A first degree equation is a polynomial equation whose degree or maximum exponential numerical value of the unknown or unknowns is 1, at the moment in which such polynomial, reduced to normal form, or rather, after having added all the similar monomials that made it up, is set equal to 0.
If the unknown exponential appears to be of a numerical value greater than 1 in an equation reduced to normal form, then such equation will be defined of a higher degree than the first.
The equations with one or more unknowns or variables (if they result to be reduced to normal form and of first degree) are called linear equations because their representation on the Cartesian plane graphically corresponds to straight lines; therefore linear geometric figures.
The normal form in which a one variable (or unknown) first degree equation is reduced is the following: ax + b = 0, which solved in x = -b / a
it is discussed in these terms:
if a is different from 0 and b is different from 0, the equation allows a solution
if a is different from 0 and b is equal to 0, the equation is 0
if a is equal to 0 and b is not equal to 0, the equation is impossible
if a is equal to 0 and b is equal to 0,
the equation is indeterminate
Instead, a first degree inequality, therefore linear, is a polynomial inequality with one or more unknowns or variables of maximum exponential degree = 1, where instead of the mathematical symbol “=” that testifies to a polynomial equivalence ( typical of equations) the symbols “>”, “<“, “> =”, “<=” are introduced, testifying a polynomial non-equivalence.
In one-dimensional space, therefore along a line or an infinite axis, the result of a linear equation is always and only a coordinate of a point graphically
while the result of a linear inequality is a segment.
In two-dimensional space, therefore on a classic Cartesian plane, the result of a linear equation is a line
while the result of a linear inequality is an area
The symbol of inequality “ > “or “<” combined with the symbol of equality “=” in the algebraic result of the inequation indicates how the point in question is included in the result and therefore, projecting the latter on a one-dimensional line, it belongs to the segment of the result of the inequation, while when projecting it on a two-dimensional Cartesian plane, it belongs to the area of the result of the inequality itself.
First degree equations reduced to two-variable normal form will always represent incident lines with the Cartesian axes.
y = +mx + q
y = -mx + q
First degree inequalities reduced to two-variable normal form will always represent infinite areas for which their side of origin will be incident with the Cartesian axes.
y> = mx + q
y <= mx + q
First degree equations reduced to one-variable normal form will always represent orthogonal lines or lines coincident to the Cartesian axes.
y = q
x = q
with q other than 0
y = 0
x = 0
First degree inequalities reduced to one-variable normal form will always represent infinite areas for which their side of origin will be orthogonal or coincident to the Cartesian axes.
y> = q
y <= q
x> = q
x <= q
with q other than 0
y> = 0
y <= 0
x> = 0
x <= 0
However, when considering the y-axis starting from the explicit equation of the generic straight line, a different graphical development is observed: the explicit equation of a generic straight line is: y = mx + q
As far as “q” is concerned, it corresponds (for the ordinates) to the point of intersection of the line with the y-axis. If the analyzed line is the y-axis, we can see that it has no points of intersection with itself, but it is coincident with itself. Therefore, considering the y-axis as a straight line consisting of an infinite set of points, it will coincide with itself in infinitive points and consequently, the value of “q” in its explicit equation will be “infinite”.
As for the coefficient “m”:
If “m” results to be the value indicating the measurement of the vertical cathetus of the right triangle built from q in the x direction and if the horizontal cathetus of this triangle results to be a segment of length “1 unit”, then:
If the straight line turns out to be vertical, a case that we want to analyze now in order to study the value of “m” in the equation of the Cartesian y-axis , the following would happen:
the more the hypothetical straight line that we want to coincide with the y-axis will increase its inclination and will consequently get closer to the y-axis, the more its slope will have an ever greater value and therefore, in the case of the y-axis or of a vertical straight line which is by definition a straight line with a maximum slope, the “m” slope will have the largest value that can be assumed; therefore it will beinfinite.
If the maximum inclination of a straight line is given by the ratio between the infinite value of the vertical cathetus of the triangle and the value 1 for the horizontal cathetus (although the numerator is an infinite value, and so its inclination, being maximum with respect to the x-axis, will generate a vertical straight line), there will, however, be an infinitesimal starting displacement with respect to the y-axis; a unitary displacement that will not allow the perfect coincidence of the explicit equation with m equal to infinity with the y-axis.
So, the following coincident line is representative of the y-axis for the equation y = oox + oo: if m is equal to + infinity, it will rotate by one infinitesimal clockwise with respect to the y-axis, and if m is equal to – infinity, it will rotate by one infinitesimal counterclockwise with respect to the y-axis.
Therefore, analyzing the redefinition of the respective areas of influence generated by the linear inequality to a variable corresponding to the y-axis on the Cartesian plane, considering the axis of the ordinates starting from the explicit equation of the straight line, the results will be as follows:
For y = oox + oo, understood as x = 0, with respect to the x intervals
the inequality y> = + oox + oo will generate the third quadrant of interval [-oo, 0-] and the second quadrant of interval [-oo, o +], and the inequality y <= + oox + oo will generate the first quadrant of interval [o +, + oo] and the fourth quadrant of interval [o -, + oo].
the inequality y> = – oox + oo will generate the first quadrant of interval [0 -, + oo] and the fourth quadrant of interval [0 +, + oo] and the inequality y <= – oox + oo will generate the third quadrant of interval [-oo, 0 +] and the second quadrant of interval [-oo, 0-].