For Each Value Of $v$, Determine Whether It Is A Solution To $21 = 9v + 3$. \[ \begin{tabular}{|c|c|c|} \hline \multirow{2}{*}{ V$} & \multicolumn{2}{|c|}{Is It A Solution?} \ \cline{2-3} & Yes & No \ \hline -6 & &

Introduction

In mathematics, solving linear equations is a fundamental concept that helps us determine the value of unknown variables. In this article, we will focus on solving the linear equation $21 = 9v + 3$ for each value of $v$. We will explore the concept of linear equations, understand the steps involved in solving them, and apply this knowledge to determine whether each value of $v$ is a solution to the given equation.

What are Linear Equations?

A linear equation is an equation in which the highest power of the variable(s) is 1. In other words, it is an equation that can be written in the form $ax + b = c$, where $a$, $b$, and $c$ are constants, and $x$ is the variable. Linear equations can be solved using various methods, including algebraic manipulation, graphing, and substitution.

The Given Equation

The given equation is $21 = 9v + 3$. To determine whether each value of $v$ is a solution to this equation, we need to isolate the variable $v$ on one side of the equation. We can do this by subtracting 3 from both sides of the equation, which gives us $18 = 9v$.

Solving for $v$

To solve for $v$, we need to isolate the variable on one side of the equation. We can do this by dividing both sides of the equation by 9, which gives us $v = 2$. Therefore, the value of $v$ that satisfies the equation is $v = 2$.

Determining Whether Each Value of $v$ is a Solution

Now that we have found the value of $v$ that satisfies the equation, we need to determine whether each value of $v$ is a solution to the equation. We can do this by substituting each value of $v$ into the equation and checking whether the equation holds true.

Substituting $v = -6$ into the Equation

If we substitute $v = -6$ into the equation, we get:

$21 = 9(-6) + 3$

Expanding the equation, we get:

$21 = -54 + 3$

Simplifying the equation, we get:

$21 = -51$

Since the equation does not hold true, we can conclude that $v = -6$ is not a solution to the equation.

Substituting $v = 2$ into the Equation

If we substitute $v = 2$ into the equation, we get:

$21 = 9(2) + 3$

Expanding the equation, we get:

$21 = 18 + 3$

Simplifying the equation, we get:

$21 = 21$

Since the equation holds true, we can conclude that $v = 2$ is a solution to the equation.

Substituting $v = 0$ into the Equation

If we substitute $v = 0$ into the equation, we get:

$21 = 9(0) + 3$

Expanding the equation, we get:

$21 = 0 + 3$

Simplifying the equation, we get:

$21 = 3$

Since the equation does not hold true, we can conclude that $v = 0$ is not a solution to the equation.

Substituting $v = 1$ into the Equation

If we substitute $v = 1$ into the equation, we get:

$21 = 9(1) + 3$

Expanding the equation, we get:

$21 = 9 + 3$

Simplifying the equation, we get:

$21 = 12$

Since the equation does not hold true, we can conclude that $v = 1$ is not a solution to the equation.

Substituting $v = 3$ into the Equation

If we substitute $v = 3$ into the equation, we get:

$21 = 9(3) + 3$

Expanding the equation, we get:

$21 = 27 + 3$

Simplifying the equation, we get:

$21 = 30$

Since the equation does not hold true, we can conclude that $v = 3$ is not a solution to the equation.

Substituting $v = 4$ into the Equation

If we substitute $v = 4$ into the equation, we get:

$21 = 9(4) + 3$

Expanding the equation, we get:

$21 = 36 + 3$

Simplifying the equation, we get:

$21 = 39$

Since the equation does not hold true, we can conclude that $v = 4$ is not a solution to the equation.

Substituting $v = 5$ into the Equation

If we substitute $v = 5$ into the equation, we get:

$21 = 9(5) + 3$

Expanding the equation, we get:

$21 = 45 + 3$

Simplifying the equation, we get:

$21 = 48$

Since the equation does not hold true, we can conclude that $v = 5$ is not a solution to the equation.

Substituting $v = 6$ into the Equation

If we substitute $v = 6$ into the equation, we get:

$21 = 9(6) + 3$

Expanding the equation, we get:

$21 = 54 + 3$

Simplifying the equation, we get:

$21 = 57$

Since the equation does not hold true, we can conclude that $v = 6$ is not a solution to the equation.

Substituting $v = 7$ into the Equation

If we substitute $v = 7$ into the equation, we get:

$21 = 9(7) + 3$

Expanding the equation, we get:

$21 = 63 + 3$

Simplifying the equation, we get:

$21 = 66$

Since the equation does not hold true, we can conclude that $v = 7$ is not a solution to the equation.

Substituting $v = 8$ into the Equation

If we substitute $v = 8$ into the equation, we get:

$21 = 9(8) + 3$

Expanding the equation, we get:

$21 = 72 + 3$

Simplifying the equation, we get:

$21 = 75$

Since the equation does not hold true, we can conclude that $v = 8$ is not a solution to the equation.

Substituting $v = 9$ into the Equation

If we substitute $v = 9$ into the equation, we get:

$21 = 9(9) + 3$

Expanding the equation, we get:

$21 = 81 + 3$

Simplifying the equation, we get:

$21 = 84$

Since the equation does not hold true, we can conclude that $v = 9$ is not a solution to the equation.

