Fractions, Rational Numbers, And Equations Practice Problems And Solutions
In this comprehensive guide, we delve into various mathematical concepts, including fractions, rational numbers, and equations. We will address specific problems related to these topics, providing detailed explanations and solutions. This exploration aims to enhance your understanding of fundamental mathematical principles and equip you with the skills to tackle similar problems effectively. Let's embark on this journey of mathematical discovery together.
1) Subtracting from the Sum of Fractions
Fractions form a cornerstone of mathematical understanding, and mastering operations with fractions is crucial for various mathematical applications. This section focuses on the concept of subtracting a value from the sum of two fractions to obtain a desired result. We'll break down the problem step-by-step, ensuring clarity and comprehension. Understanding fractions involves several key concepts, including finding a common denominator, adding and subtracting fractions, and simplifying the resulting fractions. The ability to manipulate fractions is not just essential for academic mathematics but also for real-world applications such as cooking, measuring, and financial calculations. Mastering these skills builds a solid foundation for more advanced mathematical topics. In this first problem, we address the question: What number should be subtracted from the sum of 7/8 and 4/15 to obtain 9/40? To solve this, we need to first find the sum of 7/8 and 4/15, and then determine the value that needs to be subtracted from this sum to result in 9/40. The process involves finding a common denominator for the fractions, adding them, and then setting up an equation to find the unknown value. This exercise reinforces the fundamental principles of fraction arithmetic and algebraic problem-solving. By working through this problem, we not only arrive at the solution but also strengthen our understanding of the underlying mathematical concepts. This foundational knowledge is crucial for tackling more complex problems in algebra and other branches of mathematics.
Solution
To find the number that should be subtracted from the sum of 7/8 and 4/15 to get 9/40, we first need to add 7/8 and 4/15. To add these fractions, we need to find a common denominator. The least common multiple (LCM) of 8 and 15 is 120. So, we convert both fractions to have a denominator of 120:
(7/8) * (15/15) = 105/120
(4/15) * (8/8) = 32/120
Now, we add the fractions:
105/120 + 32/120 = 137/120
Let's denote the number to be subtracted as x. We have the equation:
137/120 - x = 9/40
To solve for x, we rearrange the equation:
x = 137/120 - 9/40
We need a common denominator to subtract these fractions. The LCM of 120 and 40 is 120. So, we convert 9/40 to have a denominator of 120:
(9/40) * (3/3) = 27/120
Now, we subtract:
x = 137/120 - 27/120
x = 110/120
We simplify the fraction by dividing both the numerator and denominator by their greatest common divisor (GCD), which is 10:
x = (110 ÷ 10) / (120 ÷ 10)
x = 11/12
Therefore, 11/12 should be subtracted from the sum of 7/8 and 4/15 to get 9/40.
2) Multiplying by the Reciprocal
Multiplication and reciprocals are fundamental concepts in mathematics, particularly when dealing with fractions. Understanding how to multiply fractions and the role of reciprocals is essential for solving a wide range of mathematical problems. The reciprocal of a number is simply 1 divided by that number, and multiplying a number by its reciprocal always results in 1. This concept is particularly useful when dividing fractions, as dividing by a fraction is the same as multiplying by its reciprocal. The ability to work with reciprocals extends beyond basic arithmetic and is crucial in algebra, trigonometry, and calculus. It allows us to simplify complex expressions and solve equations more efficiently. In this section, we will focus on multiplying a fraction by the reciprocal of another fraction. This exercise reinforces the rules of fraction multiplication and the concept of reciprocals. We will explore how flipping a fraction (finding its reciprocal) transforms the multiplication problem and makes it solvable. Understanding this relationship is key to mastering fraction operations and building a strong foundation in mathematical problem-solving. The application of reciprocals is not limited to numerical calculations; it also appears in various areas of mathematics, such as solving linear equations and dealing with inverse trigonometric functions. By grasping this concept, we open doors to a more profound understanding of mathematical principles.
Solution
To multiply 6/13 by the reciprocal of -7/16, we first need to find the reciprocal of -7/16. The reciprocal of a fraction is obtained by swapping its numerator and denominator. So, the reciprocal of -7/16 is -16/7.
Now, we multiply 6/13 by -16/7:
(6/13) * (-16/7) = (6 * -16) / (13 * 7)
Multiply the numerators and the denominators:
= -96 / 91
The result is -96/91, which is an improper fraction. We can leave it in this form or convert it to a mixed number. However, for most purposes, leaving it as an improper fraction is perfectly acceptable.
Therefore, the product of 6/13 and the reciprocal of -7/16 is -96/91.
3) Finding Rational Numbers Between Two Given Numbers
Rational numbers, a fundamental part of the number system, are numbers that can be expressed as a fraction p/q, where p and q are integers and q is not zero. Finding rational numbers between two given rational numbers is a common exercise that highlights the density property of rational numbers, which states that between any two distinct rational numbers, there exists an infinite number of other rational numbers. This concept is crucial for understanding the structure of the number line and the nature of rational numbers. The process of finding rational numbers between two given numbers typically involves finding a common denominator and then identifying fractions with numerators between the two original fractions. This exercise not only reinforces the understanding of fractions and their representation but also introduces the idea of the infinite nature of rational numbers. The ability to identify and manipulate rational numbers is essential for various mathematical applications, including algebra, calculus, and real analysis. In this section, we will explore the technique of finding rational numbers between -2/3 and -1/3. This specific example provides a clear illustration of the density property and allows us to practice the skills of finding common denominators and identifying intermediate fractions. By working through this problem, we solidify our understanding of rational numbers and their properties.
