Solving equations and inequalities: Linear inequalities in one unknown
Reduction to a linear inequality
In some cases, you can reduce complicated inequalities to linear inequalities.
We note first that division by zero is not allowed and that for this reason \(6x+6\) may not be equal to zero and that therefore \(x=-1\) is not a solution.
We now distinguish two cases, namely \(6x+6>0\) and \(6x+6<0\).
In both cases we multiply the inequality on both sides by \(6x+6\) because we then get a linear inequality, for which we know there is a solution method.
Suppose \(6x+6>0\), i.e. \(x> -1\). Then we get \(9<5(6x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-30x<21\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{7}\over{10}}\).
So we have the following system of inequalities: \(x> -1\,\wedge\; x > -{{7}\over{10}}\)
and this simplifies to \(x\gt-{{7}\over{10}}\).
Suppose \(6x+6<0\), i.e. \(x< -1\). Then we get \(9>5(6x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-30x>21\).
Then, division by the coefficient of \(x\) gives \(x < -{{7}\over{10}}\).
So we have the following system of inequalities: \(x< -1\,\wedge\; x < -{{7}\over{10}}\)
and this simplifies to \(x\lt -1\).
The solution of the original inequality is \(x\lt -1\;\vee\;x\gt-{{7}\over{10}}\).