9
LINEAR EQUATIONS
A logical sequence of statements
Transposing versus exchanging sides
AN EQUATION is an algebraic statement in which the verb is "equals" = . An equation involves an unknown number, typically called x. For example,
2x + 5 = 27.
"Two times some number, plus 5, equals 27."
In what is called a linear equation, x appears only to the first power, as in the equation above. A linear equation is also called an equation of the first degree. (The degree of any equation is the highest exponent that appears on the unknown number.)
Now, the statement -- the equation -- will become true only when x has a certain value, which is called the solution of the equation. The question is: How do we find that value?
Consider this very simple equation:
x − 8 = 2.
"Some number subtract 8 is equal to 2."
The solution is x = 10, because that is the only value of x for which the statement will be true. We say that x = 10 satisfies the equation.
Not all equations of course will be so simple to solve, therefore let us observe the following:
According to the arithmetical relationship between addition and subtraction, to subtract 8 from 10 --
| 10 − 8 | ||||||
| -- we must name the number which when added to 8 will result in 10. | ||||||
| 10 − 8 | = | 2 | ||||
| because | 10 | = | 2 + 8. | |||
And so 8, which originally appears subtracted on the left-hand side of that equation, finally appears added on the right-hand side. And 10 is all alone on the left.
In algebra, which is abstracted from -- drawn from -- arithmetic, we imitate that sequence of statements:
| x − 8 | = | 2 | ||||
| if and only if | x | = | 2 + 8. | |||
As far as how things look, then, we will know that we have solved an equation when we have isolated x on the left-hand side.
Why the left? Because that is how we read, from left to right. x on the left is the style that is observed in all books.
What is more, in the standard form of a linear equation -- ax + b = 0 -- x appears on the left, not the right.
The problem, then, was how to get −8 over to the other side and, in that way, isolate x. We saw that it goes to the other side as +8.
We therefore have one of the basic rules for solving an equation. If any equation looks like this:
| x − a | = | b, | |
| then the solution will look like this: | |||
| x | = | b + a. | |
Here is another simple equation:
2x = 46.
"2 times some number is equal to 46."
In this case, we know from the arithmetical relationship between multiplication and division:
| x | = | 46 2 |
"x equals 46 divided by 2."
x = 23.
In this case, 2, which multiplies on the left side of the original equation, divides on the right side. And note that division is the inverse of multiplication. In the first example, addition is the inverse of subtraction
Inverse operations
There are two pairs of inverse operations. Addition and subtraction, multiplication and division.
Again, to solve an equation we must isolate the unknown (typically x) on the left-hand side of the equation.
ax − b + c = d
To solve that equation, then, we must get a, b, c over to the right, so that x alone is on the left.
The question is: How do we shift a number from one side of an equation to the other?
Answer:
We may shift a number from one side of an equation to the other
by writing it on the other side with the inverse operation.
For, that preserves the arithmetical relationship on the one hand between addition and subtraction; and on the other, between multiplication and division.
Thus, to solve
| ax − b + c | = | d |
| then since b is subtracted on the left, we will add it on the right: | ||
| ax + c | = | d + b |
| Since c is added on the left, we will subtract it on the right: | ||
| ax | = | d + b − c |
| And finally, since a multiplies on the left, we will divide it on the right: | ||
| x | = | d + b − c a |
We have solved the equation.
The four forms of equations
Solving any linear equation, then, will fall into four forms, corresponding to the four operations of arithmetic. The following constitute the basic rules for solving any linear equation. In each case, we will shift a to the other side.
1. If x + a = b, then x = b − a.
"If a number is added on one side of an equation,
we may subtract it on the other side."
2. If x − a = b, then x = b + a.
"If a number is subtracted on one side of an equation,
we may add it on the other side."
| 3. If ax = b, then x = | b a |
. |
"If a number multiplies one side of an equation,
we may divide it on the other side."
| 4. If | x a |
= b, then x = ab. |
"If a number divides one side of an equation,
we may multiply it on the other side."
In every case, a is shifted to the other side by means of the inverse operation. Every linear equation can be solved by combining those four formal rules.
Transposing
When the operations are addition or subtraction (Forms 1 and 2 above), that is called transposing.
We may shift a term to the other side of an equation
by changing its sign.
+ a goes to the other side as − a.
− a goes to the other side as + a.
Transposing is one of the most characteristic operations of algebra, and it is thought to be the meaning of the word algebra, which is of Arabic origin. (Arabic mathematicians learned algebra in India, from where they introduced it into Europe.) Transposing is the technique of those who actually use algebra in science and mathematics -- because it is skillful. And as we are about to see, it maintains the clear, logical sequence of statements. Moreover, it emphasizes that we do algebra with our eyes. That means that when we see
| x + a | = | b, |
| then we immediately see that +a goes to the other side as −a: | ||
| x | = | b − a. |
We can illustrate transposing by applying the algebraic rule that allows us to add or subtract the same number from both sides. (Lesson 6.)
| x − a | = | b | ||
| implies | x − a + a | = | b + a; | |
| that is, | x | = | b + a. | |
But the only reason to add a to both sides was to transpose it. Therefore the student should simply learn to transpose
A logical sequence of statements
In an algebraic sentence, the verb is typically the equal sign = .
ax − b + c = d.
