Inequation of the first degree

A inequation of the first degree to an unknown factor is a inequation where unknown factor X appears with degree 1 or 0. More generally, one calls inequation of the first degree any inequation being brought back to an inequation of the preceding type by simple algebraic operations (see Inéquation (elementary mathematics)

Examples of inequations of the first degree:

$3a + 2 \ Leq 5$ of unknown factor has
$\ frac {u-3} {2} + 4 > 5u$ of unknown factor U
$x^2 + 3x - 5 \ geq (X - 9) (x+3)$ of unknown factor X

Resolution

An inequation of the first degree is solved by isolating the unknown factor in one from the members of the inequality using the elementary rules

Example

A club of sport proposes 3 types of payment:

the fixed monthly price of 50 euros
a chart of 10 entries for 35 euros and the entries additional cost 4 euros
a price the entry of 4 euros

1) For which number of meetings in the month, the fixed monthly price is it more advantageous than the entries with the unit?

is X the number of entries, the price of X entered to the unit is 4 X . It is a question of solving 50 < 4 X . This inequation is equivalent to

\ frac {50} {4} < x

one divided the two members of the inequality by 4.

Donc the fixed monthly price is more advantageous than the entries with the unit starting from 13 entries and beyond

2) For which number of entries is it more advantageous to buy a chart of 10 entries than to pay the month's subscription?

For less than 10 entries, it is obvious that the chart is more advantageous than the subscription, that is to say X the number of entries (X larger than 10), the price of X entered with a chart is then of 35 + 4 (X - 10). It is a question of solving 35 + 4 (X - 10) < 50. This inequation is successivment equivalent to the inequation following:

4x - 5 < 50 algebraic operations on the first member of the inequation

4x < 55 one adds 5 to the two members of the inequation

X < 13,75 one divides each member of the inequation by 4

the purchase of a chart is more advantageous than the month's subscription if the number of entries does not exceed 13.

General case

An inequation of the first degree is always brought back to one of the following cases

$ax + B < 0 \ or \ ax + B \ Leq 0$

if has > 0, the inequation is equivalent to $X < \ frac {- B} {has} \ or \ X \ Leq \ frac {- B} {has}$
if has < 0, the inequation is equivalent to $X > \ frac {- B} {has} \ or \ X \ geq \ frac {- B} {has}$
if has = 0, the inequation becomes: $b < 0 \ or \ B \ Leq 0$ . Inequality independent of X and which

or is always true then the whole of the solutions is R
or is always false then the inequation does not have a solution.

Together solutions

One often presents the whole of the solutions in the form of a interval of R .

X < C is characteristic of the interval $] - \ infty; C
* X ≤ C is characteristic of the interval] - \ infty; C]$
X > C is characteristic of the interval $] C; + \ infty
* X ≥ C is characteristic of the interval the two other intervals which one is brought to meet are {} or \ emptyset, and$ R

Graphic interpretation and study of sign

For any reality not no one has, the chart of the right-hand side of equation there = ax + B confirms and illustrates the preceding results.

If has > 0, the line assembles and ax + B is initially negative (for X < - b/a ) then positive
If has < 0, the line descends and ax + B is initially positive (for X < - b/a ) then negative
ax + B is cancelled for X = - b/a which is called the root of ax + B

What makes say that, for all has not no one, ax + B is sign of has after its root

One summarizes this result in a table of sign which indicates, according to the values of X , the sign of ax + B . The first line of the table positions X on the line of realities, the second informs about the sign of ax + B .

for has > 0

|--- |sign ax + B | |}

for has < 0

|--- |sign ax + B | |}

System of several inequations of the first degree

A system of two inequations of the first degree can be reduced to the following form:

$\ left \ {\ begin {matrix} ax+b <0 \ \ and \ \ cx + D < 0 \ end {matrix} \ right.$

To solve this system is to find the whole of realities X checking at the same time the first inequation and the second inequation.

Method : it is enough to solve each inequation separately. One then obtains for each inequation a interval solution $I_1$ for the first inequation, $I_2$ for the second inequation. the solution unit of the system is the Intersection of the two intervals, it is an interval.

Example: To solve the system $\ left \ {\ begin {matrix} 2x + 3 < X + 50 \ \ and \ \ x^2 +3x \ Leq (x+4) (x+5) \ end {matrix} \ right.$

the first inequation is equivalent to the following inequations:

2x < X + 47 one cut off 3 with each member from the inequation

X < 47 one cut off X with each member from the inequation

the unit from the solutions from the first inequation is

I_1 =] - \ infty; 47

the second inequation is equivalent to the following inequations:

x ^2 + 3x \ Leq x^2 + 9x + 20

one developed and reduces the second member

0 \ Leq 6x + 20

one cut off $x^2 + 3x$ with each member

X \ geq -10/3

one used the rule of the sign of ax + B

the whole of the solutions of the second inequation is

I_2 =; + \ infty [

The whole of the solutions of the system is the intervalle $I_1 \ course I_2 =; 47 [$

Notice 1: there exist sometimes systems of inequations: $\ left \ {\ begin {matrix} ax+b <0 \ \ or \ \ cx + D < 0 \ end {matrix} \ right.$ whose solutions are realities X checking one or the other of both inequation (it is enough that one at least of the inequations is checked). The whole of the solutions will be then the Union units solutions.
The first form of system being most current, it is frequent that the word and do not appear any more. On the other hand, for a system of the second form, the word or is essential.

Notice 2: one can conceive according to the same principle a system of three, four,… N inequations.

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