Find the largest integer such that 8 more than six times the integer is less than -42.?

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Answer 1 · 2017-07-01T20:09:21

The largest integer that satisfies this is $- 9$ $-9$ .

Translating this into an equation would be:

"Find the largest integer $x$ $x$ such that $6 x + 8 < - 42$ $6x+8<-42$ ."

Simplifying the inequality:

$6 x + 8 < - 42$ $6x+8<-42$

$6 x < - 50$ $6x<-50$

$x < - \frac{50}{6}$ $x<-\frac{50}{6}$

$x < - 8.333$ $x<-8.333$

So, the largest integer less than $- 8.333$ $-8.333$ would be $- 9$ $-9$ .