Question

Solve the system of inequalities: \[\dfrac{{x + 7}}{{x - 8}} > 2,\dfrac{{2x + 1}}{{7x - 1}} > 5\]
(A) \[\left( { - \dfrac{{16}}{7},\dfrac{5}{7}} \right)\]
(B) \[\left( { - \dfrac{7}{8},\dfrac{2}{7}} \right)\]
(C) \[(2,5)\]
(D) No solution

Answer 1

Hint: Here we are asked to find the solution of the given inequalities. First, we will try to simplify the first inequality by using the fact that in division the denominator has to be greater than zero. On doing this we will end up getting two inequalities for the value of $x$ using that we will write the value of $x$ in range or limit. The same process is repeated for the second given inequality. After that we will compare the solutions of both inequalities to check whether they are the same or not. If they are not the same then the given inequalities have no solution.

Complete step by step solution:
A linear inequality is a topic in mathematics where two values are compared by the inequality symbols which are \[' > ',' < ',' \geqslant ',' \leqslant '\]. These values can be of two types. They can be a numerical value or they can be an algebraic value. Linear inequality equations are the same as linear equations, the only difference is that, in linear equations, the inequality sign is replaced by the equal to sign.
Two equations are given to us in the above question and the equation is as follows.
\[\dfrac{{x + 7}}{{x - 8}} > 2\]……. $\left( 1 \right)$
\[\dfrac{{2x + 7}}{{7x - 1}} > 5\]……. $\left( 2 \right)$
First, we will solve $\left( 1 \right)$ which is as follows.
\[\dfrac{{x + 7}}{{x - 8}} > 2\]
The denominator of the fraction should always be greater than zero otherwise the fraction will be not defined because the denominator cannot be zero and the division of any number by zero results in not defined.
\[x - 8 > 0\]
\[ \Rightarrow x > 8\]
Now we will solve our equation further.
\[\dfrac{{x + 7}}{{x - 8}} > 2\]
Now we will do cross multiplication in the above equation.
\[x + 7 > 2(x - 8)\]
\[ \Rightarrow x + 7 > 2x - 16\]
Now we will take the terms with variable $x$ at one side and the constant term on another side.
\[16 + 7 > 2x - x\]
\[ \Rightarrow 23 > x\]
So after solving the equation we come to know that $x$ is greater than \[23\].
So after solving the first equation, we come to know that \[x \in (8,23)\].
Now we will solve $\left( 2 \right)$ which is as follows.
\[\dfrac{{2x + 1}}{{7x - 1}} > 5\]
The denominator should be greater than zero otherwise the result will be not defined. so,
\[
7x - 1 > 0 \\
\Rightarrow 7x > 1 \\
\Rightarrow x > \dfrac{1}{7} \\
\]
On solving $\left( 2 \right)$, the following results will be obtained.
\[\dfrac{{2x + 1}}{{7x - 1}} > 5\]
Now we will cross multiply the equation.
\[
2x + 1 > 5(7x - 1) \\
\Rightarrow 2x + 1 > 35x - 5 \\
\]
The terms with variable x will be on one side and the constant term will be on the other side.
\[
6 > 33x \\
\Rightarrow \dfrac{2}{{11}} > x \\
\]
So for $\left( 2 \right)$, \[x \in \left( {\dfrac{1}{7},\dfrac{2}{{11}}} \right)\]
So after solving $\left( 1 \right)$ and $\left( 2 \right)$, we come to know that there is no common solution for both equations.
Hence the correct answer is (D) no solution

So, the correct answer is “Option D”.

Note: There are different ways to solve the inequality equation. The first method is to add and subtract the same number on both sides of the equation. The second method is to multiply the same number on both sides. If we change the sign of the number then their inequality sign will also be changed.