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Question Number 25088 by naziri2013 last updated on 03/Dec/17

Q...((x+7)/(x+4))>1, x∈R

$$ \\ $$ $$ \\ $$ $$ \\ $$ $${Q}...\frac{{x}+\mathrm{7}}{{x}+\mathrm{4}}>\mathrm{1},\:\:\:\:\:{x}\in{R} \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

Commented bymoxhix last updated on 03/Dec/17

∴x+4≠0 ((x+7)/(x+4))>1 (↓×(x+4)^2 >0) (x+7)(x+4)>(x+4)^2 x^2 +11x+28>x^2 +8x+16 3x>−12 ∴x>−4

$$\therefore{x}+\mathrm{4}\neq\mathrm{0} \\ $$ $$\frac{{x}+\mathrm{7}}{{x}+\mathrm{4}}>\mathrm{1}\:\:\left(\downarrow×\left({x}+\mathrm{4}\right)^{\mathrm{2}} >\mathrm{0}\right) \\ $$ $$\left({x}+\mathrm{7}\right)\left({x}+\mathrm{4}\right)>\left({x}+\mathrm{4}\right)^{\mathrm{2}} \\ $$ $${x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{28}>{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{16} \\ $$ $$\mathrm{3}{x}>−\mathrm{12} \\ $$ $$\therefore{x}>−\mathrm{4} \\ $$

Answered by Rasheed.Sindhi last updated on 04/Dec/17

((x+7)/(x+4))>1 , x∈R x+4≠0 C-1: x+4>0 ((x+7)/(x+4))>1⇒x+7>x+4⇒7>4 This means for x+4>0 , the inequality always satisfied. So the solution is x>−4 C-2: x+4<0 ((x+7)/(x+4))>1⇒x+7<x+4⇒7<4 false So for x+4<0 the inequality is always false. Hence x+4≮0⇒x≮−4 Finally x>−4

$$\:\:\:\:\:\frac{\mathrm{x}+\mathrm{7}}{\mathrm{x}+\mathrm{4}}>\mathrm{1}\:,\:\mathrm{x}\in\mathbb{R} \\ $$ $$\mathrm{x}+\mathrm{4}\neq\mathrm{0} \\ $$ $$\mathrm{C}-\mathrm{1}:\:\mathrm{x}+\mathrm{4}>\mathrm{0} \\ $$ $$\:\:\frac{\mathrm{x}+\mathrm{7}}{\mathrm{x}+\mathrm{4}}>\mathrm{1}\Rightarrow\mathrm{x}+\mathrm{7}>\mathrm{x}+\mathrm{4}\Rightarrow\mathrm{7}>\mathrm{4} \\ $$ $$\mathrm{This}\:\mathrm{means}\:\mathrm{for}\:\mathrm{x}+\mathrm{4}>\mathrm{0}\:,\:\mathrm{the} \\ $$ $$\mathrm{inequality}\:\mathrm{always}\:\mathrm{satisfied}. \\ $$ $$\mathrm{So}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{x}>−\mathrm{4} \\ $$ $$\mathrm{C}-\mathrm{2}:\:\mathrm{x}+\mathrm{4}<\mathrm{0} \\ $$ $$\frac{\mathrm{x}+\mathrm{7}}{\mathrm{x}+\mathrm{4}}>\mathrm{1}\Rightarrow\mathrm{x}+\mathrm{7}<\mathrm{x}+\mathrm{4}\Rightarrow\mathrm{7}<\mathrm{4}\:\mathrm{false} \\ $$ $$\mathrm{So}\:\mathrm{for}\:\mathrm{x}+\mathrm{4}<\mathrm{0}\:\mathrm{the}\:\mathrm{inequality} \\ $$ $$\mathrm{is}\:\mathrm{always}\:\mathrm{false}. \\ $$ $$\mathrm{Hence}\:\:\mathrm{x}+\mathrm{4}\nless\mathrm{0}\Rightarrow\mathrm{x}\nless−\mathrm{4} \\ $$ $$\mathrm{Finally}\:\mathrm{x}>−\mathrm{4} \\ $$

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