Solution
To find 5 rational numbers between -2/3 and -1/3, we first need to find a common denominator. In this case, the denominators are already the same, which is 3. However, to find 5 rational numbers, we need to increase the gap between the two fractions. We can do this by multiplying both fractions by a number greater than 5. Let's multiply both fractions by 6/6:
(-2/3) * (6/6) = -12/18
(-1/3) * (6/6) = -6/18
Now, we need to find 5 rational numbers between -12/18 and -6/18. We can easily list these by finding fractions with numerators between -12 and -6:
- -11/18
- -10/18 (which simplifies to -5/9)
- -9/18 (which simplifies to -1/2)
- -8/18 (which simplifies to -4/9)
- -7/18
So, 5 rational numbers between -2/3 and -1/3 are -11/18, -5/9, -1/2, -4/9, and -7/18.
4) Simplifying Algebraic Expressions
Simplifying algebraic expressions is a fundamental skill in algebra. It involves using the order of operations, the distributive property, and combining like terms to rewrite an expression in its simplest form. This process is crucial for solving equations, evaluating expressions, and manipulating formulas. Simplifying expressions makes them easier to understand and work with. The distributive property, which states that a(b + c) = ab + ac, is a key tool in simplifying expressions that involve parentheses. Combining like terms, which involves adding or subtracting terms with the same variable and exponent, is another essential step in simplification. The ability to simplify algebraic expressions is not just important for academic mathematics but also for various fields such as engineering, physics, and computer science. In this section, we will tackle the problem of simplifying the expression 2(3x-1) - 5x = -7/2 - 2(2x-7). This equation involves both the distributive property and combining like terms. We will break down the steps involved in simplifying the equation, ensuring a clear understanding of the process. By working through this problem, we reinforce our skills in algebraic manipulation and build a solid foundation for more advanced algebraic concepts.
Solution
To simplify the expression 2(3x - 1) - 5x = -7/2 - 2(2x - 7), we need to follow the order of operations and use the distributive property.
First, distribute the 2 in the first term and the -2 in the last term:
6x - 2 - 5x = -7/2 - 4x + 14
Now, combine like terms on both sides of the equation:
(6x - 5x) - 2 = -4x + (14 - 7/2)
x - 2 = -4x + (28/2 - 7/2)
x - 2 = -4x + 21/2
Next, we want to isolate the variable x. Add 4x to both sides:
x + 4x - 2 = 21/2
5x - 2 = 21/2
Now, add 2 to both sides. To do this, we need to express 2 as a fraction with a denominator of 2, which is 4/2:
5x = 21/2 + 4/2
5x = 25/2
Finally, divide both sides by 5. To do this, we can multiply both sides by 1/5:
x = (25/2) * (1/5)
x = 25/10
Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor (GCD), which is 5:
x = (25 ÷ 5) / (10 ÷ 5)
x = 5/2
Therefore, the solution to the equation is x = 5/2.
5) Solving Linear Equations
Solving linear equations is a core skill in algebra, serving as a foundation for more advanced mathematical concepts. A linear equation is an equation in which the highest power of the variable is 1. Solving such equations involves isolating the variable on one side of the equation by performing the same operations on both sides. This process typically involves adding, subtracting, multiplying, or dividing both sides of the equation by constants or terms involving the variable. The goal is to manipulate the equation until the variable is alone on one side, revealing its value. Linear equations appear in various contexts, from simple word problems to complex scientific and engineering applications. The ability to solve them efficiently and accurately is crucial for problem-solving in mathematics and related fields. In this section, we will focus on solving the equation (5x-2)/7 = (5/14). This equation involves fractions and requires a careful application of algebraic principles to isolate the variable x. We will walk through the steps involved in solving the equation, highlighting the key techniques used in manipulating linear equations. By mastering this type of problem, we strengthen our algebraic skills and prepare for more complex mathematical challenges.
Solution
To solve the equation (5x - 2)/7 = 5/14, we first want to eliminate the fractions. We can do this by multiplying both sides of the equation by the least common multiple (LCM) of the denominators, which is 14.
14 * [(5x - 2)/7] = 14 * (5/14)
On the left side, 14 divided by 7 is 2:
2(5x - 2) = 14 * (5/14)
On the right side, 14 cancels out:
2(5x - 2) = 5
Now, distribute the 2 on the left side:
10x - 4 = 5
Next, add 4 to both sides of the equation:
10x = 5 + 4
10x = 9
Finally, divide both sides by 10 to solve for x:
x = 9/10
Therefore, the solution to the equation is x = 9/10.
In this comprehensive guide, we've explored several key mathematical concepts, including operations with fractions, finding rational numbers, and solving algebraic equations. Each problem has been broken down step-by-step, providing a clear understanding of the underlying principles and techniques. By mastering these fundamental skills, you'll be well-equipped to tackle more complex mathematical challenges and apply these concepts in various real-world scenarios. Continuous practice and a solid understanding of these basics are the keys to success in mathematics.
