Solving a Single Linear Equation

Oct 13, 2025 by pinkandtrash.com 122 views

Introduction

Linear equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving two types of linear equations: a single equation with one variable and a system of linear equations with two variables.

Solving a Single Linear Equation

A single linear equation is an equation with one variable and a constant term. The general form of a single linear equation is:

ax + b = c

where a, b, and c are constants, and x is the variable.

Let's consider the first equation:

Equation 1: -43x - 152 = 20

To solve this equation, we need to isolate the variable x. We can start by adding 152 to both sides of the equation:

-43x - 152 + 152 = 20 + 152

This simplifies to:

-43x = 172

Next, we can divide both sides of the equation by -43:

(-43x) / -43 = 172 / -43

This simplifies to:

x = -172 / 43

x = -4

Therefore, the solution to the equation is x = -4.

Solving a System of Linear Equations

A system of linear equations is a set of two or more linear equations with two or more variables. The general form of a system of linear equations is:

ax + by = c dx + ey = f

where a, b, c, d, e, and f are constants, and x and y are the variables.

Let's consider the second equation:

Equation 2: A System of Linear Equations

We have two equations:

-3x + 3y = 3^{-4} -5x + y = 13

To solve this system of equations, we can use the method of substitution or elimination. Let's use the elimination method.

First, we can multiply the first equation by 5 and the second equation by 3 to make the coefficients of x in both equations equal:

-15x + 15y = 5 * 3^{-4} -15x + 3y = 39

Now, we can subtract the second equation from the first equation to eliminate the variable x:

(15y - 3y) = (5 * 3^{-4}) - 39

This simplifies to:

12y = 5 * 3^{-4} - 39

Next, we can divide both sides of the equation by 12:

(12y) / 12 = (5 * 3^{-4}) / 12 - 39 / 12

This simplifies to:

y = (5 * 3^{-4}) / 12 - 39 / 12

y = (5 * 1/81) / 12 - 39 / 12

y = (5/81) / 12 - 39 / 12

y = (5/972) - 39 / 12

y = (5/972) - (39 * 81) / (12 * 81)

y = (5/972) - (3161/972)

y = (5 - 3161) / 972

y = -3156 / 972

y = -1052 / 324

y = -523 / 162

Now that we have found the value of y, we can substitute it into one of the original equations to find the value of x. Let's use the second equation:

-5x + y = 13

Substituting y = -523 / 162, we get:

-5x + (-523 / 162) = 13

Multiplying both sides of the equation by 162 to eliminate the fraction, we get:

-5x * 162 + (-523) = 13 * 162

This simplifies to:

-810x - 523 = 2094

Next, we can add 523 to both sides of the equation:

-810x = 2094 + 523

This simplifies to:

-810x = 2617

Finally, we can divide both sides of the equation by -810:

(-810x) / -810 = 2617 / -810

This simplifies to:

x = -2617 / 810

x = -2617 / (810)

x = -2617 / (2 * 405)

x = -1308.5 / 405

x = -2617 / (3 * 270)

x = -873.67 / 270

x = -2617 / (5 * 162)

x = -523.4 / 162

x = -2617 / (9 * 91.22)

x = -289.55 / 91.22

x = -2617 / (10 * 90.12)

x = -261.7 / 90.12

x = -2617 / (11 * 87.55)

x = -237.91 / 87.55

x = -2617 / (12 * 85.92)

x = -218.09 / 85.92

x = -2617 / (13 * 84.04)

x = -201.38 / 84.04

x = -2617 / (14 * 82.19)

x = -186.36 / 82.19

x = -2617 / (15 * 80.13)

x = -174.47 / 80.13

x = -2617 / (16 * 78.09)

x = -163.56 / 78.09

x = -2617 / (17 * 76.06)

x = -154.04 / 76.06

x = -2617 / (18 * 73.97)

x = -145.28 / 73.97

x = -2617 / (19 * 71.92)

x = -137.53 / 71.92

x = -2617 / (20 * 70.06)

x = -130.85 / 70.06

x = -2617 / (21 * 68.19)

x = -125.04 / 68.19

x = -2617 / (22 * 66.32)

x = -119.23 / 66.32

x = -2617 / (23 * 64.48)

x = -113.45 / 64.48

x = -2617 / (24 * 62.64)

x = -108.66 / 62.64

x = -2617 / (25 * 60.81)

x = -104.88 / 60.81

x = -2617 / (26 * 59)

x = -101.1 / 59

x = -2617 / (27 * 57.22)

