Simplifying Complex Fractions And Solving Linear Equations A Comprehensive Guide

Let's dive into simplifying a complex fraction, a common challenge in mathematics. Complex fractions, which look intimidating at first glance, are essentially fractions within fractions. To solve them efficiently, a step-by-step approach is key. In this section, we will methodically break down the given expression, ensuring clarity at each stage. The given expression is:

$$\frac{\left(1 \frac{3}{7}-\frac{5}{8}\right)+\frac{2}{3} \text { of } 1 \frac{1}{5}}{\frac{3}{4}+1 \frac{5}{7} \div \frac{4}{7} \text { of } 2 \frac{1}{3}}$$

To simplify this complex fraction, we will follow the order of operations (PEMDAS/BODMAS), which stands for Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). This ensures that we tackle the expression in a logical sequence, avoiding common pitfalls.

Step 1: Convert Mixed Numbers to Improper Fractions

The initial step involves converting mixed numbers into improper fractions. This conversion is crucial for making the subsequent arithmetic operations more manageable. A mixed number combines a whole number and a fraction, while an improper fraction has a numerator larger than its denominator. For instance, $1 \frac{3}{7}$ can be converted by multiplying the whole number (1) by the denominator (7) and adding the numerator (3), then placing the result over the original denominator. So, $1 \frac{3}{7} = \frac{(1 \times 7) + 3}{7} = \frac{10}{7}$. Similarly, $1 \frac{1}{5}$ becomes $\frac{6}{5}$, and $1 \frac{5}{7}$ transforms into $\frac{12}{7}$, and $2 \frac{1}{3}$ becomes $\frac{7}{3}$. Replacing the mixed numbers with their improper fraction equivalents, the expression now looks like this:

$$\frac{\left(\frac{10}{7}-\frac{5}{8}\right)+\frac{2}{3} \text { of } \frac{6}{5}}{\frac{3}{4}+ \frac{12}{7} \div \frac{4}{7} \text { of } \frac{7}{3}}$$

Step 2: Simplify Within Parentheses

Next, we focus on simplifying the expressions within the parentheses in both the numerator and the denominator. Starting with the numerator, we need to subtract $\frac{5}{8}$ from $\frac{10}{7}$. To do this, we find a common denominator, which in this case is the least common multiple (LCM) of 7 and 8, which is 56. We then convert both fractions to have this denominator: $\frac{10}{7} = \frac{10 \times 8}{7 \times 8} = \frac{80}{56}$ and $\frac{5}{8} = \frac{5 \times 7}{8 \times 7} = \frac{35}{56}$. Subtracting these gives us $\frac{80}{56} - \frac{35}{56} = \frac{45}{56}$.

Moving to the denominator, we address the operation “$\frac{4}{7} \text{ of } \frac{7}{3}$”. The word “of” in mathematics often indicates multiplication. Thus, we multiply these two fractions: $\frac{4}{7} \times \frac{7}{3} = \frac{4 \times 7}{7 \times 3} = \frac{28}{21}$. This fraction can be simplified by dividing both the numerator and denominator by their greatest common divisor (GCD), which is 7, resulting in $\frac{4}{3}$. Now our expression looks like:

$$\frac{\frac{45}{56}+\frac{2}{3} \text { of } \frac{6}{5}}{\frac{3}{4}+ \frac{12}{7} \div \frac{4}{3}}$$

Step 3: Perform Multiplication

Following the order of operations, we perform the multiplication operations next. In the numerator, we have “$\frac{2}{3} \text{ of } \frac{6}{5}$”, which translates to $\frac{2}{3} \times \frac{6}{5}$. Multiplying these fractions gives us $\frac{2 \times 6}{3 \times 5} = \frac{12}{15}$. This fraction can be simplified by dividing both numerator and denominator by their GCD, which is 3, giving us $\frac{4}{5}$.

In the denominator, we have $\frac{12}{7} \div \frac{4}{3}$. Dividing by a fraction is the same as multiplying by its reciprocal. Therefore, we multiply $\frac{12}{7}$ by $\frac{3}{4}$: $\frac{12}{7} \times \frac{3}{4} = \frac{12 \times 3}{7 \times 4} = \frac{36}{28}$. This fraction can be simplified by dividing both numerator and denominator by their GCD, which is 4, resulting in $\frac{9}{7}$. Our expression is now:

$$\frac{\frac{45}{56}+\frac{4}{5}}{\frac{3}{4}+\frac{9}{7}}$$

Step 4: Perform Addition

Now we perform the addition in both the numerator and the denominator. In the numerator, we add $\frac{45}{56}$ and $\frac{4}{5}$. The LCM of 56 and 5 is 280. Converting the fractions, we get $\frac{45}{56} = \frac{45 \times 5}{56 \times 5} = \frac{225}{280}$ and $\frac{4}{5} = \frac{4 \times 56}{5 \times 56} = \frac{224}{280}$. Adding these gives us $\frac{225}{280} + \frac{224}{280} = \frac{449}{280}$.

