91Ó°ÊÓ

Understanding the concept of the reciprocal is vital in solving linear equations, especially when they involve fractions. A reciprocal is a flipped version of a given fraction, meaning if you have a fraction like \( \frac{a}{b} \), its reciprocal is \( \frac{b}{a} \). The idea is based on the fact that multiplying a number by its reciprocal gives you 1. In equation solving, this property is used to eliminate fractions and isolate the variable.

For instance, if you have \( \frac{2}{3}t = -10 \), the coefficient of \( t \) is \( \frac{2}{3} \). The reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \) because when multiplied together, \( \frac{2}{3} \times \frac{3}{2} = 1 \). By multiplying both sides of the equation by this reciprocal, you effectively remove the fraction as a coefficient of \( t \), setting the stage for easier manipulation and solution of the variable.

When it comes to solving equations that involve fractions, knowing how to multiply fractions is a fundamental skill. To multiply fractions, you simply multiply the numerators (top numbers) and the denominators (bottom numbers) with each other. The product forms a new fraction that is the result of the multiplication.

Here's a quick example:

Multiplying \( \frac{1}{4} \) by \( \frac{2}{3} \) would involve multiplying the numerators 1 and 2 to get 2, and the denominators 4 and 3 to get 12, resulting in \( \frac{1 \times 2}{4 \times 3} = \frac{2}{12} \), which simplifies to \( \frac{1}{6} \). This method directly applies to solving linear equations where you multiply both sides of the equation by a fraction, as seen in the exercise with \( t = -10 \times \frac{3}{2} \).

Algebraic problem solving can be seen as a multi-step process that requires understanding of concepts and application of various strategies to find a solution. In solving the exercise \( \frac{2}{3} t = -10 \), the problem-solving process can be broken down into discernible steps. First, identify the operations affecting the variable, which, in this case, is the multiplication of \( t \) by a fraction.

Next, determine the most efficient strategy to isolate the variable—in our scenario, it's using the reciprocal to cancel out the fraction. Lastly, simplify the equation by performing arithmetic operations, like the multiplication of a whole number and a fraction. The goal is to simplify the equation step by step, making it possible to solve for \( t \) and reach the conclusion that \( t = -15 \). Problem-solving in algebra often hinges on these strategies, which simplify and streamline the process.