How do you solve $- 3 (2 x + 5) \geq 3 x + 3$ $-3( 2x + 5) \geq 3x + 3$ ?

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How do you solve $- 3 (2 x + 5) \geq 3 x + 3$ $-3( 2x + 5) \geq 3x + 3$ ?

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Answer 1 · 2018-03-26T21:53:58

$x \leq - 2$ $x <=-2$

Explanation:

Start out by using the distributive property and multiply $- 3$ $-3$ by the numbers in the parentheses.

$- 3 (2 x + 5) \geq 3 x + 3$ $-3(2x + 5) >= 3x + 3$

$- 6 x - 15 \geq 3 x + 3$ $-6x - 15 >= 3x + 3$

Now finish simplifying:

$- 6 x - 15 \geq 3 x + 3$ $-6x - 15 >= 3x + 3$

$- 9 x - 15 \geq 3$ $-9x - 15 >= 3$

$- 9 x \geq 18$ $-9x >= 18$

$- x \geq 2$ $-x >= 2$

We want $x$ $x$ to be positive, not negative, so we will multiply both sides by $- 1$ $-1$ . But, there's a catch! We need to switch the greater than sign ( $>$ $>$ ) to a lesser than sign ( $<$ $<$ ) because we did this.

$- x (- 1) \geq 2 (- 1)$ $-x(-1)color(red)>= 2(-1)$

$x \leq - 2$ $x color(red)<=-2$

This means that x is less than or equal to -2.

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How do you solve $- 3 (2 x + 5) \geq 3 x + 3$ $-3( 2x + 5) \geq 3x + 3$ ?

1 Answer

Explanation:

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How do you solve −3(2x+5)≥3x+3-3( 2x + 5) \geq 3x + 3?

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How do you solve $- 3 (2 x + 5) \geq 3 x + 3$ $-3( 2x + 5) \geq 3x + 3$ ?