In the denominator, we add $\frac{3}{4}$ and $\frac{9}{7}$. The LCM of 4 and 7 is 28. Converting the fractions, we have $\frac{3}{4} = \frac{3 \times 7}{4 \times 7} = \frac{21}{28}$ and $\frac{9}{7} = \frac{9 \times 4}{7 \times 4} = \frac{36}{28}$. Adding these gives us $\frac{21}{28} + \frac{36}{28} = \frac{57}{28}$. The expression is now:

$$\frac{\frac{449}{280}}{\frac{57}{28}}$$

Step 5: Divide the Fractions

Finally, we divide the fraction in the numerator by the fraction in the denominator. Dividing by a fraction is equivalent to multiplying by its reciprocal. So, we multiply $\frac{449}{280}$ by $\frac{28}{57}$: $\frac{449}{280} \times \frac{28}{57} = \frac{449 \times 28}{280 \times 57}$. We can simplify this by canceling out common factors. 28 is a factor of both 28 and 280 (280 = 28 * 10), so we can simplify the expression to $\frac{449}{10 \times 57} = \frac{449}{570}$.

Thus, the simplified form of the complex fraction is $\frac{449}{570}$.

Let's shift our focus to another fundamental concept in mathematics: solving linear equations. A linear equation represents a straight line on a graph and can be written in the general form of $ax + by = c$, where a, b, and c are constants, and x and y are variables. Solving linear equations is a cornerstone of algebra and has applications in various fields, including physics, engineering, and economics. The problem presented here involves finding the equation of a line given two points it passes through. This is a classic problem that illustrates the connection between geometry and algebra.

The problem states: A straight line $ax + by = 16$ passes through points $A(2, 5)$ and $B(3, 7)$. Find the values of $a$ and $b$.

Step 1: Substitute the Points into the Equation

Since the line $ax + by = 16$ passes through points $A(2, 5)$ and $B(3, 7)$, the coordinates of these points must satisfy the equation. This means that when we substitute the x and y values of each point into the equation, it should hold true. For point A(2, 5), we substitute $x = 2$ and $y = 5$ into the equation:

$a(2) + b(5) = 16$

This simplifies to:

$2a + 5b = 16$ (Equation 1)

Similarly, for point B(3, 7), we substitute $x = 3$ and $y = 7$ into the equation:

$a(3) + b(7) = 16$

This simplifies to:

$3a + 7b = 16$ (Equation 2)

Now we have a system of two linear equations with two variables, a and b. This system can be solved using various methods, such as substitution, elimination, or matrix methods. In this explanation, we will use the elimination method, which is particularly effective for this type of problem.

Step 2: Solve the System of Equations

The elimination method involves manipulating the equations so that when they are added or subtracted, one of the variables is eliminated. To eliminate the variable 'a', we need to make the coefficients of 'a' in both equations the same or additive inverses of each other. We can do this by multiplying Equation 1 by 3 and Equation 2 by 2:

Multiplying Equation 1 by 3:

$3(2a + 5b) = 3(16)$

$6a + 15b = 48$ (Equation 3)

Multiplying Equation 2 by 2:

$2(3a + 7b) = 2(16)$

$6a + 14b = 32$ (Equation 4)

Now, we subtract Equation 4 from Equation 3 to eliminate 'a':

$(6a + 15b) - (6a + 14b) = 48 - 32$

This simplifies to:

$b = 16$

Now that we have found the value of b, we can substitute it back into either Equation 1 or Equation 2 to solve for a. Let's use Equation 1:

$2a + 5(16) = 16$

$2a + 80 = 16$

Subtract 80 from both sides:

$2a = 16 - 80$

$2a = -64$

Divide by 2:

$a = -32$

Therefore, the values of $a$ and $b$ are -32 and 16, respectively.

In this article, we tackled two distinct yet fundamental problems in mathematics. First, we meticulously simplified a complex fraction, emphasizing the importance of following the order of operations and breaking down the problem into manageable steps. We converted mixed numbers to improper fractions, simplified expressions within parentheses, performed multiplication and division, and finally, addition and subtraction. This process highlights the step-by-step approach required for complex mathematical problems.

Secondly, we solved a problem involving linear equations. We demonstrated how to find the equation of a line given two points it passes through. This involved substituting the coordinates of the points into the equation, setting up a system of linear equations, and solving for the unknown variables using the elimination method. This problem underscores the connection between algebraic equations and geometric concepts.

Both types of problems are crucial in mathematics and showcase the importance of a structured, methodical approach. Mastering these concepts is essential for further studies in mathematics and its applications in various scientific and engineering disciplines. Whether it's simplifying fractions or solving linear equations, a clear understanding of the underlying principles and a systematic approach are key to success.