9
LINEAR EQUATIONS
A logical sequence of statements
Transposing versus exchanging sides
AN EQUATION is a statement involving an unknown number, typically called x. (In what is called a "linear" equation, x appears only to the first power.) The statement -- the equation -- will become true only when x has a certain value. The question is: How do we find that value?
For example,
x + 36 = 100.
"Some number plus 36 is equal to 100."
Now, we know from arithmetic that when we are given the sum (100) and one term (36), then to find the other term, we subtract:
| x | = | 100 − 36 | |
| x | = | 64. | |
We have solved the equation. That is, formally, to solve an equation we must isolate x on the left of the equal sign.
Why the left? Because we read from left to right. A standard form of a linear equation is ax + b = 0, not 0 = ax + b. If we are asked, "What is the solution?", we would answer, "x equals 2," for example, not "2 equals x." x on the left -- x = 2 -- is the style that is observed in all books.
(Compare ax² + bx + c = 0, which is called a quadratic equation in standard form. That form and ax + b = 0 are in the the general form of a polynomial equation of their respective degrees. See Topic 7 of Precalculus.)
Notice: 36, which was added on the left-hand side of the original equation, appears subtracted on the right.
Here is another example:
5x = 80
"5 times some number is equal to 80."
Again, we know from arithmetic that, to find that number, we divide:
| x | = | 80 5 |
| x | = | 16. |
Notice: 5, which multiplies on the left, divides on the right.
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 on the left-hand side of the equal sign.
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.
(We will prove this below.)
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.
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 resolved into some combination of those four forms.
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. It is skillful, and it maintains the clear, logical sequence of statements, as we are about to see.
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 logically transforming it -- changing its form -- statement by statement, line by line, 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
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.
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 that standard form, namely ax + b = 0.
| Problem 6 . Solve: | ax + b | = | 0. | |
| x | = | − | b a |
|
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 6). 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 5, Problem 8.)
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 |
| x | = | 8 |
Problem 12. Solve for x:
| 3x − 15 2x + 1 |
= 0 |
(Hint: Compare Problem 17, Lesson 6.)
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 |
Here are the basic theorems:
| Theorem 1. | If | |||||
| x + a | = | b, | ||||
| then | ||||||
| x | = | b − a. | ||||
For, upon adding −a to both sides:
Which is what we wanted to prove.
Having proved this, then from now on we may write immediately:
| x + a | = | b | |
| implies | x | = | b − a. |
That is, we may transpose.
Similarly, by adding a to both sides, we could prove:
| If | ||||
| x − a | = | b, | ||
| then | ||||
| x | = | b + a. | ||
| Theorem 2. | If | ||||||
| ax | = | b, | |||||
| then | |||||||
| x | = | b a |
. | ||||
| For, on multiplying both sides by | 1 a |
: |
| 1 a |
· ax | = | 1 a |
· b |
| 1· x | = | b a | (Definition of division) | |
| x | = | b a | . | |
Which is what we wanted to prove.
Having proved this, then from now on we may write immediately:
| ax | = | b | ||
| implies | x | = | b a |
. |
Similarly, by multiplying both sides by a, we could prove:
| If | ||||
| x a |
= | b, | ||
| then | ||||
| x | = | ab. | ||
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