Problem 36 - Algebra 1

Question

Solve the proportion. Check for extraneous solutions. $$\frac{5}{m+1}=\frac{4 m}{m}$$

Step-by-Step Solution

Answer

The solution to the proportion is \(m = 0\) and \(m = 1/4\).

Step 1: Cross-multiplication

Cross-multiply the numerators with denominators. This results in \(5m = 4m*(m+1)\)

Step 2: Simplify the equation

Simplify the equation by performing the multiplication on the right side, which yields \(5m = 4m^2 + 4m\).

Step 3: Rearranging terms

Rearrange all terms to one side to allow factoring. Subtract \(5m\) from both sides which gives \(0 = 4m^2 - m\).

Step 4: Solve for m

Solve for \(m\)by factoring out \(m\) which gives \(0 = m(4m - 1)\). Setting \(m = 0\) and \(4m - 1 = 0\) and solving for \(m\) yields \(m = 0\) and \(m = 1/4\).

Step 5: Check for extraneous solutions

Substitute \(m = 0\) and \(m = 1/4\) into the original equation to see if they satisfy the equation. After substitution, both solutions hold true, therefore there are no extraneous solutions.

Key Concepts

Cross-Multiplication

Cross-multiplication is an essential method used to solve proportions involving fractions. A proportion is an equation that states two ratios are equivalent. For example, in the exercise \( \frac{5}{m+1} = \frac{4m}{m} \), we treat each side as a fraction. To solve this proportion using cross-multiplication, multiply the numerator of one fraction by the denominator of the other. Do this for both fractions. This technique is useful because it eliminates the fractions and simplifies the equation to a linear or polynomial equation. Here’s how it works:

• Multiply 5 by \( m \).
• Multiply \( 4m \) by \( m+1 \).

These steps transform our proportion into the equation \( 5m = 4m(m+1) \). By cross-multiplying, you reduce the complexity of handling fractions. This is a powerful technique for students to harness when dealing with rational expressions.

Factoring

Factoring is the process of breaking down an expression into a product of simpler expressions. It's a crucial skill when dealing with polynomials and is often needed to solve equations. After cross-multiplication in our example, we obtained the equation \( 0 = 4m^2 - m \). The goal of factoring is to transform the polynomial into a set of simpler, multiplied terms. This process makes it easier to find the solutions of the equation, as it sets the stage for using the zero-product property.To factor \( 0 = 4m^2 - m \):

• Identify that \( m \) is a common factor in both terms.
• Factor \( m \) out to get \( m(4m - 1) = 0 \).

Once the equation is in this factored form, you can solve for \( m \) by applying the zero-product property. This states that if the product of two expressions is zero, then at least one of the expressions must be zero. Hence, either \( m = 0 \) or \( 4m - 1 = 0 \). Solving \( 4m - 1 = 0 \) gives us \( m = \frac{1}{4} \). Factoring is a key tool for simplifying and solving polynomial equations.

Extraneous Solutions

Extraneous solutions are solutions derived from the algebraic manipulation of an equation that do not satisfy the original equation. Checking for these solutions is essential because some algebraic strategies, like squaring both sides of an equation, can introduce solutions that are not true to the original problem.In the exercise we're discussing, once we found the solutions \( m = 0 \) and \( m = \frac{1}{4} \), we must verify them. Verification involves substituting these values back into the original proportion:

• Substitute \( m = 0 \) into \( \frac{5}{m+1} = \frac{4m}{m} \). After simplification, both sides equal zero, confirming this as a true solution.
• Substitute \( m = \frac{1}{4} \) and verify through simplification that the original equation holds true.

Both substitutions confirm the validity of the solutions, indicating there are no extraneous solutions. Always remember, an extraneous solution often arises from the additional steps necessary to solve equations, especially when dealing with rational expressions.