Problem 46 - Algebra for College Students

Question

Set up an equation and solve each problem. Find two numbers such that their sum is 6 and their product is \(7 .\)

Step-by-Step Solution

Answer

The numbers are \(3 + \sqrt{2}\) and \(3 - \sqrt{2}\).

Step 1: Define the Variables

Let's define the two numbers we need to find. Let the first number be \( x \) and the second number be \( y \). We'll set up equations based on these variables.

Step 2: Set Up the Equations

According to the problem, the sum of the two numbers is 6. This can be written as an equation: \( x + y = 6 \).The problem also states that the product of the two numbers is 7. This is another equation: \( x \cdot y = 7 \).

Step 3: Express One Variable in Terms of the Other

From the equation \( x + y = 6 \), solve for one variable in terms of the other. For example, express \( y \) in terms of \( x \):\[ y = 6 - x \]

Step 4: Substitute Into the Second Equation

Substitute the expression for \( y \) from Step 3 into the second equation \( x \cdot y = 7 \):\[ x(6 - x) = 7 \]

Step 5: Form a Quadratic Equation

Expand the equation from Step 4 and move all terms to one side to form a quadratic equation:\[ x(6 - x) = 7 \]\[ 6x - x^2 = 7 \]\[ -x^2 + 6x - 7 = 0 \]Rearrange to make it more standard:\[ x^2 - 6x + 7 = 0 \]

Step 6: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve the equation \( x^2 - 6x + 7 = 0 \).Here, \( a = 1 \), \( b = -6 \), \( c = 7 \):\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} \]\[ x = \frac{6 \pm \sqrt{36 - 28}}{2} \]\[ x = \frac{6 \pm \sqrt{8}}{2} \]\[ x = \frac{6 \pm 2\sqrt{2}}{2} \]\[ x = 3 \pm \sqrt{2} \]

Step 7: Determine the Values of the Two Numbers

The values for \( x \) are \( 3 + \sqrt{2} \) and \( 3 - \sqrt{2} \).Substitute back into \( y = 6 - x \) to find \( y \):If \( x = 3 + \sqrt{2} \), then \( y = 3 - \sqrt{2} \).If \( x = 3 - \sqrt{2} \), then \( y = 3 + \sqrt{2} \).

Key Concepts

Simultaneous Equations

Simultaneous equations are a set of equations with multiple variables which are solved together. To solve them, we look for values of the variables that satisfy all given equations at the same time.
In our original exercise, we have two key conditions forming our simultaneous equations:

• The sum of two numbers is 6: \( x + y = 6 \).
• The product of the same two numbers is 7: \( x \cdot y = 7 \).

These equations need to be solved together because they describe the same pair of numbers. This means we are looking for values of \( x \) and \( y \) that make both equations true simultaneously.
To find these values, we use substitution or elimination methods. In our solution, we use substitution: solve one equation for a variable and substitute it into the other equation to find a solution that satisfies both.

Equation Solving

Equation solving involves finding the value of unknown variables that make an equation true. In the context of our problem, solving the equation solves the puzzle of finding the mystery numbers.
In this problem, we encounter a quadratic equation:

• \( x^2 - 6x + 7 = 0 \)

Before reaching the quadratic form, we set up the known equations and solved one for a variable in terms of the other, leading to a substitution. The resulting equation, \( x(6 - x) = 7 \), expands into a quadratic equation,
which are equations of the second degree of the form \( ax^2 + bx + c = 0 \).
Quadratic equations like this one are often solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a \), \( b \), and \( c \) are coefficients from the equation. For our specific equation, \( a = 1 \), \( b = -6 \), and \( c = 7 \). This formula helps us find the unknown variables \( x \) and \( y \) of our simultaneous equations.

Algebraic Expressions

Algebraic expressions are mathematical phrases that include numbers, variables, and operation symbols. They are used to formulate equations and solve mathematical problems.
In our problem, variables \( x \) and \( y \) represent the unknown numbers. Our given situation translates into algebraic expressions and equations: the sum \( x + y = 6 \) and the product \( x \cdot y = 7 \).
To solve the quadratic equation, manipulating these expressions is key. You’ll see transformations of expressions as we derive \( y = 6 - x \) or \( x (6 - x) = 7 \). These transformations are based on arithmetic operations including addition, subtraction, multiplication, and expanding terms like \( 6x - x^2 = 7 \). Such expressions help create the quadratic equation we eventually solve: \( x^2 - 6x + 7 = 0 \). Understanding how to manipulate these expressions is crucial as they lead us step by step to find the solutions for \( x \) and \( y \), which in turn solve the initial problem.