Question

In the expression $\frac{\frac{6}{5}}{\frac{12}{2y}-\frac{5}{y}}=1$, what is the value of $y$ ?

Short answer

The value of $y$ in the given expression is $5$.

Step-by-step solution

Rewrite the equation without the complex fraction structure

Rewrite the given expression as a product, multiplying both sides of the equation by the common denominator, which is $(5y)$, to eliminate the complex fraction structure: $$\frac{\frac{6}{5}}{\frac{12}{2y}-\frac{5}{y}}(5y)=1(5y)$$

Simplify the product

Compute the product on both sides of the equation, taking into account the rules of fractions when dealing with multiplication. $$\frac{6}{5}\cdot 5y-\frac{12}{2y}\cdot 5y+\frac{5}{y}\cdot 5y=5y$$

Evaluate the computed products

Evaluate each of the 3 computed products on the left side of the equation, as well as the right side: $$6y-30+25=5y$$

Isolate the variable y

Add 30 to both sides and subtract 5y from both sides to isolate the variable y: $$6y-5y=30-25$$

Simplify and find the value of y

Simplify both sides of the equation and solve for y: $$y=5$$ The value of y in the given expression is 5.

Key concepts

Simplifying Algebraic Expressions

Understanding how to simplify algebraic expressions is crucial when tackling complex fractions and other algebraic problems. Simplifying can involve various steps, such as combining like terms, factoring, expanding expressions, and canceling common factors. In the context of complex fractions, the process typically requires finding a common denominator to combine fractions, as seen in our exercise example. To simplify the given complex fraction, we multiplied both the numerator and the denominator by the least common denominator, in this case, $5y$. This removed the complex fraction structure, allowing us to proceed with further simplification.

To ensure you're following the correct steps, remember to apply the distributive property correctly when multiplying terms together and always simplify the resulting expressions as much as possible. For example, if a variable is present on both sides of the equation, as in the final steps of our solution, you should combine them to isolate the variable you're solving for. Correct simplification is pivotal not only to find the correct answer but to do so efficiently and accurately.

SAT Math Preparation

When preparing for the SAT math section, familiarity with algebraic concepts is essential. Problems like the one presented often appear on the SAT, testing your ability to manipulate and simplify expressions. To master these types of questions, practice is key. Work through various exercises that challenge you to simplify expressions, solve equations, and understand the properties of numbers.

Key Strategies for SAT Math Success

• Learn the foundational concepts: Make sure your understanding of algebra, geometry, and basic trigonometry is solid.
• Work on timing: The SAT is timed, so practice completing questions quickly and accurately.
• Review SAT-specific strategies: Learn the types of questions asked on the SAT and the best approaches to them.
• Take full-length practice tests: Simulate the testing experience to build stamina and reduce anxiety.
• Analyze your mistakes: Go over your incorrect answers to understand where you need to improve.

By incorporating these strategies into your study plan, complex fractions and related algebraic questions will become much more approachable on the SAT.

Solving Linear Equations

Linear equations are fundamental to algebra and appear in various forms, requiring different solving techniques. The general goal when solving linear equations is to isolate the variable to one side of the equation, resulting in a statement like $y=[somenumber]$.

In the solution to our complex fraction problem, once the fraction structure was removed, we were left with a simplified linear equation. The process involved combining like terms and appropriately adding or subtracting terms on both sides of the equation. It's important to perform the same operation on both sides to maintain the equation's balance.

Remember that when faced with linear equations:

• Always simplify both sides of the equation as much as possible.
• Use inverse operations to move terms across the equals sign (e.g., if a term is subtracted, add it to both sides).
• Combining like terms can often make the process of isolating the variable easier.
• Check your solution by substituting the found value back into the original equation to verify it satisfies the equation.

Applying these steps with practice will make the solving of linear equations more intuitive.