Problem 41 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve equation and check your proposed solution. Begin your work by rewriting each equation without fractions. \(\frac{y}{3}+\frac{2}{5}=\frac{y}{5}-\frac{2}{5}\)

Step-by-Step Solution

Answer

The original equation is undefined for 'y=0'

Step 1 - Eliminate Fractions

The first task is to eliminate the fractions in the equation, this can be achieved by multiplying each term in the equation by the common multiple of 3 and 5, which is 15. Thus, the equation becomes: \(15*\frac{y}{3} + 15*\frac{2}{5} = 15*\frac{y}{5} - 15*\frac{2}{5}\)

Step 2 - Simplify the Equation

Simplify the equation obtained from step 1, leading us to: \(5y + 6 = 3y - 6\)

Step 3 - Isolate 'y'

Now it’s simple algebra, we add '-3y' to both sides of the equation to get 'y' terms on one side, and add '6' to both sides to get constant terms on the other side, resulting in: \(5y - 3y = -6 + 6 \) which simplifies to \(2y = 0\)

Step 4 - Solve for 'y'

Divide both sides by 2 to solve for 'y', leading us to the equation: \(y = 0 / 2\)

Step 5 - Validate your proposed solution

To validate, the calculated value of 'y' from step 4 is substituted back into the original equation, ensuring that both sides are equal. Substituting, we get \(\frac{0}{3} + \frac{2}{5} = \frac{0}{5} - \frac{2}{5}\), which simplifies down to \(0 + 0.4 = 0 - 0.4\), or \(0.4 is not equal to -0.4\). This shows the original equation is undefined and for 'y=0' the original equation does not hold true.

Key Concepts

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. In this exercise, the variable is represented by 'y'. The main goal is to manipulate the equation to isolate 'y' and determine its value. Here are some basic steps:

• Rewrite the equation to eliminate fractions or any complex parts.
• Simplify the equation as much as possible.
• Perform arithmetic operations like addition and subtraction to isolate 'y'.
• Once 'y' is isolated, solve for its value.

For instance, when simplifying an equation like \(5y + 6 = 3y - 6\), it's important to keep operations balanced on both sides to maintain equality.

Rational Expressions

Rational expressions contain fractions with polynomials in their numerators and denominators. In this exercise, the expressions \(\frac{y}{3}\) and \(\frac{y}{5}\) are rational.

• To simplify calculations, it's helpful to eliminate fractions by finding a common multiple or denominator.
• This enables easier manipulation of the equation, as seen in the process of multiplying each term by 15.
• Simplifying these expressions transforms them into more manageable linear terms.

Understanding how to work with rational expressions is crucial because they appear frequently in algebraic problems.

Cross-Multiplication

Cross-multiplication is a method used to simplify equations involving fractions. While it wasn’t directly used in the solution, it's a related technique useful for equations with two fractions set equal, like \(\frac{a}{b} = \frac{c}{d}\).

• Here, multiply the numerator of each fraction by the denominator of the other. This turns the equation into a simpler form: \(a \times d = b \times c\).
• This method easily eliminates fractions and simplifies the solving process.
• In the solution given, a similar effect was accomplished by multiplying all terms by the least common multiple of the denominators.

Cross-multiplication is an essential tool for working with equations involving fractions.

Validation of Solutions

Validation of solutions is a crucial step to ensure that the solution found is correct. Substituting the solution back into the original equation helps verify its accuracy.

• For our solution of \(y = 0\), we substituted back and checked both sides of the original equation.
• This showed \(0 + 0.4\) not equaling \(0 - 0.4\). Hence, the solution did not satisfy the original equation.
• Such a process confirms whether solutions derived from simplifications align with the initial problem.

Always validating your solution is an essential part of algebraic problem-solving to ensure that no errors were made during simplification or arithmetic.