# Linear equations

A linear equation is an equation whose sides are both linear expressions. Similarly, a linear inequality is an equality shose sides are both linear expressions.

Linear equations and inequalities in one variable can always be solved using this method:

• Simplify both sides (if necessary).
• If there is a term on the right side with the variable in it, then subtract this term from both sides (and simplify).
• If there is a constant term on the left side, then subtract this term from both sides (and simplify).
• If there is now a negative coefficient on the variable on the left side, then take the opposite of both sides (and simplify).
• If there is now a positive coefficient on the variable on the left side, then divide both sides by that coefficient (and simplify).
At this point, you should have the answer, with the variable equal to (or less than, whatever) a constant. Failing that, you might have a constant statement (with no variable in it), which will be either true or false; then that is your answer.

## Examples

Here is a straightforward example:
• 5x + 2 = 3x + 6 — original equation;
• 2x + 2 = 6 — subtract 3x from both sides;
• 2x = 4 — subtract 2 from both sides;
• x = 2 — divide both sides by 2.
I'll check this solution too: 5(2) + 2 = 12, and 3(2) + 6 = 12, so this checks. In other words, the solution set for x is {2}.

The next example needs to be simplified first:

• 3(2a + 4) = 2a + 2(a + 9) — original equation;
• 6a + 12 = 4a + 18 — simplify both sides;
• 2a + 12 = 18 — subtract 2a from both sides;
• 2a + 12 = 18 — subtract 2a from both sides;
• 2a = 6 — subtract 12 from both sides;
• a = 3 — divide both sides by 2.
Checking, 3(2(3) + 4) = 3(10) = 30, and 2(3) + 2(3 + 9) = 6 + 2(12) = 30, so this checks; the solution set for a is {3}.

Here's an example with some negative numbers:

• 2t − 8 = 5t − 5 — original equation;
• −3t − 8 = −5 — subtract 5t from both sides;
• −3t = 3 — add 8 (same as subtracting −8) from both sides;
• 3t = −3 — take the opposite of both sides;
• t = −1 — divide both sides by 3.
Checking, 2(−1) −8 = −10, and 5(−1) − 5 = −10, so this checks; the solution set for t is {−1}.

Here's an example where you can skip one step:

• 4x = 2x + 6 — original equation;
• 2x = 6 — subtract 2x from both sides;
• x = 3 — divide both sides by 2.
To check, 4(3) = 12, and 2(3) + 6 = 12, so this checks; the solution set for x is {3}.

Here's a linear inequality; the basic technique is the same:

• 5y + 4 < 2y − 2 — original inequality;
• 3y + 4 < −2 — subtract 2y from both sides;
• 3y < −6 — subtract 4 from both sides;
• y < −2 — divide both sides by 3.
To check, 5(−2) + 4 = −6, and 2(−2) − 2 = −6, so the original inequality is just barely false when y := −2 (as it should be). To check the direction of the inequality, I'll try y := −3; then 5(−3) + 4 = −11, and 2(−3) − 2 = −8, and −11 < −8. So this checks. In other words, the solution set for y is (−∞, −2).

Here's an example where I have to change the direction of the inequality:

• 2c + 5 ≥ 3c + 7 — original inequality;
• c + 5 ≥ 7 — subtract 3c from both sides;
• c ≥ 2 — subtract 5 from both sides;
• c ≤ −2 — take the opposite of both sides.
To check, 2(−2) + 5 = 1, and 3(−2) + 7 = 1, so the original inequality is just barely true when c := −2. To check the direction, try c := −4 this time; then 2(−4) + 5 = −3, and 3(−4) + 7 = −5, and −3 ≥ −5. So this checks; the solution set for c is (−∞, −2].

## Degenerate cases

Sometimes, when you subtract a variable term, you'll find that the variable disappears after you simplify. (Actually, it could disappear even earlier, when you first simplify the original expressions.) In this case, there is not much point in continuing with the remaining steps; you will have a statement that is simply either true or false.

Here is an example of a false statement:

• 2n + 4 = 2(n + 4) — original expression;
• 2n + 4 = 2n + 8 — simplify both sides;
• 4 = 8 — subtract 2n from both sides;
• False — 4 < 8 in fact.
So this equation is never true; it has no solutions. In other words, the solution set for n is ∅.

Here is an example of a true statement:

• 3p + 6 = 3(p + 2) — original expression;
• 3p + 6 = 3p + 6 — simplify both sides;
• 6 = 6 — subtract 3p from both sides;
• True — 6 = 6 in fact.
So this equation is always true; every real number is a solution for p. (Actually, you could see this immediately after the expressions were simplified; since then both sides are exactly the same!) In other words, the solution set for p is (−∞, ∞).
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