How do solve $\frac{x + 2}{x + 5} \geq 1$ $(x+2)/(x+5)>=1$ algebraically?

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Answer 1 · 2017-01-23T12:09:53

The answer is $x < - 5$ $x<-5$

Multiply LHS and RHS by ${(x + 5)}^{2}$ $(x+5)^2$

Therefore,

$\frac{x + 2}{x + 5} \cdot {(x + 5)}^{2} \geq 1 \cdot {(x + 5)}^{2}$ $(x+2) / (x+5) * (x+5)^2>=1*(x+5)^2$

$(x + 2) (x + 5) \geq {(x + 5)}^{2}$ $(x+2)(x+5)>=(x+5)^2$

$(x + 2) (x + 5) - {(x + 5)}^{2} \geq 0$ $(x+2)(x+5)-(x+5)^2>=0$

$(x^{2} + 7 x + 10) - (x^{2} + 10 x + 25) \geq 0$ $(x^2+7x+10)-(x^2+10x+25)>=0$

$7 x + 10 - 10 x - 25 \geq 0$ $7x+10-10x-25>=0$

$- 3 x - 15 \geq 0$ $-3x-15>=0$

$3 x \leq - 15$ $3x<=-15$

$x \leq - 5$ $x<=-5$

We have to remove the equal sign, as we divide by $0$ $0$

So,

$x < - 5$ $x<-5$

How do solve (x+2)/(x+5)>=1x+2x+5≥1(x+2)/(x+5)>=1 algebraically?