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:
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
Therefore
Hence
The manipulation process here creates new equalities from previous ones until an
expression
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
to get a new one of the form
The first step is to subtract 9 from both sides. This gives
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
Therefore
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
to get a new one of the form
A first step is to subtract 6 from both sides. This gives
A next step to further isolate x is to divide by 5 (or multiply by [1/5])
since [(5x)/5] = x. This manipulation gives
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.
|