Problem 75 - Algebra A Combined Function

Question

$$ \begin{array}{|l|c|c|c|c|} \hline \text { Monomials } & \text { Add } & \text { Subtract } & \text { Multiply } & \text { Divide } \\ \hline 6 x, 3 x & 6 x+3 x=9 x & 6 x-3 x=3 x & 6 x \cdot 3 x=18 x^{2} & \frac{6 x}{3 x}=2 \\ \hline-12 x^{2}, 2 x & -12 x^{2}+2 x ; \text { can't be simplified } & -12 x^{2}-2 x ; \text { can't be simplified } & -12 x^{2} \cdot 2 x=-24 x^{3} & \frac{-12 x^{2}}{2 x}=-6 x \\ \hline 5 a, 15 a & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Addition: \(20a\), Subtraction: \(10a\), Multiplication: \(75a^2\), Division: \(\frac{1}{3}\).

Step 1: Add 5a and 15a

To add the monomials, combine them by adding their coefficients. The expression is \(5a + 15a\). Add the coefficients: \(5 + 15 = 20\). Thus, the result of the addition is \(20a\).

Step 2: Subtract 5a from 15a

To subtract the monomials, subtract the coefficients. The expression is \(15a - 5a\). Subtract the coefficients: \(15 - 5 = 10\). Thus, the result of the subtraction is \(10a\).

Step 3: Multiply 5a and 15a

To multiply the monomials, multiply their coefficients and add the exponents of like variables. The expression is \(5a \cdot 15a\). Multiply the coefficients: \(5 \times 15 = 75\). For the variable \(a\), add the exponents: \(a^1 \cdot a^1 = a^{1+1} = a^2\). Thus, the product is \(75a^2\).

Step 4: Divide 5a by 15a

To divide one monomial by another, divide the coefficients and subtract the exponents of like variables. The expression is \(\frac{5a}{15a}\). Divide the coefficients: \(\frac{5}{15} = \frac{1}{3}\). For the variable \(a\), subtract the exponents: \(a^1 / a^1 = a^{1-1} = a^0 = 1\). Thus, the quotient is \(\frac{1}{3}\).

Key Concepts

Addition of monomials

Adding monomials is a straightforward process. When you add monomials, you are essentially combining like terms. Like terms are terms that have the same variable raised to the same power. For instance, consider the monomials \(5a\) and \(15a\). Both are like terms because they contain \(a^1\). To add them:

• Simplify by adding the coefficients while keeping the variable unchanged.
• In our example, you add: \(5 + 15 = 20\), which gives you \(20a\).
• The variable \(a\) remains the same because it is part of each monomial.

Remember, you can only add monomials that have the exact same variables and exponents, such as \(4x^2\) and \(7x^2\). If the variables or exponents differ, the monomials cannot be combined further.

Subtraction of monomials

Subtraction works on the same principle as addition of monomials, involving only the subtraction of coefficients. Let’s take \(15a\) and \(5a\) as our example for subtraction.

• Begin by subtracting the coefficients: \(15 - 5 = 10\).
• This results in the new monomial: \(10a\).

Again, ensure that the terms you're subtracting are like terms. This means they must share the same variable and exponent. If they do not match, as in cases like \(7x^2\) and \(4x\), they cannot be combined through subtraction.

Multiplication of monomials

When you multiply monomials, you combine their coefficients and sum the exponents of like variables. Take the multiplication of \(5a\) and \(15a\) as an example:

• First, multiply the coefficients: \(5 \times 15 = 75\).
• Then, deal with the variable by adding the exponents: \(a^1 \cdot a^1 = a^{1+1} = a^2\).
• The result of the multiplication is \(75a^2\).

This method applies universally to any monomials with the same variable structure, making multiplication an essential skill in polynomial algebra. Remember, you should add exponents only when the bases (variables) are the same.

Division of monomials

Dividing monomials involves dividing their coefficients and subtracting the exponents of like variables. Consider dividing \(5a\) by \(15a\):

• First, divide the coefficients: \(\frac{5}{15} = \frac{1}{3}\).
• Next, handle the variable by subtracting the exponents: for \(a\), it becomes \(a^{1-1} = a^0 = 1\).
• Thus, the quotient is simply \(\frac{1}{3}\).

Remember, dividing monomials with the same base means you subtract the exponents, and any variable to the power of zero equals one. Division of monomials is a key concept, especially when simplifying expressions in algebra.