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Solve the following compound inequality: −2(x + 4) + 10 < x − 7 or −2x + 9 > 3(x + 8)
A: x < 3 or x > −3
B: x > 3 or x < −3
C: x < 7 or x > −3
D: x > 7 or x < −3

Respuesta :

#1

[tex]\\ \sf{:}\!\implies -2(x+4)+10<x-7[/tex]

[tex]\\ \sf{:}\!\implies -2x-8+10<x-7[/tex]

[tex]\\ \sf{:}\!\implies x+2x>2+7[/tex]

[tex]\\ \sf{:}\!\implies 3x>9[/tex]

[tex]\\ \sf{:}\!\implies x>\dfrac{9}{3}[/tex]

[tex]\\ \sf{:}\!\implies x>3[/tex]

#2

[tex]\\ \sf{:}\!\implies -2x+9>3(x+8)[/tex]

[tex]\\ \sf{:}\!\implies -2x+9>3x+24[/tex]

[tex]\\ \sf{:}\!\implies -2x-3x>24-9[/tex]

[tex]\\ \sf{:}\!\implies -5x>15[/tex]

[tex]\\ \sf{:}\!\implies x>\dfrac{15}{-5}[/tex]

[tex]\\ \sf{:}\!\implies x>-3[/tex]

Hi1315

Answer:

Answer B is correct

Step-by-step explanation:

[tex]1) - 2(x + 4) + 10 < x - 7 \\ - 2x - 8 + 10 < x - 7 \\ - 2x - x - 8 + 10 < - 7 \\ - 3x + 2 < - 7 \\ - 3x < - 7 - 2 \\ - 3x < - 9 \\ \frac{ - 3x}{ - 3} < \frac{ - 9}{ - 3} \\ x > 3[/tex]

[tex]2) - 2x + 9 > 3(x + 8) \\ - 2x + 9 > 3x + 24\\ - 2x - 3x > - 9 + 24\\ - 5x > 15 \\ \frac{ - 5x}{ - 5} > \frac{15}{ - 5} \\ x < - 3[/tex]

hope this helps you.

let me know if you have any other questions :-)

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