Problem 3 - Elementary Algebra

Question

For Problems \(1-10\), find the indicated products by applying the distributive property; for example, $$ \begin{aligned} (x+1)(y+5) &=x(y)+x(5)+1(y)+1(5) \\ &=x y+5 x+y+5 \end{aligned} $$ $$ (x-4)(y+1) $$

Step-by-Step Solution

Answer

\((x-4)(y+1) = xy - 4y + x - 4\).

Step 1: Apply the Distributive Property

To solve the expression \((x-4)(y+1)\), we will apply the distributive property. This means we distribute each term in the first binomial \((x-4)\) to each term in the second binomial \((y+1)\).

Step 2: Distribute x

First, distribute \(x\) from \((x-4)\) to each term in the second binomial \((y+1)\).So, calculate:\(x(y) + x(1) = xy + x\).

Step 3: Distribute -4

Next, distribute \(-4\) from \((x-4)\) to each term in the second binomial \((y+1)\).So, calculate:\(-4(y) + (-4)(1) = -4y - 4\).

Step 4: Combine Like Terms

Now, combine the results from distributing both \(x\) and \(-4\). The intermediate step gives us:\(xy + x - 4y - 4\).Combine like terms to finalize:\(xy - 4y + x - 4\).

Key Concepts

Binomial Multiplication

When you encounter a problem involving the multiplication of two binomials like \((x-4)(y+1)\), the key is to apply the distributive property effectively. Binomial multiplication involves multiplying every term in one binomial by every term in the other. Here's how it works:

• Recognize the Binomials: Identify the two binomials you need to multiply. In this case, they are \((x-4)\) and \((y+1)\).
• Apply the Distributive Property: To use the distributive property, start by taking each term in the first binomial and multiply it by every term in the second binomial. So, step by step, you do:
  • First term in the first binomial \(x\) times each term in the other binomial - \(x(y) + x(1)\).
  • Second term in the first binomial \(-4\) times each term in the other binomial - \(-4(y) + (-4)(1)\).

The process ensures that every combination of terms between the two binomials is considered. This ensures we don't miss any necessary multiplications in our solution.

Combining Like Terms

After you have distributed all the terms in your binomials, you'll often end up with an expression that has several terms, like \(xy + x - 4y - 4\). These terms need to be simplified. This is where combining like terms comes into play. Simply put, like terms are terms that have the same variable raised to the same power.
Here's how you can actively combine them:

• Identify Like Terms: Look for terms that have the same variable component. In the expression \(xy + x - 4y - 4\), like terms to consider might be those with the variable \(y\) or constants.
• Combine Them: Add or subtract the coefficients of these like terms. For instance, if you had terms \(5x\) and \(3x\), their combination would be \(8x\). In our case, however, the goal is to arrange all terms conveniently since they don't combine further.

By organizing similar terms together and simplifying where possible, the expression becomes cleaner and easier to understand, especially in a lengthy problem.

Algebraic Expressions

At the core of distributing and combining terms are algebraic expressions, which are a combination of numbers, variables, and mathematical operations. They do not have an equality; hence, they differ from equations.
Here’s how to handle algebraic expressions effectively:

• Components of Expressions: Recognize the different parts, which include constants, variables (like \(x\) and \(y\)), coefficients (numbers in front of variables), and operators (such as \(+\), \(-\)).
• Decoding Variables: Variables are letters that stand in for numbers we don’t know yet. Using variables allows for generalization, meaning we can apply formulas to multiple numbers.
• Language of Math: An algebraic expression translates a real-world scenario into a mathematical form. This makes them essential in solving problems and finding unknown quantities.

Understanding algebraic expressions means appreciating how they offer a universal method for expressing a wide range of mathematical ideas and problems. They give structure to problems and pave the way for finding solutions.