Problem 16 - Algebra 2 and Trigonometry

Question

In \(3-20,\) solve each equation and check. $$ \frac{x}{2}=\frac{3}{2 x+1} $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{3}{2}\) and \(x = -2\).

Step 1: Identify the Cross-Multiplication Need

The given equation is \( \frac{x}{2} = \frac{3}{2x + 1} \). This is a proportion, which can be solved by cross-multiplying to eliminate the fractions.

Step 2: Cross Multiply

Cross-multiply the terms to get rid of the fractions: \[ x(2x + 1) = 3 \times 2 \] This simplifies to: \[ 2x^2 + x = 6 \]

Step 3: Rearrange into a Quadratic Equation

Subtract 6 from both sides to form a standard quadratic equation: \[ 2x^2 + x - 6 = 0 \]

Step 4: Factor the Quadratic Equation

We factor the quadratic equation \( 2x^2 + x - 6 = 0 \) into \[ (2x - 3)(x + 2) = 0 \] Here, we directly identify factors that satisfy the equation.

Step 5: Solve for the Roots

Set each factor equal to zero and solve for \(x \): For \(2x - 3 = 0\), we get \(x = \frac{3}{2} \). For \(x + 2 = 0\), we get \(x = -2 \).

Step 6: Check the Solutions

Substitute \(x = \frac{3}{2} \) into the original equation to check: \[ \frac{3/2}{2} = \frac{3}{4} \] and \[ \frac{3}{2 \times \frac{3}{2} + 1} = \frac{3}{4} \] Both sides are equal, so \(x = \frac{3}{2} \) is correct. Substitute \(x = -2\): \[ \frac{-2}{2} = -1 \] and \[ \frac{3}{2 \times -2 + 1} = 3/(-3) = -1 \] Both sides are equal, so \(x = -2\) is also correct.

Key Concepts

Cross-Multiplication

Cross-multiplication is a powerful tool when dealing with equations that present a proportion. This technique allows us to eliminate fractions by multiplying across the equal sign. For example, in the equation \( \frac{x}{2} = \frac{3}{2x + 1} \), cross-multiplication involves multiplying the numerator of one side by the denominator of the other:

• Multiply \( x \) by \( 2x + 1 \) on one side.
• Multiply \( 3 \) by \( 2 \) on the other side.

This results in the equation \( x(2x + 1) = 3 \times 2 \), effectively removing the denominators and simplifying the solving process. The main benefit is the simplification of the equation, allowing you to work with integers instead of fractions, which can make further algebraic manipulation more straightforward.
To apply this method, ensure that the equation forms a proportion and cross-multiply the terms to clear the fraction, producing a much simpler equation to manage.

Quadratic Equations

After applying cross-multiplication to our equation, we arrive at a quadratic equation: \( 2x^2 + x = 6 \). This form, characterized by an \( x^2 \) term, signifies we are dealing with a quadratic equation. Quadratic equations are polynomial equations of degree 2, and their standard form is \( ax^2 + bx + c = 0 \). In our case, the equation becomes \( 2x^2 + x - 6 = 0 \) after moving all terms to one side.
Understanding quadratic equations involves knowing their components:

• \( x^2 \) is the quadratic term.
• \( x \) is the linear term.
• The constant term adjusts the equation to equal zero.

By setting them to zero, we aim to find the "roots" or solutions of the equation. These roots can be found using various methods, the most common being factoring, completing the square, or using the quadratic formula.

Factoring

Factoring is an important strategy for solving quadratic equations like \( 2x^2 + x - 6 = 0 \). When we factor, we're looking for two binomials whose product equals the quadratic equation. Here, the equation factors to \[ (2x - 3)(x + 2) = 0 \]Finding such factors often involves:

• Identifying two numbers that multiply to give the constant term \(-6\).
• Those numbers must also add to give the linear coefficient \(+1\).

Through trial, error, or identity recognition, we determine suitable factors. Once factored, setting each binomial equal to zero gives us the solutions of the equation, namely:

• \( 2x - 3 = 0 \), yielding \( x = \frac{3}{2} \).
• \( x + 2 = 0 \), yielding \( x = -2 \).

Factoring provides an efficient and straightforward method for solving quadratic equations, especially when they can be split neatly into relatable binomials.

Checking Solutions

After finding solutions for an equation, it's essential to check that they satisfy the original equation. For \( \frac{x}{2} = \frac{3}{2x + 1} \), checking each solution involves substituting back into the initial equation and ensuring both sides match.

• For \( x = \frac{3}{2} \), both fractions simplify to \( \frac{3}{4} \), confirming our solution is correct by both sides equating similarly.
• For \( x = -2 \), the left side simplifies to \(-1\) and the right to \(-1\), verifying this solution too, as both sides equate.

Checking solutions is crucial as it verifies the accuracy of the solutions in light of the initial condition. Mistakes in calculation, especially during factoring, can result in erroneous solutions, so this verification step acts as a safety net to ensure consistent results.