For example, since *x* := 5 is the only assignment
that will make

2true, and it's also (obviously) the only assignment that will makex− 4 = 3x− 9

true, these two equations are equivalent. Similarly, the equationx= 5

is equivalent to the compound statementt= 1/t

because in each case the only solutions aret= 1 ort= −1,

2is equivalent to the inequalityx− 4 > 3x− 9

because in each case, the solutions are given by assigningx< 5,

The book doesn't introduce a symbol to describe equivalence of statements, but you can use ‘⇔’ if you wish. (This symbol is usually read aloud as ‘if and only if’.) For example, the equivalences above can be symbolised thus:

- 2
*x*− 4 = 3*x*− 9 ⇔*x*= 5; *t*= 1/*t*⇔*t*= 1 or*t*= −1;- 2
*x*− 4 > 3*x*− 9 ⇔*x*< 5.

Now, you don't really need this symbol;
as you solve an equation,
you usually just list equivalent equations in a column,
from your original equation to the final answer.
So as far as I'm concerned, you don't have to learn this symbol.
However, I do want to make the point
that you should *not* use an equals sign!
For example, this would be quite wrong:

2In this problem, you dox− 4 = 3x− 9 =x= 5. (WRONG!)

2(is equivalent to the equationx+y) = 7x

2(and of course there's nothing special about the 7x+ 2y= 7x

If an expression in an equation
has a side condition (like *x* ≠ 0),
then you can remove that condition from the equation
if you make a compound statement with ‘and’
that includes the side condition.
For example, consider the equation

Sincex^{3}/x=x.

Notice thatx^{2}=xandx≠ 0.

The tricks above have nothing to do with *equations* as such,
and they work just as well for inequalities
(or any other statement one might make about real numbers).
The big idea for solving equations specifically is this:
*Do the same invertible operation to both sides.*
An **invertible operation**,
is any operation on a real number
that is always defined
and that has another (inverse) operation
that will always turn the result back into the number you started with.

For example, you can add 5 to any number,
and you get back where you started if you then subtract 5,
so adding 5 is an invertible operation.
Or you can add *x* to any number,
and you get back where you started if you then subtract *x*,
so adding *x* is an invertible operation.
In fact, you can add any defined expression you like,
and you get back where you started if you then subtract that same expression,
so adding any defined expression is always an invertible operation.

Other invertible operations include subtracting any expression,
multiplying by any expression known to be nonzero,
and dividing by any expression known to be nonzero.
For multiplication and division,
it's important that you know that the expression is nonzero;
usually, this means that you can only multiply or divide by constants.
In general, you can't divide by *x*,
because if *x* is 0, then this operation is not defined.
And you can't multiply by *x*,
because if *x* is 0, then this operation is not invertible.
(However, sometimes you're in a situation
where you know that *x* can't be 0;
then it's OK to multiply or divide by *x*.)

Finally, you can always turn an equation into an equivalent equation by swapping the two sides. This may seem silly, but it's sometimes nice.

In summary, here's a list of techniques for solving equations:

- Replace either side (or both) with an equivalent expression;
- Add the same expression to both sides (and then simplify them);
- Subtract the same expression from both sides (and simplify);
- Multiply both sides by the same nonzero expression (and simplify);
- Divide both sides by the same nonzero expression (and simplify);
- Swap the sides.

I'll discuss later another technique useful when you have an absolute value in an expression. You'll learn some others in Intermediate Algebra. There are always more techniques, some of which are still being discovered.

To solve an *order* inequality, in contrast,
is a little trickier.
The first technique
—replacing one expression by an equivalent expression—
is exactly the same.
The last technique —swapping the sides— is almost the same,
but you also have to reverse the direction on the inequality.
(For example, 2 < *x* is equivalent to *x* > 2.)
But the main technique
—doing the same invertible operation to each side—
is more complicated;
the operation must *also* be preserve order.

An **order-preserving** operation
is any operation on a real number
that always takes smaller numbers to smaller numbers
and larger numbers to larger numbers.
Notice that this is *relative*;
adding a million may take small numbers to large numbers,
but as long as one number is small*er* than another,
then it will still be smaller after you add a million to both of them.

Addition and subtraction always preserve order
(as long as you are subtracting *from* the two sides of the inequality).
Multiplication or division by a *positive* quantity
is also order-preserving.
But multiplication or division by a *negative* quantity
does *not* preserve order!
For example, start with 2 and 3; notice that 2 < 3.
After you multiply these by the negative number −4,
you get (−4)2 = −8 and (−4)3 = −12,
but −8 > −12.

In fact, multiplication or division by a negative quantity
*reverses* order.
An **order-reversing operation**
is any operation on a real number
that always takes smaller numbers to *larger* numbers
and larger numbers to *smaller* numbers.
You can still use such an operation to solve order inequalities,
but you must reverse the direction of the inequality when you do so!

So here's a summary of techniques for solving order inequalities:

- Replace either side (or both) with an equivalent expression;
- Add the same expression to both sides (and then simplify them);
- Subtract the same expression from both sides (and simplify);
- Multiply both sides by the same positive expression (and simplify);
- Divide both sides by the same positive expression (and simplify);
- Multiply both sides by the same negative expression and reverse the inequality;
- Divide both sides by the same negative expression and reverse the inequality;
- Swap the sides and reverse the inequality.

I'd like to divide both sides of the first equation byx^{2}=xandx≠ 0.

which obviously equivalent to justx= 1 andx≠ 0,

In summary:

*x*^{3}/*x*=*x*;*x*^{2}=*x*and*x*≠ 0;*x*= 1 and*x*≠ 0;*x*= 1.

If you have a compound inequality, like

2 <then instead of doing the same order-preserving operation tox+ 1 < 5,

1 <so that's the answer.x< 4,

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