That sentence -- that statement -- will logically imply other statements. Let us follow the logical sequence that leads to the final statement, which is the solution.
| (1) | ax − b + c | = | d | |
| implies | (2) | ax | = | d + b − c |
| implies | (3) | x | = | d + b − c . a |
The original equation (1) is "transformed" by first transposing the terms (Lesson 1). Statement (1) implies statement (2).
That statement is then transformed by dividing by a. Statement (2) implies statement (3), which is the solution.
Thus we solve an equation by transforming it -- changing its form -- statement by statement, line by line according to the rules of algebra, until x finally is isolated on the left. That is how books on mathematics are written (but unfortunately not books that teach algebra!). Each line is its own readable statement that follows from the line above -- with no crossings out
In other words, What is a calculation? It is a discrete transformation of symbols. In arithmetic we transform "19 + 5" into "24". In algebra we transform "x + a = b" into "x = b − a."
Problem 1. Write the logical sequence of statements that will solve this equation for x :
abcx − d + e − f = 0
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Do the problem yourself first!
| (1) | abcx − d + e − f | = | 0 | |
| implies | (2) | abcx | = | d − e + f |
| implies | (3) | x | = | d − e + f . abc |
First, transpose the terms. Line (2).
It is not necessary to write the term 0 on the right.
Then divide by the coefficient of x.
Problem 2. Write the logical sequence of statements that will solve this equation for x :
| (1) | 2x + 5 | = | 27 | |
| implies | (2) | 2x | = | 27 − 5 = 22 |
| implies | (3) | x | = | 22 2 |
| implies | (4) | x | = | 11. |
Problem 3. Solve for x : (p − q)x + r = s
| x = | s − r p − q |
Problem 4. Solve for x : ab(c + d)x − e + f = 0
| x = | e − f ab(c + d) |
Problem 5. Solve for x : 2x + 1= 0
x = −½
This equation, incidentally, is in the standard form, namely ax + b = 0.
| Problem 6 . Solve: | ax + b | = | 0. | |
| x | = | − | b a |
|
Each of these is an example of doing algebra with your eyes. That is, the student should see the solution immediately. In the example above, the student should see that b will go to the other side as −b and that a will divide.
That is skill in algebra.
Problem 7. Solve for x : ax = 0 (a
0).
Now, when the product of two numbers is 0, then at least one of them must be 0 (Lesson 5). Therefore, any equation with that form has the solution,
x = 0.
We could solve that formally, of course, by dividing by a.
| x = | 0 a |
= 0. |
Problem 8. Solve for x :
| 4x − 2 | = | −2 |
| 4x | = | −2 + 2 = 0 |
| x | = | 0. |
Problem 9. Write the sequence of statements that will solve this equation:
| (1) | 6 − x | = | 9 |
| (2) | −x | = | 9 − 6 |
| (3) | −x | = | 3 |
| (4) | x | = | −3. |
When we go from line (1) to line (2), −x remains on the left. For, the terms in line (1) are 6 and −x.
We have "solved" the equation when we have isolated x -- not −x -- on the left. Therefore we go from line (3) to line (4) by changing the signs on both sides. (Lesson 6.)
Alternatively, we could have eliminated −x on the left by changing all the signs immediately:
| (1) | 6 − x | = | 9 |
| (2) | −6 + x | = | −9 |
| (3) | x | = | −9 + 6 = −3 |
| Problem 10. Solve for x : 3 − x | = | −5 | |
| x | = | 8 | |
Problem 11. Solve for x :
| 5 − 2x | = | −11 |
| −2x | = | −11 − 5 |
| 2x | = | 16 Lesson 6 |
| x | = | 8 |
Problem 12. Solve for x:
| 3x − 15 2x + 1 |
= 0 |
(Hint: Compare Lesson 5, Problem 18.)
x = 5.
Transposing versus exchanging sides
| Example 1. | a + b = c − x |
We can easily solve this -- in one line -- simply by transposing x to the left, and what is on the left, to the right:
x = c −a − b.
| Example 2. | a + b = c + x |
In this Example, +x is on the right. Since we want +x on the left, we can achieve that by exchanging sides:
c + x = a + b
Note: When we exchange sides, no signs change.
The solution easily follows:
c + x = a + b − c
In summary, when −x is on the right, it is skillful simply to transpose it. But when +x is on the right, we may exchange the sides.
Problem 13. Solve for x :
| p + q | = | r − x − s | |
| Transpose: | |||
| x | = | r − s − p − q | |
Problem 14. Solve for x :
| p − q + r | = | s + x | |
| Exchange sides: | |||
| s + x | = | p − q + r | |
| x | = | p − q + r − s | |
Problem 15. Solve for x :
| 0 | = | px + q | |
| px + q | = | 0 | |
| px | = | − | q |
| x | = | − | q p |
Problem 16. Solve for x :
| −2 | = | −5x + 1 |
| 5x | = | 1 + 2 = 3 |
| x | = | 3 5 |
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