91Ó°ÊÓ

Cross-multiplication is a technique used to solve equations that involve two fractions set equal to each other.

Essentially, this method involves multiplying the numerator of one fraction by the denominator of the other fraction and vice versa. This creates a new equation without fractions and makes it easier to solve for the unknown variable.

For example, consider the equation \( \frac{12.5}{y} = \frac{102}{55} \). Here, we multiply 12.5 (the numerator of the first fraction) by 55 (the denominator of the second fraction). Then, we multiply \(y\) (the denominator of the first fraction) by 102 (the numerator of the second fraction). This results in the new equation \(12.5 \times 55 = 102 \times y\), which can then be simplified and solved for \(y\).

It's a valuable approach when dealing with proportions or problems involving ratios, as it quickly eliminates the fraction and reduces complexity.

After cross-multiplying, the next step in solving linear equations is to isolate the variable. This means we want to move everything else to the other side of the equation, leaving the variable alone.

To isolate the variable, perform operations that will 'undo' whatever is being done to the variable. If the variable is being multiplied by a number, divide by that number to cancel out the multiplication. If it's being divided by a number, multiply to cancel out the division, and so on for addition and subtraction.

In our equation, \(102y = 687.5\), we want to isolate \(y\), which means we need to divide both sides of the equation by 102. Doing so gives us \(y = \frac{687.5}{102}\), thus effectively isolating \(y\) on one side of the equation.

The final step in our problem-solving process involves rounding decimals, which is essential when the question specifies the accuracy required for the answer.

To round to the nearest hundredth, which is the second decimal place, look at the third decimal place. If it's 5 or greater, increase the second decimal place by one. If it's less than 5, leave the second decimal place as it is.

In our calculation for \(y\), after dividing 687.5 by 102, we get an answer that may have more than two decimal places. To meet the requirement of rounding to the nearest hundredth, we might have to adjust the last digit of the calculation. Let's say our division gave us \(y = 6.73529...\), we would round up the second decimal place from 3 to 4 because the third decimal place (5) is equal or greater than 5. Thus, we would report \(y = 6.74\) as the final, rounded answer.