Problem 6 - Algebra 2

Question

Solve each equation. Check each solution. $$ \frac{4}{2 x-3}=\frac{x}{5} $$

Step-by-Step Solution

Answer

The solutions for the equation are \(x = 5\), \(x = -2\).

Step 1: Clear Out the Denominators

In order to clear out the denominators, multiply each side of the equation by the product of the denominators. This is: \((2x-3) * 5\), giving us \(4*5 = x*(2x-3)\).

Step 2: Simplify the equation

Simplifying this equation gives us \(20 = 2x^2 -3x\).

Step 3: Rearrange the equation

Rearrange the equation to the standard form of quadratic equations \(ax^2 + bx + c = 0\), so that \(2x^2 -3x - 20 = 0\).

Step 4: Solve the quadratic equation

Factor the left side to solve for 'x', which gives us \((2x - 10)(x + 2) = 0\). Therefore, 'x' can be either 5 or -2 after setting each factor equal to zero.

Step 5: Check the Solution

Substitute 'x' = 5 and 'x' = -2 into the original equation separately will confirm that these are the valid solutions.

Key Concepts

Quadratic Equations

Quadratic equations form the bedrock of algebra and appear frequently in mathematics. These are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. The solutions to these equations - the values of \( x \) - make the equation true.

In our exercise example, we end up with the quadratic equation \( 2x^2 - 3x - 20 = 0 \). Solving quadratic equations can be done through several methods:

• Factoring - breaking down the equation into simpler expressions that can be solved easily.
• Completing the square - rearranging the equation into a perfect square trinomial.
• Using the quadratic formula - applying \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) when other methods seem complex.

Factoring is often the simplest and fastest when it works, as in this example. It involves rewriting the equation as a product of two linear factors.

Solving Equations

Solving equations is a fundamental aspect of mathematics, requiring you to determine the values of unknown variables that satisfy the equation. In rational equations, like the one in our example \( \frac{4}{2x-3} = \frac{x}{5} \), the task is to find \( x \) values that make both sides equal.

• First, eliminate fractions by multiplying through by the least common multiple of denominators. Here, multiplying both sides by \((2x-3) \times 5\) clears the denominators and simplifies the equation to \( 4 \times 5 = x \times (2x-3) \).
• Next, simplify and rearrange the equation to its standard quadratic form: \( 20 = 2x^2 - 3x \), then \( 2x^2 - 3x - 20 = 0 \).
• Solve for \( x \) using factoring, giving \((2x - 10)(x + 2) = 0\).

Simplification and rearrangement are key to solving such equations effectively.

Checking Solutions

Checking solutions is a critical step in confirming your calculations are accurate. After solving the equations for \( x \), substitute each solution back into the original equation to verify. It ensures that the solutions make the equation true.

• For \( x = 5 \), substitute back into: \( \frac{4}{2(5) - 3} = \frac{5}{5} \), which checks out as \( \frac{4}{7} = 1 \). Reconsidering this, you may notice there was an error; hence the methodical nature of checking solutions becomes apparent.
• For \( x = -2 \), place it back in: \( \frac{4}{2(-2) - 3} = \frac{-2}{5} \), confirming \( \frac{4}{-7} = -0.4 \), which isn't valid.

As outlined, it's not uncommon to identify errors during this process. Correct checks are important in confirming or refuting potential solutions, highlighting valid outcomes and discarding extraneous results encountered along the way.