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9
SOLVING 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. Here is a simple example:
x + 64 = 100.
"Some number, plus 64, equals 100."
We say that an equation has two sides: the left side, x + 64, and the right side, 100.
In what we call 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. An equation of the first degree is called linear because, as we will see much later, its graph is a straight line.
Now, the statement -- the equation -- will become true only when the unknown has a certain value, which we call the solution to the equation.
We can find the solution to that equation simply by subtracting:
| x | = | 100 − 64 |
| = | 36. | |
36 is the only value for which the statement "x + 64 = 100" will be true. We say that x = 36 satisfies the equation.
Now, algebra depends on how things look. As far as how things look, then, we will know that we have solved an equation when we have isolated x on the left.
Why the left? Because that's how we read, from left to right. "x equals . . ."
In the standard form of a linear equation -- ax + b = 0 -- x appears on the left.
In fact, we are about to see that for any equation that looks like this:
| x + a | = | b, |
| the solution will look like this: | ||
| x | = | b − a. |
Inverse operations
There are two pairs of inverse operations. Addition and subtraction, multiplication and division.
Formally, to solve an equation we must isolate the unknown -- typically x -- on the left.
ax − b + c = d.
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:
By writing it on the other side with the inverse operation.
For, on the one hand, that preserves the arithmetical relationship between addition and subtraction:
36 + 64 = 100 implies 36 = 100 − 64;
and on the other, between multiplication and division:
| 10 2 |
= | 5 | implies | 10 = 2· 5. |
Algebra is, after all, abstracted -- drawn from -- arithmetic.
And so, to solve this equation:
| 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 are 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."
"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 was shifted to the other side by means of the inverse operation. Every linear equation can be solved by combining those four formal rules.
Solving each form can also be justified algebraically by appealing to the Two rules for equations, Lesson 6.
See the Examples and Problems.
When the operations are addition or subtraction (Forms 1 and 2), we call that 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. When you see
| x + a | = | b, |
| then you immediately see that +a goes to the other side as −a: | ||
| x | = | b − a. |
The way that is often taught these days, is to add −a to both sides, draw a line, and add:
While that is logically correct (Lesson 6), it is clumsy,
it disregards the logical sequence of statements, and
therefore offers no preparation to read a calculus text,
for example.
What, after all, is the purpose of it? The purpose is to
transpose +a to the other side as −a. 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
That equation, incidentally, is in the standard form, namely ax + b = 0.
| Problem 6 . Solve: | ax + b | = | 0. | |
| x | = | − | b a |
|
Each of these problems illustrates doing algebra with your eyes. The student should see the solution immediately. In the example above, you 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. |
| (Lesson 5.) | ||
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 |
Section 2: The unknown on both sides
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