Question

The sum of two numbers is the same as their product, and the difference of their reciprocals is 3 . Find the numbers.

Short answer

The numbers are \( \frac{1}{2} \) and -1.

Step-by-step solution

Set Up the Equations

First, let's denote the two numbers as x and y. According to the problem, we have two key pieces of information: 1) The sum of x and y is the same as their product: \[ x + y = xy \] 2) The difference of their reciprocals is 3: \[ \frac{1}{x} - \frac{1}{y} = 3 \]

Simplify the Reciprocal Equation

Rewrite the second equation: \[ \frac{1}{x} - \frac{1}{y} = 3 \] Combine the fractions: \[ \frac{y - x}{xy} = 3 \] Since we know from the first equation that \[ xy = x + y \], substitute this into the simplified equation: \[ \frac{y - x}{x + y} = 3 \]

Solve for y in Terms of x

Cross-multiply to eliminate the fraction: \[ y - x = 3(x + y) \] Simplify and group like terms: \[ y - x = 3x + 3y \] \[ y - 3y = 3x + x \] \[ -2y = 4x \] Divide both sides by -2: \[ y = -2x \]

Substitute y in the Sum/Product Equation

Now substitute \[ y = -2x \] into our first equation \[ x + y = xy \]: \[ x + (-2x) = x(-2x) \] Simplify: \[ -x = -2x^2 \] Divide both sides by -1: \[ x = 2x^2 \] Rearrange to form a quadratic equation: \[ 2x^2 - x = 0 \] Factor out x: \[ x(2x - 1) = 0 \]

Find the Values of x

Solve the factored equation: \[ x = 0 \] or \[ 2x - 1 = 0 \] For \[ 2x - 1 = 0 \]: \[ 2x = 1 \] \[ x = \frac{1}{2} \] Since \[ x = 0 \] would make a reciprocal undefined, we discard it. Therefore, only \[ x = \frac{1}{2} \] is valid.

Find the Corresponding y Value

Substitute \[ x = \frac{1}{2} \] back into \[ y = -2x \]: \[ y = -2\frac{1}{2} \] \[ y = -1 \]

Key concepts

System of Equations

A system of equations consists of two or more equations that share the same set of unknowns. In our problem, we have two equations involving the variables x and y. The first equation is that the sum of the numbers equals their product: \( x + y = xy \)The second equation states that the difference of their reciprocals is 3: \( \frac{1}{x} - \frac{1}{y} = 3 \). To solve this, we need to find values for x and y that satisfy both equations simultaneously.Working with systems of equations often requires using various algebraic techniques such as substitution or elimination. Each technique helps to simplify and solve the equations step by step.

Reciprocals

The reciprocal of a number is 1 divided by that number. For instance, the reciprocal of x is \( \frac{1}{x} \). In the given problem, the difference of the reciprocals of x and y is given as: \( \frac{1}{x} - \frac{1}{y} = 3 \). Reciprocals are useful in algebra because they can transform division into multiplication. By rewriting and combining the fractions into a single fraction, we simplify our equations: \( \frac{y - x}{xy} = 3 \). This simplification often helps to eventually find the values of the unknowns.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2, usually in the form \(ax^2 + bx + c = 0 \). In our problem, after substitution and simplification, we derive a quadratic equation: \( 2x^2 - x = 0 \). Solving quadratic equations can be done by factoring, using the quadratic formula, or by completing the square. Factoring in our case gives: \( x(2x - 1) = 0 \). From here, we set each factor equal to zero to find the solutions: \( x = 0 \) or \( x = \frac{1}{2} \). Since \( x = 0 \) would lead to a reciprocal which is undefined, we discard it, leaving us with \( x = \frac{1}{2} \) as a valid solution.

Factoring

Factoring is breaking down a complex expression into simpler components, known as factors, that when multiplied together give the original expression. In our problem’s quadratic equation: \( 2x^2 - x = 0 \), we factor by pulling out the common term x: \( x(2x - 1) = 0 \). Factoring is critical in solving quadratic equations because it allows the equation to be broken down into simpler linear equations. Once factored, we set each factor to zero: \( x = 0 \) or \( 2x - 1 = 0 \). Solving these simpler equations gives us the possible values for x, which we then test and validate.