Question

. Solve the following problem from tablet YBC \(6967: \mathrm{A}\) number exceeds its reciprocal by \(7 .\) Find the number and the reciprocal. (In this case, that two numbers are "reciprocals" means that their product is \(60 .\) )

Short answer

Answer: The two possible solutions for the number and its reciprocal are (-5, -12) and (12, 5).

Step-by-step solution

Translate Problem Statement into Equations

We can let the number be x and its reciprocal be y. The problem statement tells us the following two pieces of information: 1. A number (x) exceeds its reciprocal (y) by 7, which can be translated to: x - y = 7 2. The product of the number and its reciprocal is 60, which can be translated to: xy = 60 Now we have a system of two equations with two unknowns, which we can solve.

Solve for One Unknown in Terms of the Other

We will solve the first equation to find x in terms of y, and then substitute the expression we find into the second equation. From equation 1, we have: x = y + 7

Substitute the Expression Found in Step 2 into the Second Equation

Substitute x = y + 7 in the second equation: (y + 7)y = 60

Solve the Quadratic Equation

We now have a quadratic equation to solve for y: y^2 + 7y = 60 Next, bring 60 to the left side to convert it to standard form: y^2 + 7y - 60 = 0

Factor the Quadratic Equation

Factor the quadratic equation to find the roots: (y + 12)(y - 5) = 0

Solve for y

Set each factor equal to zero and solve for y: 1. y + 12 = 0 y = -12 2. y - 5 = 0 y = 5

Solve for x

We have two possible values for y, so we will now solve for x for both values using the expression x = y + 7: 1. When y = -12, x = -12 + 7 = -5 2. When y = 5, x = 5 + 7 = 12

Verify the Solutions

Check if the solutions are valid: 1. For x = -5 and y = -12, xy = (-5)(-12) = 60, which is a valid solution. 2. For x = 12 and y = 5, xy = (12)(5) = 60, which is also a valid solution. Therefore, the two solutions for the number and its reciprocal are (-5, -12) and (12, 5).

Key concepts

Reciprocal Numbers

Understanding reciprocal numbers is a foundational aspect of algebra, particularly when dealing with rational numbers and their properties.

Reciprocal numbers, also referred to as multiplicative inverses, are numbers that when multiplied together yield a product of 1. The reciprocal of a number \( x \) is often denoted as \( \frac{1}{x} \), or simply \( x^{-1} \). An important point to note is that every nonzero number has a reciprocal, and finding a reciprocal is as simple as dividing 1 by the number in question.

For instance, the reciprocal of 10 is \( \frac{1}{10} \) or 0.1, and when you multiply 10 by its reciprocal, the result is 1 (\(10 \times 0.1 = 1\)). In the context of the given problem, reciprocal numbers have a special significance as the product of the number and its reciprocal is given to be 60. Not all reciprocal relations are as straightforward as the basic definitions, especially when it involves the multiplication of two reciprocal numbers equalling a number other than 1.

Algebraic Problem Solving

Algebraic problem solving is a systematic process that involves the use of algebraic expressions, equations, and functions to find unknown quantities. When faced with an algebraic problem, one key step is to translate the words of the problem statement into mathematical language. This typically involves identifying what represents the unknowns, often assigning them variables like \( x \) and \( y \) — as we have in our given exercise.

To solve algebraic problems, it’s important to follow a structured approach:

• Begin by defining variables.
• Translate the problem statement into one or more equations.
• Manipulate these equations to isolate the unknowns.
• Apply mathematical operations to find solutions.
• Verify the solutions to ensure they make sense within the context of the original problem.

This methodical process is quite evident in the solution steps for the exercise, where each step was carefully executed to find the values of \( x \) and its reciprocal \( y \) that meet the given criteria.

System of Equations

A system of equations consists of two or more equations with a set of variables that you aim to solve simultaneously. Solutions to these systems represent where the equations intersect when plotted graphically. Here are some methods often used to solve systems of equations:

• Substitution: Solve one equation for one variable and substitute that expression into the other equation(s).
• Elimination (or addition): Add or subtract equations to eliminate one variable, making it easier to solve for the other.
• Graphing: Plot each equation on a coordinate grid and look for points of intersection.
• Matrix methods: Use matrix operations to solve more complex systems of equations.

In the case of our problem, the substitution method was employed. The first equation was rearranged to express one variable in terms of the other, and this expression was substituted into the second equation. This process turned a system of equations into a much more manageable single equation to solve. Once a value was found for one variable, it was substituted back into one of the original equations to find the corresponding value of the second variable. Systems of equations are a crucial concept in algebra as they have wide applications in various fields, including economics, physics, and computer science.