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15 Linear Equations [5]
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Chapters and Appendices

Book Entrance
15 Algebra Solutions
15 Triangular Systems
15 Making Triangular
15 With 3 Unknowns
15 Rules and Advice

Foreword
1. Introduction
2. Implication Rules [4]
3. Chains of Reason [3]
4. Induction Mathematical
4. Romeo and Juliet
6  Old Language
5 Knowledge Islands [2]
7  Arith Skill Check [4 X 2]
Arith Webvideos
7. The Next Chapters
8 The Three Skills
8 VNR-Concise-Encyclopedia
PS. What is a Variable [8]
9. Algebra Talk [7]
10 Two More Skills[5]
11 Why Shorthand
12 Shorthand Usage [10]
13 What's Next
PS: The 4-th Skill For Algebra
14 Compound Interest [6]
15 Linear Equations [5]
16 Painless Proofs
17 Pythagoras
PS I.  Distributive Law
PS II. Polynomials
18 Rules of Algebra [20]
19  Functions & Sets
20 Degrees & Radians
21 What's Next
22. Arith & Geometric Sums [2]
23 Summation Notation
24 Your Money [3]
25 Induction & Recursion [4]
26 What's Next
27 Pronouns in Logic
28 Occurrence Tables
29 Contrapositive
30 Truth Tables
31 Indirect Reason
Pathways for Learning

Book Entrance

Would you like to show yourself or others how to be  algebra power users?

What is a Variable?
Introduction
Variation between Examples
Variation of Letters
A letter denotes a variable
Cases of Double Variation
Three Notions of a Variable
Constants, Parameters
& Variables
Talking about numbers
Dependent or Independent
Variable, a Matter of Choice

Chapter 15
Solving Linear Equations

Previous Chapter: Algebra versus Arithmetic Solutions in forward and backward use of the Compound Interest Formula

Here are some more examples in which we solve equations. Our aim is to become familiar or at ease with handling and manipulating equations. So we look at the algebraic solution of equations containing one or more unknown numbers.

1  One Unknown

1.1  First Example

When we let x = 5, we have 2x = 10 and 4x ¹ 15. Suppose now we forgot the value of x which made 2x = 10, could we find the value of x from the equation 2x = 10? The answer is yes. We can solve for the unknown or forgotten value of x as follows:

x = 2x 2 = 10 2 = 5

In this solution, we used the property [(ab)/(b)] = b with the role of a played by x and the role of b played by 2. This gives the first equality. The second equality follows from assumption that 2x = 10. The latter allows 2x to be replaced by its value 10. Another way to look at this solution is to say

2x = 10

Therefore

2x 2 = 10 2

Hence

x = 5

The manipulation process here creates new equalities from previous ones until an expression

x = a numerical value

appears. How we get find the value of x from an equation involving x or other unknowns is a matter of taste.¹

1.2  Second Example

Problem:   Find the value of x which satisfies the equation 7x+9 = 65.

Solution:   The aim is to manipulate (or change or massage) the given equation

7x+9 = 65

to get a new one of the form

x = a numerical value.

The first step is to subtract 9 from both sides. This gives

7x = 65-9

Some of you may know that 65-9 = 56. We could write 56 instead of 65-9. A next step to further isolate x is to divide by 7 (or multiply by [1/7]) since [(7x)/7] = x. This manipulation gives

7x 7 = (65-9) 7

Therefore

x = (65-9) 7 = 56 7 = 8

The isolation of x is complete. The solution is x = 8. To check this, just in case we made a mistake, observe when x = 8, we have 7x+9 = 7·8+9 = 56+9 = 65. So the original the equation 7x+9 = 65 holds (is satisfied, is true) when x = 8.

1.3  Third Example

Problem:   Find the value of x which satisfies 5x+6 = 117.

Solution:   The aim is to manipulate the given equation

5x+6 = 117

to get a new one of the form

x = a numerical value

A first step is to subtract 6 from both sides. This gives

5x = 117-6

A next step to further isolate x is to divide by 5 (or multiply by [1/5]) since [(5x)/5] = x. This manipulation gives

x = 5x 5 = (117-6) 5

Therefore

x = 111 5 = 22+ 1 5 = 22 1 5

The isolation of x is done. The solution is x = [111/5]. To check this, just in case we made a mistake, observe when x = [111/5], we get 5x+6 = 111+6 = 117.

The solution

x = 111 5 =  111 ×2 5 ×2 =  222 10 = 22.2 = 22 1 5

can be written in several ways. Which way we prefer is a matter of taste.

Further sections of this chapter are given by the web pages

[15 Algebra Solutions] [15 Triangular Systems] [15 Making Triangular] [15 Equations With 3 Unknowns] [15 Rules and Advice]

Use the next and back buttons in these links to move between these sections or to return this page.

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