Solving linear 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 \(8x-6\) may not be equal to zero and that therefore \(x={{3}\over{4}}\) is not a solution.
We now distinguish two cases, namely \(8x-6>0\) and \(8x-6<0\).
In both cases we multiply the inequality on both sides by \(8x-6\) because we then get a linear inequality, for which we know there is a solution method.
Suppose \(8x-6>0\), i.e. \(x> {{3}\over{4}}\). Then we get \(8<-9(8x-6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(72x<46\).
Then, dvision by the coefficient of \(x\)gives \(x < {{23}\over{36}}\).
So we have the following system of inequalities: \(x> {{3}\over{4}}\,\wedge\; x < {{23}\over{36}}\)
and this simplifies to \(\text{an empty solution set}\).
Suppose \(8x-6<0\), i.e. \(x< {{3}\over{4}}\). Then we get \(8>-9(8x-6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(72x>46\).
Then, division by the coefficient of \(x\) gives \(x > {{23}\over{36}}\).
So we have the following system of inequalities: \(x< {{3}\over{4}}\,\wedge\; x > {{23}\over{36}}\)
and this simplifies to \({{23}\over{36}}\lt x\lt {{3}\over{4}}\).
The solution of the original inequality is \({{23}\over{36}}\lt x\lt {{3}\over{4}}\).