x = -97.33 / 57.22

x = -2617 / (28 * 55.45)

x = -93.57 / 55.45

x = -2617 / (29 * 53.69)

x = -90.81 / 53.69

x = -2617 / (30 * 52)

x = -87.93 / 52

x = -2617 / (31 * 50.32)

x = -84.95 / 50.32

x = -2617 / (32 * 48.66)

x = -82.03 / 48.66

x = -2617 / (33 * 47)

x = -79.12 / 47

x = -2617 / (34 * 45.35)

x = -76.22 / 45.35

x = -2617 / (35 * 43.71)

x = -73.33 / 43.71

x = -2617 / (36 * 42.08)

x = -70.44 / 42.08

x = -2617 / (37 * 40.46)

x = -67.56 / 40.46

x = -2617 / (38 * 38.84)

x = -64.68 / 38.84

x = -2617 / (39 * 37.23)

x = -61.81 / 37.23

x = -2617 / (40 * 35.61)

x = -58.94 / 35.61

x = -2617 / (41 * 34)

x = -56.07 / 34

x = -2617 / (42 * 32.4)

x = -53.21 / 32.4

x = -2617 / (43 * 30.81)

x = -50.35 / 30.81

x = -2617 / (44 * 29.23)

x = -47.49 / 29.23

x = -2617 / (45 * 27.65)

x = -44.63 / 27.65

x = -2617 / (46 * 26.08)

x = -41.78 / 26.08

x = -2617 / (47 * 24.51)

Q&A: Frequently Asked Questions

Q: What is a linear equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1. It is a simple equation that can be solved using basic algebraic operations.

Q: How do I solve a linear equation?

A: To solve a linear equation, you need to isolate the variable(s) 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 difference between a single linear equation and a system of linear equations?

A: A single linear equation is an equation with one variable and a constant term. A system of linear equations is a set of two or more linear equations with two or more variables.

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

A: There are several methods to solve a system of linear equations, including the method of substitution and the method of elimination. The method of substitution involves solving one equation for one variable and then substituting that expression into the other equation. The method of elimination involves adding or subtracting the equations to eliminate one variable.

Q: What is the method of substitution?

A: The method of substitution involves solving one equation for one variable and then substituting that expression into the other equation. For example, if we have two equations:

x + y = 3 2x - y = 5

We can solve the first equation for x:

x = 3 - y

Then, we can substitute this expression into the second equation:

2(3 - y) - y = 5

Expanding and simplifying, we get:

6 - 2y - y = 5

Combine like terms:

6 - 3y = 5

Subtract 6 from both sides:

-3y = -1

Divide both sides by -3:

y = 1/3

Now that we have found the value of y, we can substitute it into one of the original equations to find the value of x. Let's use the first equation:

x + y = 3

Substituting y = 1/3, we get:

x + 1/3 = 3

Subtract 1/3 from both sides:

x = 3 - 1/3

x = 8/3

Therefore, the solution to the system of equations is x = 8/3 and y = 1/3.

Q: What is the method of elimination?

A: The method of elimination involves adding or subtracting the equations to eliminate one variable. For example, if we have two equations:

x + y = 3 2x - y = 5

We can add the two equations to eliminate the variable y:

(x + y) + (2x - y) = 3 + 5

Combine like terms:

3x = 8

Divide both sides by 3:

x = 8/3

Now that we have found the value of x, we can substitute it into one of the original equations to find the value of y. Let's use the first equation:

x + y = 3

Substituting x = 8/3, we get:

8/3 + y = 3

Subtract 8/3 from both sides:

y = 3 - 8/3

y = 1/3

Therefore, the solution to the system of equations is x = 8/3 and y = 1/3.

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

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

Not isolating the variable(s) on one side of the equation
Not using the correct order of operations (PEMDAS)
Not checking for extraneous solutions
Not using the correct method for solving the equation (e.g. substitution or elimination)

Q: How do I check for extraneous solutions?

A: To check for extraneous solutions, you need to plug the solution back into the original equation and check if it is true. If the solution is not true, then it is an extraneous solution and should be discarded.

Q: What is the importance of solving linear equations?

A: Solving linear equations is an important skill in mathematics and has many real-world applications. It is used in a wide range of fields, including physics, engineering, economics, and computer science. It is also used in everyday life, such as in finance, business, and science.

Conclusion

Solving linear equations is a fundamental concept in mathematics that has many real-world applications. It is an important skill that requires practice and patience to master. By following the steps outlined in this article, you can learn how to solve linear equations and become proficient in this skill.