3 years ago

12x+7<-11 or 5x-8>40

Mathematics

1 answer:

puteri [66]3 years ago

7 0

Answer:

$\large\boxed{x9\dfrac{3}{5}\to x\in\left(-\infty,\ -1\dfrac{1}{2}\right)\ \cup\ \left(9\dfrac{3}{5},\ \infty\right)}$

Step-by-step explanation:

$12x+7$

$5x-8>40\qquad\text{add 8 to both sides}\\\\5x-8+8>40+8\\\\5x>48\qquad\text{divide both sides by 5}\\\\\dfrac{5x}{5}>\dfrac{48}{5}\\\\x>\dfrac{48}{5}\\\\x>9\dfrac{3}{5}\\===========================$

$x9\dfrac{3}{5}$

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