What is the value of x/3?

Understanding the value of $frac{x}{3}$ is a fundamental concept in algebra that involves division of a variable. Let’s break this down step by step to ensure a thorough understanding.

Basic Concept of Division

Division is one of the four basic operations in arithmetic, alongside addition, subtraction, and multiplication. To divide means to split a number into equal parts. For example, if you have 12 apples and you want to divide them among 3 friends equally, each friend would get 4 apples because $12 div 3 = 4$

Variables in Algebra

In algebra, a variable is a symbol (often a letter) that represents a number. For instance, in the expression $x + 5 = 10$, $x$ is a variable. Variables allow us to write expressions and equations that can represent many different situations.

Dividing a Variable

When we talk about $frac{x}{3}$, we are essentially dividing the variable $x$ by 3. This means we are splitting the value that $x$ represents into three equal parts. The result is one-third of whatever $x$ is.

Example 1: Simple Division

Suppose $x = 9$. To find $frac{x}{3}$, we divide 9 by 3:

$frac{9}{3} = 3$

So, if $x = 9$, then $frac{x}{3} = 3$

Example 2: Another Simple Division

Now, let’s say $x = 15$. To find $frac{x}{3}$, we divide 15 by 3:

$frac{15}{3} = 5$

Thus, if $x = 15$, then $frac{x}{3} = 5$

General Formula

The general formula for dividing any variable $x$ by a number $n$ is:

$frac{x}{n}$

In our specific case, $n = 3$, so the formula becomes:

$frac{x}{3}$

Real-World Applications

Understanding how to divide variables is crucial in many real-world scenarios. For example, if you are budgeting and you have $x$ amount of money to spend over 3 months, $frac{x}{3}$ will tell you how much you can spend each month.

Example 3: Budgeting

Imagine you have $x = 300$ dollars to spend over 3 months. To find out how much you can spend each month, you calculate:

$frac{300}{3} = 100$

So, you can spend $100 each month.

Example 4: Sharing Resources

Let’s say you have $x = 12$ liters of water and you need to share it equally among 3 people. Each person gets:

$frac{12}{3} = 4$

So, each person receives 4 liters of water.

Algebraic Manipulation

Sometimes, you might need to manipulate the expression $frac{x}{3}$ in algebraic equations. For instance, if you have the equation $frac{x}{3} = 7$, you can solve for $x$ by multiplying both sides by 3:

$x = 7 times 3$

$x = 21$

Thus, $x = 21$

Example 5: Solving Equations

Consider the equation $frac{x}{3} = 4$. To find $x$, multiply both sides by 3:

$x = 4 times 3$

$x = 12$

So, $x = 12$

Graphical Representation

You can also represent $frac{x}{3}$ graphically. If you plot $y = frac{x}{3}$ on a graph, you will get a straight line that passes through the origin (0,0) with a slope of $frac{1}{3}$. This line shows how $y$ changes as $x$ changes.

Example 6: Plotting the Graph

To plot $y = frac{x}{3}$, you can create a table of values:

$x$$y = frac{x}{3}$
00
31
62
93

Plot these points on a graph and draw a line through them. This line represents the equation $y = frac{x}{3}$

Conclusion

Understanding the value of $frac{x}{3}$ is a simple yet essential concept in algebra. It involves dividing a variable by a number, which can be applied in various real-world scenarios and algebraic manipulations. Whether you’re budgeting, sharing resources, or solving equations, the ability to divide variables is a valuable skill.

Citations

  1. 1. Khan Academy – Basic Algebra
  2. 2. Purplemath – Division of Variables
  3. 3. Math is Fun – Division

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