How do you solve the compound inequalities \[6b<42\] or \[4b+12>8\]?
Answer
624k+ views
Hint: 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
\[\begin{align}
& \Rightarrow 4b+12-12>8-12 \\
& \Rightarrow 4b>-4 \\
\end{align}\]
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
\[\begin{align}
& \therefore b\in \left( -\infty ,7 \right)\bigcup \left( -1,\infty \right) \\
& \therefore b\in \left( -\infty ,\infty \right) \\
\end{align}\]
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.
\[a+k>b+k,k\in \]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
\[\begin{align}
& \Rightarrow 4b+12-12>8-12 \\
& \Rightarrow 4b>-4 \\
\end{align}\]
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
\[\begin{align}
& \therefore b\in \left( -\infty ,7 \right)\bigcup \left( -1,\infty \right) \\
& \therefore b\in \left( -\infty ,\infty \right) \\
\end{align}\]
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.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

