Question

x+4/x-3< 2 is satisfied when x satisfies {a} (-~,3)U(10,~) {b}(3,10) {c}(-~,3)U[10,~) {d}N.O.T​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{\dfrac{x+4}{x-3}\,<\,2}$

$\underline{\textsf{To find:}}$

$\textsf{Solution set of the given inequality}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{x+4}{x-3}\,<\,2}$

$\mathsf{\dfrac{x+4}{x-3}-2<\,0}$

$\mathsf{\dfrac{x+4-2(x-3)}{x-3}\,<\,0}$

$\mathsf{\dfrac{x+4-2x+6}{x-3}\,<\,0}$

$\mathsf{\dfrac{10-x}{x-3}\,<\,0}$

$\textsf{Multiplying minus on bothsides, we get}$

$\mathsf{\dfrac{x-10}{x-3}\,>\,0}$

$\left\begin{array}{|c|c|c|c|}\cline{1-4}&&&\\Interval&x-3&x-10&\dfrac{x-10}{x-3}\\&&&\\\cline{1-4}&&&\\(-\infty,3)&-&-&+\\&&&\\(3,10)&-&+&-\\&&&\\(10,\infty)&+&+&+\\&&&\\\cline{1-4}\end{array}}$

$\textsf{From the table it is clear that,}$

$\mathsf{The\,solution\;set\;is\;(-\infty,3)\,\cup\,(10,\infty)}$

$\therefore\mathsf{Option\;(a)\;is\;correct}$

Answer 2

(x+4)/(x-3)<2 is satisfied when x satisfies

(a) (-∞, 3) U (10, ∞) (b) (-∞, 3) U [10, ∞)

(c) (3,10) (d) NOT

Step-by-step explanation:

=> (x + 4)/(x - 3) < 2

=> (x + 4)/(x - 3) - 2 < 0

=> {x + 4 - 2(x - 3)}/(x - 3) < 0

=> (x + 4 - 2x + 6)/(x - 3) < 0

=> (10 - x)/(x - 3) < 0 , it means one of them must be -ve(both must be of opposite signs)

There are two possible ways to satisfy this:

Case 1:

(10 - x) > 0 then (x - 3) < 0

10 > x & 3 > x

Case 2:

(10 - x) < 0 then (x - 3) > 0

10 < x & 3 < x

Combine both the cases,

x can be greater than 10 and lesser than 3.

x belongs to (- ∞, 3) U (10, + ∞)

Option(1) is correct.