Problem 27 - Elementary Algebra

Question

Three is added to an integer and that sum is doubled. When this result is multiplied by the original integer the product is 20 . Find the integer.

Step-by-Step Solution

Answer

Answer: The integers that satisfy the given conditions are 2 and -5.

Step 1: Understand and translate the problem into an algebraic equation

Let x be the integer. The problem states that we add 3 to x and then double the sum. This can be expressed as 2(x+3). Then, we are asked to multiply this result by the original integer (x). So, the equation becomes x * 2(x+3) = 20.

Step 2: Simplify and solve for the integer x

Now let's simplify and solve the equation for x: x * 2(x+3) = 20 Distribute the 2: 2x(x+3) = 20 2x^2 + 6x = 20 Now move all terms to one side: 2x^2 + 6x - 20 = 0

Step 3: Factor and solve the quadratic equation

We can now factor the quadratic equation: 2(x^2 + 3x - 10) = 0 2(x-2)(x+5) = 0

Step 4: Find the integer solutions for x

Now set each factor to zero and solve for x: x - 2 = 0 => x = 2 x + 5 = 0 => x = -5 Therefore, the integers that satisfy the given conditions are 2 and -5.

Key Concepts

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operators that represent a value or set of values. They provide a way to express situations mathematically using letters to stand in for numbers. In the problem, the integer is represented by the variable \( x \). When you're given a problem like adding three to an integer and doubling the result, algebraic expressions allow you to write this as \( 2(x + 3) \). This expression communicates precisely what operations are happening to our unknown number. Translating word problems into algebraic expressions is a crucial skill in algebra, helping us represent and solve mathematical problems effectively.

Factoring

Factoring is the process of breaking down an equation into simpler parts that, when multiplied together, produce the original equation. It's a key technique for solving quadratic equations. In our exercise, we need to factor the quadratic equation \( 2x^2 + 6x - 20 = 0 \).To factor it, we first factor out a \( 2 \) from all terms, resulting in \( 2(x^2 + 3x - 10) = 0 \). The next step is to decompose \( x^2 + 3x - 10 \) into two binomial factors. We do this by finding two numbers that multiply to \( -10 \) and add to \( 3 \). Upon factoring, we get \( (x - 2)(x + 5) \), showing that \( 2(x-2)(x+5) = 0 \). Factoring helps simplify equations, making it easier to find the solutions.

Integer Solutions

Integer solutions are whole numbers that solve an equation. For our equation \( 2(x - 2)(x + 5) = 0 \), we find integer solutions by setting each factor equal to zero.

• First, we solve \( x - 2 = 0 \) by adding 2 to both sides, giving us \( x = 2 \).
• Second, we solve \( x + 5 = 0 \) by subtracting 5 from both sides, giving us \( x = -5 \).

This process shows that the solutions to the original equation are \( x = 2 \) and \( x = -5 \). Both solutions are integers and they satisfy the algebraic equation after substituting back, showing full validity.

Equation Solving

Equation solving involves finding the value of unknowns that make the equation true. In the given problem, we have the equation \( 2x(x+3) = 20 \), and we need to solve for \( x \).Start by simplifying the equation:

• Distribute the \( 2x \) across \( x+3 \) to obtain \( 2x^2 + 6x = 20 \).
• Bring everything to one side resulting in \( 2x^2 + 6x - 20 = 0 \).

Factor and solve the remaining quadratic equation. Solving expressions like these requires understanding both algebraic techniques and arithmetic conceptual understanding. By practicing problems like these, you hone the skills needed to tackle more complex mathematical challenges.