How do you solve $- 5 x + 5 (2 - 3 x) < 0$ $-5x + 5(2 - 3x) <0$ ?

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How do you solve $- 5 x + 5 (2 - 3 x) < 0$ $-5x + 5(2 - 3x) <0$ ?

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Answer 1 · 2018-04-08T22:26:53

$x > \frac{1}{2}$ $x>1/2$

Explanation:

$- 5 x + 5 (2 - 3 x) < 0$ $-5x+5(2-3x)<0$
Open bracket
$\Rightarrow - 5 x + 10 - 15 x < 0$ $=>-5x+10-15x<0$
Collecting like terms
$\Rightarrow - 5 x - 15 x < 0 - 10$ $=>-5x-15x<0-10$
$\Rightarrow - 20 x < - 10$ $=>-20x<-10$
$\Rightarrow x <$ $=>x<$ $cancel(-10^1)/(cancel(-20^2))$
Since both sides where divided by a negative integer the sign changes
$=>x>1/2$

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How do you solve $- 5 x + 5 (2 - 3 x) < 0$ $-5x + 5(2 - 3x) <0$ ?

1 Answer

Explanation:

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How do you solve -5x + 5(2 - 3x) <0−5x+5(2−3x)<0-5x + 5(2 - 3x) <0?

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Explanation:

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How do you solve $- 5 x + 5 (2 - 3 x) < 0$ $-5x + 5(2 - 3x) <0$ ?