Problem 139 - Algebra and Trigonometry

Question

Will help you prepare for the material covered in the next section. $$ \text { Solve: } \frac{x+2}{4 x+3}=\frac{1}{x} $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x=3\) and \(x=-1\).

Step 1: Cross Multiply

In order to clear the fraction, you need to cross multiply. The equation then becomes: \((x+2) \cdot x = (4x+3)\cdot1\), which simplifies to \(x^2+2x = 4x + 3\).

Step 2: Simplifying the equation

Next, simplify the equation and bring it to a standard quadratic equation, \(ax^2 + bx + c = 0\). By doing this, the equation turns into \(x^2+2x-4x-3=0\), which further simplifies to \(x^2 - 2x - 3 = 0\).

Step 3: Solve the quadratic equation

Now solve the quadratic equation: \(x^2 - 2x - 3 = 0\). We can solve this by factoring, which gives us: \((x-3)(x+1) = 0\).

Step 4: Find the `x` values

Finally, set each factor equal to 0 and solve for `x`. Thus, \(x-3=0\) yields \(x=3\) and \(x+1=0\) yields \(x=-1\).

Step 5: Check the solution

It's important to check that these solutions don't make the denominator of the original equation equal to zero, since that would be undefined. For \(4x + 3 = 0\), the solution \(x=-\frac{3}{4}\) would cause this, but neither of our solutions (-1 and 3) match this value, so both are valid solutions.

Key Concepts

Cross Multiplication

Understanding the technique of cross multiplication is crucial when solving equations that involve fractions. It's used to eliminate the fractions and make the equation easier to solve. Let's break down the process using the given exercise.

The exercise presents us with a fractional equation \( \frac{x+2}{4 x+3}=\frac{1}{x} \). Here, cross multiplication involves multiplying the numerator of one fraction by the denominator of the other and vice versa. Thus, \( (x+2) \cdot x = (4x+3) \cdot 1 \). This step is all about removing the fractions, making the subsequent steps of simplification and solving much more straightforward.

Simplifying Equations

After cross-multiplying, simplifying the equation is the next step. Simplification involves redistributing terms and combining like terms so that the equation is easier to solve.

Once cross multiplication is complete, we get \(x^2+2x = 4x + 3\). The goal is to simplify this into the form \(ax^2 + bx + c = 0\), which is the standard form for quadratic equations. To do this, we subtract \(4x\) and \(3\) from both sides, resulting in \(x^2+2x-4x-3=0\). This further simplifies to the cleaner expression \(x^2 - 2x - 3 = 0\), which is now primed for the next phase: factoring.

Factoring Quadratic

The factoring quadratic method is an efficient way to solve quadratics when it's possible to express the equation as a product of binomials.

The simplified equation \(x^2 - 2x - 3 = 0\) is ready for factoring. To factor, we need to find two numbers that multiply to give the constant term (-3), and add to give the middle coefficient (-2). In this case, the numbers are -3 and +1. Rewriting the quadratic as binomials, we get \( (x-3)(x+1) = 0 \).
The Zero Product Property states that if a product of two factors is zero, at least one of the factors must be zero. Therefore, to find the values of \(x\), we set each binomial equal to zero and solve: \((x - 3) = 0\) yielding \(x = 3\), and \((x + 1) = 0\) yielding \(x = -1\). These are the solutions to the original equation, as long as they do not make the denominators in any fractions equal to zero.