General solution of a linear inequality

Linear formulas and equations: Linear equations and inequalities

General solution of a linear inequality

General solution to a linear inequalityLet #a# and #b# be numbers. We will take a look at the four possible inequalities with #a \cdot x +b# on the left hand side and #0# to the right hand side. All four inequalities have four possible forms of solutions, dependent on the value of #a# and #b#.

The inequality #a \cdot x+b \gt 0#

Case	Solutions
#a \gt 0#	#x \gt -\frac{b}{a}#
#a \lt 0#	#x \lt -\frac{b}{a}#
#a=0# and #b \gt 0#	every #x#
#a=0# and #b \leq 0#	none

The inequality #a \cdot x+b \geq 0#

Case	Solutions
#a \gt 0#	#x \geq -\frac{b}{a}#
#a \lt 0#	#x \leq -\frac{b}{a}#
#a=0# and #b\geq 0#	every #x#
#a=0# and #b\lt 0#	none

The inequality #a \cdot x+b \lt 0#

Case	Solutions
#a \gt 0#	#x \lt -\frac{b}{a}#
#a \lt 0#	#x \gt -\frac{b}{a}#
#a=0# and #b \lt 0#	every #x#
#a=0# and #b \geq 0#	none

The inequality #a \cdot x+b \leq 0#

Case	Solutions
#a \gt 0#	#x \leq -\frac{b}{a}#
#a \lt 0#	#x \geq -\frac{b}{a}#
#a=0# and #b\leq 0#	every #x#
#a=0# and #b\gt 0#	none

All inequalities

Please note that the linear inequality with #a \cdot x +b# on the left hand side and #0# to the right hand side contains all linear inequalities. Because the terms on the right hand side can be subtracted from both sides, so that we get a linear inequality with #a \cdot x +b# on the left hand side, and #0# on the right hand side.

Reduction

Just as with linear equations, we can find solutions by reduction.

In the case #a \gt 0# we subtract #b# from both sides, next we divide by #a#. Since #a# is positive, the sign does not switch.

In the case #a \lt 0# we subtract #b# from both sides, next we divide by #a#. Since #a# is negative, the sign switches.

In the case #a=0# we have a number on both the right and left hand side of the inequality. We will then check if the inequality is true, and if so the inequality will hold for every possible #x#. If the inequality is not true, then the equality will not hold for any possible #x#.

In practice, we do not have to remember these tables. We can solve the linear equations by means of reduction. If the #x# is removed from the inequality and the inequality of the numbers is true, then it will hold for every #x#. If it is not true, then there is no value of #x# for which the inequality holds.

Determine all #x# for which #6\cdot x+1\leq 6\cdot x+4#.

Give your answer in one of the following forms with a suitable number for #a#. Simplify #a# as much as possible.

#x \lt a#
#x\gt a#
#x\le a#
#x\ge a#

When there are no solutions, enter "none". Or when all values of #x# are solutions, enter "all".

#all#

This can be reduced as follows:
\[\begin{array}{rcl} 6x+1 &\leq & 6x+4\\&&\phantom{xxx}\blue{\text{given}} \\
1 &\leq& 4\\ &&\phantom{xxx}\blue{6x \text{ subtracted on both sides }}\\
\end{array}\]
#x# was eliminated from the inequality. Because the remaining inequality is correct, all values of #x# are solutions.

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