Problem 102 - Beginning and Intermediate Algebra

Question

If \(Q(a)=6 a-1,\) find \(a\) so that \(Q(a)=-9\)

Step-by-Step Solution

Answer

The value of 'a' for which Q(a) equals -9 is \(a = -\frac{4}{3}\).

Step 1: Write down the given information

We are given the function Q(a) = 6a - 1, and we have to find the value of 'a' for which Q(a) = -9.

Step 2: Substitute the given value of Q(a) into the equation

In order to find the value of 'a' for which Q(a) equals -9, we need to substitute -9 for Q(a) in the equation: -9 = 6a - 1

Step 3: Solve the equation for a

Now, we need to solve this equation to find the value of 'a': 1. Start by adding 1 to both sides of the equation: \-9 + 1 = 6a - 1 + 1 \-8 = 6a 2. Next, divide both sides by 6: \(-8/6 = (6a)/6\) Simplify the fractions: \(-4/3 = a\)

Step 4: Write down the final answer

So, the value of 'a' for which Q(a) equals -9 is -4/3.

Key Concepts

Functions

Functions are one of the fundamental concepts in mathematics and are essential in understanding linear equations and algebraic expressions. A function describes a specific relationship between two sets of numbers or variables. In our exercise, the function is given by the equation \(Q(a) = 6a - 1\). This tells us that for any input \(a\), the function \(Q(a)\) describes an operation where 6 is multiplied by \(a\) and then reduced by 1.

Key characteristics of functions include:

• The input, which is often referred to as the 'domain' and in this case is the variable \(a\).
• The output, known as the 'range', which is the value that results from using the function. For instance, if \(a = 1\), then \(Q(1) = 6 \times 1 - 1 = 5\).
• Functions may also be linear, meaning they form a straight line when graphed. The function \(Q(a) = 6a - 1\) is a linear function because it can be written in the form \( f(x) = mx + c \), where \(m\) is the slope and \(c\) is the y-intercept.

Understanding functions requires grasping how these relationships work, how to manipulate them, and what they represent in terms of real-world situations or mathematical models.

Solving Equations

Solving equations is a key skill in algebra that involves finding the value of a variable that makes a given equation true. In our exercise, we needed to solve the equation \(-9 = 6a - 1\) to determine the value of \(a\). Let's break down the process:

• First, you identify the equation you need to solve. Here, we know that \(Q(a) = -9\), which gives us \(-9 = 6a - 1\).
• The next step is to isolate the variable \(a\). To do this, you perform operations on both sides of the equation. You can think of these like balancing scales — whatever you do to one side, do to the other. In this case, start by adding 1 to both sides: \(-9 + 1 = 6a - 1 + 1\), resulting in \(-8 = 6a\).
• Once simplified, you can further isolate \(a\) by dividing both sides by 6, thus \(-8/6 = a\). Simplifying this gives \(a = -4/3\).

Solving linear equations requires practice to understand how to perform these steps accurately without errors. It's about making sure each step maintains equality.

Algebraic Expressions

Algebraic expressions form the core vocabulary of algebra, allowing you to generalize mathematical situations. An expression represents a combination of variables, numbers, and operators (like addition and multiplication). In our given function \(Q(a) = 6a - 1\), \(6a - 1\) itself is an algebraic expression because it combines the variable \(a\) with numbers 6 and -1 using multiplication and subtraction.

Important points about algebraic expressions include:

• They allow you to describe equations generally. Instead of solving specific numbers, you're dealing with symbols that can represent many numbers.
• Simplifying expressions is crucial. This might include combining like terms or performing arithmetic operations.
• They are often part of larger equations, such as the one we solved here when \(6a - 1\) was set equal to -9.

Working with algebraic expressions helps develop a deeper understanding of both number operations and relationships, which is crucial for solving more complex mathematical problems.