Question

How do you solve the compound inequalities $6b<42$ or $4b+12>8$?

Answer 1

To solve this linear inequality in one variable, we have to take the variable terms to one side of the inequality, and the constant terms to the other side. Inequalities do not provide a fixed value as a solution, it gives a range. All the values in this range hold the inequality. To solve an inequality we should know some of the properties of the inequality as follows, given that$a>b$. We can state the following from this.
$a+k>b+k,k\in$Real numbers
$ak>bk,k\in$Positive real numbers
$ak < bk,k\in$Negative real numbers

Complete step by step answer:
We are asked to solve the compound inequalities $6b<42$ or $4b+12>8$. Compound inequality means we have to find the range of values that satisfy any of the inequality. To do this, we have to solve both inequality separately, and then take the union of the ranges.
The first inequality is $6b<42$, by multiplying or dividing an inequality by a positive quantity, the inequality sign does not change. Dividing both sides of the above inequality by 6, we get
$\Rightarrow \dfrac{6b}{6}<\dfrac{42}{6}$
$\therefore b<7$
$\therefore b\in \left( -\infty ,7 \right)$
The second inequality is $4b+12>8$. Subtracting 12 from both sides, we get

& \Rightarrow 4b+12-12>8-12 \\\ & \Rightarrow 4b>-4 \\\ \end{aligned}$$ Dividing both sides by 4, we get $$\Rightarrow \dfrac{4b}{4}>\dfrac{-4}{4}$$ $$\therefore b\in \left( -1,\infty \right)$$ The solution ranges of the inequalities are $$b\in \left( -\infty ,7 \right)$$ and $$b\in \left( -1,\infty \right)$$ respectively. Taking the union of the two ranges, we get $$\begin{aligned} & \therefore b\in \left( -\infty ,7 \right)\bigcup \left( -1,\infty \right) \\\ & \therefore b\in \left( -\infty ,\infty \right) \\\ \end{aligned}$$ **Note:** Here, we took the union of the ranges because the question says ‘$$6b<42$$ or $$4b+12>8$$’, it has the word ‘or’ between the inequalities. If the question was ‘$$6b<42$$ and $$4b+12>8$$’, then we have to find the range for two ranges separately and take the intersection of their ranges.