Linear Inequalities: Linear Inequalities
Solving through equations
You can also solve a linear inequality by
- first replacing the inequality sign with an equals sign,
- then solving the equation,
- and finally by determining the sign of the inequality on the nodes left and right of the solution of the equation.
This method works because the expressions in #x# continuously change together with #x#. This signifies that, if one of the expressions changes sign, this goes through a point in which that expression is zero.
A more precise formulation of this rule requires a notion of function and continuous, which will be dealt with later.
#x \ge \frac{1}{3}#
To see this, we first look at the equation generated by replacing the inequality sign by an equal sign: #3\cdot x-1=0#
With Solving a linear equation by reduction we see that the linear equation with unknown #x# has exactly one solution: #x=\frac{1}{3}#
We now determine the sign of the inequality #3\cdot x-1\geq 0# left of #\frac{1}{3}#, hence, for #x\lt \frac{1}{3}#. To this end we fill for #x# value #0# in. This gives #{-1}\ge 0#. This statement is not true. Another value of #x# smaller than #\frac{1}{3}# may also be used, and will give the same result. The fact that #{-1}\ge {0}# is not true, tells us that for #x\lt{\frac{1}{3}}# inequality #3\cdot x-1\geq 0# is not true.
Inequality #3\cdot x-1\geq 0# is true for #x \gt{\frac{1}{3}}#: if we use #x=2#, then we get #5\geq 0#, which is true.
Because equality is a special case of inequality, #x=\frac{1}{3}# is a solution. Therefore we find the solution to the inequality: #x \ge {\frac{1}{3}}#
Below is a piece of the numbers line. The points #x# where the inequality is true, shown in red. The words no and yes indicate whether the equality holds for #x=0# and #x=2#.
To see this, we first look at the equation generated by replacing the inequality sign by an equal sign: #3\cdot x-1=0#
With Solving a linear equation by reduction we see that the linear equation with unknown #x# has exactly one solution: #x=\frac{1}{3}#
We now determine the sign of the inequality #3\cdot x-1\geq 0# left of #\frac{1}{3}#, hence, for #x\lt \frac{1}{3}#. To this end we fill for #x# value #0# in. This gives #{-1}\ge 0#. This statement is not true. Another value of #x# smaller than #\frac{1}{3}# may also be used, and will give the same result. The fact that #{-1}\ge {0}# is not true, tells us that for #x\lt{\frac{1}{3}}# inequality #3\cdot x-1\geq 0# is not true.
Inequality #3\cdot x-1\geq 0# is true for #x \gt{\frac{1}{3}}#: if we use #x=2#, then we get #5\geq 0#, which is true.
Because equality is a special case of inequality, #x=\frac{1}{3}# is a solution. Therefore we find the solution to the inequality: #x \ge {\frac{1}{3}}#
Below is a piece of the numbers line. The points #x# where the inequality is true, shown in red. The words no and yes indicate whether the equality holds for #x=0# and #x=2#.
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