Substituting $v = 10$ into the Equation

If we substitute $v = 10$ into the equation, we get:

$21 = 9(10) + 3$

Expanding the equation, we get:

$21 = 90 + 3$

Simplifying the equation, we get:

$21 = 93$

Since the equation does not hold true, we can conclude that $v = 10$ is not a solution to the equation.

Substituting $v = 11$ into the Equation

If we substitute $v = 11$ into the equation, we get:

$21 = 9(11) + 3$

Expanding the equation, we get:

$21 = 99 + 3$

Simplifying the equation, we get:

$21 = 102$

Since the equation does not hold true, we can conclude that $v = 11$ is not a solution to the equation.

Substituting $v = 12$ into the Equation

If we substitute $v = 12$ into the equation, we get:

$21 = 9(12) + 3$

Expanding the equation, we get:

$21 = 108 + 3$

Simplifying the equation, we get:

$21 = 111$

Since the equation does not hold true, we can conclude that $v = 12$ is not a solution to the equation.

Substituting $v = 13$ into the Equation

If we substitute $v = 13$ into the equation, we get:

$21 = 9(13) + 3$

Expanding the equation, we get:

$21 = 117 + 3$

Simplifying the equation, we get:

$21 = 120$

Since the equation does not hold true, we can conclude that $v = 13$ is not a solution to the equation.

Substituting $v = 14$ into the Equation

If we substitute $v = 14$ into the equation, we get:

$21 = 9(14) + 3$

Expanding the equation, we get:

$21 = 126 + 3$

Simplifying the equation, we get:

$21 = 129$

Since the equation does not hold true, we can conclude that $v = 14$ is not a solution to the equation.

**Substituting

Q: What is a linear equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1. In other words, it is an equation that can be written in the form $ax + b = c$, where $a$, $b$, and $c$ are constants, and $x$ is the variable.

Q: How do I solve a linear equation?

A: To solve a linear equation, you need to isolate the variable on one side of the equation. You can do this by adding, subtracting, multiplying, or dividing both sides of the equation by the same value.

Q: What is the first step in solving a linear equation?

A: The first step in solving a linear equation is to simplify the equation by combining like terms. This involves adding or subtracting the same value to both sides of the equation.

Q: How do I determine whether a value is a solution to a linear equation?

A: To determine whether a value is a solution to a linear equation, you need to substitute the value into the equation and check whether the equation holds true. If the equation holds true, then the value is a solution to the equation.

Q: What is the difference between a solution and a value that is not a solution?

A: A solution is a value that satisfies the equation, while a value that is not a solution is a value that does not satisfy the equation.

Q: Can a value be both a solution and a value that is not a solution?

A: No, a value cannot be both a solution and a value that is not a solution. If a value is a solution, then it satisfies the equation, and if it is not a solution, then it does not satisfy the equation.

Q: How do I know whether a value is a solution to a linear equation?

A: To determine whether a value is a solution to a linear equation, you need to substitute the value into the equation and check whether the equation holds true. If the equation holds true, then the value is a solution to the equation.

Q: What is the importance of solving linear equations?

A: Solving linear equations is an important skill in mathematics because it helps us to determine the value of unknown variables. It is also used in many real-world applications, such as physics, engineering, and economics.

Q: Can linear equations be used to model real-world situations?

A: Yes, linear equations can be used to model real-world situations. For example, a linear equation can be used to model the cost of producing a certain number of items, or the distance traveled by an object.

Q: How do I apply linear equations to real-world situations?

A: To apply linear equations to real-world situations, you need to identify the variables and constants in the equation, and then use the equation to solve for the unknown variable.

Q: What are some common applications of linear equations?

A: Some common applications of linear equations include:

• Modeling the cost of producing a certain number of items
• Determining the distance traveled by an object
• Calculating the interest on a loan
• Finding the area of a rectangle

Q: Can linear equations be used to solve systems of equations?

A: Yes, linear equations can be used to solve systems of equations. A system of equations is a set of two or more equations that are solved simultaneously.

Q: How do I solve a system of linear equations?

A: To solve a system of linear equations, you need to use the method of substitution or elimination to find the values of the variables.

Q: What is the method of substitution?

A: The method of substitution involves substituting the value of one variable into the other equation to solve for the other variable.

Q: What is the method of elimination?

A: The method of elimination involves adding or subtracting the two equations to eliminate one of the variables.

Q: Can linear equations be used to solve quadratic equations?

A: Yes, linear equations can be used to solve quadratic equations. A quadratic equation is a polynomial equation of degree two.

Q: How do I solve a quadratic equation using linear equations?

A: To solve a quadratic equation using linear equations, you need to use the method of substitution or elimination to find the values of the variables.

Q: What is the difference between a linear equation and a quadratic equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1, while a quadratic equation is a polynomial equation of degree two.

Q: Can linear equations be used to solve polynomial equations of degree three or higher?

A: Yes, linear equations can be used to solve polynomial equations of degree three or higher. However, the method of solution may be more complex and may involve the use of advanced mathematical techniques.

Q: What are some common mistakes to avoid when solving linear equations?

A: Some common mistakes to avoid when solving linear equations include:

• Not simplifying the equation before solving
• Not isolating the variable on one side of the equation
• Not checking whether the equation holds true for the solution
• Not using the correct method of solution (such as substitution or elimination)

Q: How do I check whether a solution is correct?

A: To check whether a solution is correct, you need to substitute the value into the equation and check whether the equation holds true. If the equation holds true, then the solution is correct.