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Question

Question: Solve the following linear equation in R: \(\dfrac{x}{x-5}>\dfrac{1}{2}\)...

Solve the following linear equation in R:
xx5>12\dfrac{x}{x-5}>\dfrac{1}{2}

Explanation

Solution

We have given an inequality as follows: xx5>12\dfrac{x}{x-5}>\dfrac{1}{2}. Now, subtract 12\dfrac{1}{2} on both the sides of the given equation then solve the subtraction and then find the solutions in x such that the solutions of x are satisfying this inequality. While considering the solutions of x, make sure you will exclude the solution when x is 5 because when x is 5 then the denominator becomes 0 and we know that when the denominator is 0 then the solution becomes not defined.

Complete step-by-step answer:
We have given the following inequality:
xx5>12\dfrac{x}{x-5}>\dfrac{1}{2}
We have to solve this inequality and find the values of x.
Now, subtracting 12\dfrac{1}{2} on both the sides of this inequality which will give us:
xx512>0\dfrac{x}{x-5}-\dfrac{1}{2}>0
Taking 2 and (x5)\left( x-5 \right) as L.C.M on the L.H.S of the above equation we get,
2x(x5)2(x5)>0 x+52(x5)>0 \begin{aligned} & \dfrac{2x-\left( x-5 \right)}{2\left( x-5 \right)}>0 \\\ & \Rightarrow \dfrac{x+5}{2\left( x-5 \right)}>0 \\\ \end{aligned}
The critical points of the above inequality are:
When (x+5=0)&(x5=0)\left( x+5=0 \right)\And \left( x-5=0 \right) so solving these inequalities we get,
x+5=0 x=5 x5=0 x=5 \begin{aligned} & x+5=0 \\\ & \Rightarrow x=-5 \\\ & x-5=0 \\\ & \Rightarrow x=5 \\\ \end{aligned}
Now, we got the critical points as -5, 5.
Plotting these points on the number line we get,

Now, if we substitute the values which are greater than 5 then the expression becomes positive because both (x+5)&(x5)\left( x+5 \right)\And \left( x-5 \right) becomes positive. And if we put values which are less than -5 then also we will get the whole term positive because (x+5)&(x5)\left( x+5 \right)\And \left( x-5 \right) both becomes negative and division of two negative terms is positive.
Hence, the range of values of x where the given expression holds true is:
x(,5)(5,)x\in \left( -\infty ,-5 \right)\bigcup \left( 5,\infty \right)
In the above expression, (,5)\left( -\infty ,-5 \right) means the x can take values less than -5 and (5,)\left( 5,\infty \right) means that the x can take values greater than 5. And \bigcup means union of the two ranges of x that we have just shown.

Note: The point to be noted here is that in the range of values of x that we have written above, don’t include the critical points 5 and -5. There are two reasons for that:
First is if x equals 5 then the denominator becomes 0 and the solution becomes not defined and the other reason is that we have given the inequality “>” not an equality so we cannot include the values.
For e.g. if instead of inequality “>” ''\ge '' sign is given then the expression becomes:
x+52(x5)0\dfrac{x+5}{2\left( x-5 \right)}\ge 0
Then we are going to include -5 also but again, we are not including 5 because then the solution becomes not defined. Then the range of solutions for the above inequality is:
x(,5](5,)x\in (-\infty ,-5]\bigcup \left( 5,\infty \right)