Problem 18 - Algebra 1: Concepts and Skills

Question

Solve the equation by cross multiplying. Check your solutions. \(\frac{1}{y}=\frac{2}{y-3}\)

Step-by-Step Solution

Answer

There is no solution for the given equation, as the possible solution produces an undefined term in the original equation.

Step 1: Cross multiplication

Cross multiply to get rid of the fractions. Multiply \(1 * (y - 3)\) and \(2 * y\), which results in \(y - 3 = 2y\).

Step 2: Simplify the Equation

Now, simplify the resulting equation. Move the term with y to one side of the equation, and move the constant term to the other side. This results in \(-y = -3\).

Step 3: Solve for y

Solving \(-y = -3\) for y gives \(y = 3\).

Step 4: Check the Solution

Go back to the original equation and substitute \(y = 3\) to see if both sides are equal. When substituting \(y = 3\), we end up with undefined terms (division by zero) on the right-hand side, which indicates that the solution \(y = 3\) is extraneous, and is therefore rejected.

Key Concepts

Extraneous Solutions

When solving equations, particularly rational equations, students may encounter extraneous solutions. These are solutions that emerge during the solving process but do not actually satisfy the original equation. This can happen for a variety of reasons, including errors in algebraic manipulation or, more commonly, when the process used to solve the equation involves steps that might assume conditions that aren't true (such as multiplying or dividing by zero).

In the equation we have here \(\frac{1}{y} = \frac{2}{y-3}\), by cross-multiplying, we introduced possibilities of solutions that are not valid for the initial equation. Specifically, when we solve for \(y = 3\), substituting back into the original equation results in division by zero, thus making it an extraneous solution.

• Always verify solutions by substituting back into the original equation.
• Be mindful of potential restrictions like division by zero in rational equations.

Rational Equations

Rational equations are equations that involve fractions with polynomials in the numerator, the denominator, or both. These equations are common in algebra and require specific techniques to solve because of their unique characteristics.

For example, our equation \(\frac{1}{y} = \frac{2}{y-3}\) involves rational expressions on both sides of the equation. The goal when working with rational equations is often to eliminate the fractions by multiplying all terms by a common denominator, which results in a simpler, more manageable equation.

• Identify the common denominator among fractions.
• Use cross multiplication as a tool to eliminate the fractions.
• Check for extraneous solutions by substituting back the derived solutions.

Solving Equations

Solving equations involves finding the values of the variables that make the equation true. The process can vary greatly depending on the type of equation being solved. For linear equations, like the one derived from our rational equation, the main methods involve isolating the variable to one side to solve for its value.

In our problem, once the fractions were eliminated through cross multiplication, we simplified to \(y - 3 = 2y\). The next step involved using basic algebraic manipulation - moving terms from one side of the equation to the other - to isolate \(y\). From \(-y = -3\), it was straightforward to solve for \(y = 3\).

• Rearrange the equation to isolate the variable.
• Use inverse operations to solve the equation.
• Verify each step to avoid introducing errors or extraneous solutions.