x\) can be rewritten as \(-2(2x-1)\). By factoring and identifying relationships, we set the stage for combining all fractions into a single denominator: \((4x-2)(1-2x)(3x+6)\). This step is crucial as it paves the way to further simplification and solution.

factoring

Factoring is a fundamental skill in algebra that simplifies expressions and equations, making them easier to solve. It involves expressing an equation as a product of its factors. In our example, we observe that each denominator can be factored:

• \(4x-2 = 2(2x-1)\)
• \(1-2x = -2(2x-1)\)
• \(3x+6 = 3(x+2)\)

Recognizing these factors allows us to find a common denominator and rewrite the equation. Factoring also helps in simplifying complex expressions, as we'll see later when we expand and combine terms. Remember, always look for common factors and rewrite them to make your equations manageable.

simplifying expressions

Simplifying expressions involves reducing them to their simplest form. This makes equations easier to handle and solve. In our example, once we've rewritten each term with a common denominator, we face complex numerators: \[ \frac{5 \times (1-2x)(3x+6) - 1 \times (4x-2)(3x+6) = 7 \times (4x-2)(1-2x)}{(4x-2)(1-2x)(3x+6)} \] We need to expand the numerators:

• \(5(3x + 6 - 6x^2 - 12x)\)
• \(- (12x^2 + 24x - 6x - 12)\)

After expanding, we combine like terms to simplify:

• \(5(-6x^2 -12x + 3x + 6) - (12x^2 + 24x - 6x - 12)\)
• \(7(-56x^2 + 28x - 14)\)

This process reduces the equation to a simpler form, easier to isolate and solve for the variable.

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations when they cannot be factored easily. It states: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our example, after simplifying and combining terms, we might end up with a quadratic equation of the form \(ax^2 + bx + c = 0\). If factoring is difficult or impossible, we use the quadratic formula:

• Identify \(a\), \(b\), and \(c\) from the equation.
• Substitute these values into the formula.
• Solve for \(x\) to find the roots of the equation.

This formula guarantees a solution (including complex ones) for any quadratic equation, ensuring we can always find the values of \(x\).