How do you solve $7x+8<=3x-16$ ?

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Answer 1 · 2018-03-04T16:02:34

$x<=-6$

$7x+8<=3x-16$

Let start by subtracting $3x$ from both sides

$7x + 8 - 3x <= cancel(3x) - 16cancel(-3x)$

$7x + 8 -3x <= -16$

Then subtract $8$ from both sides

$7x cancel(+8) - 3x cancel(-8) <=-16 - 8$

$7x - 3x <=-24$

$4x <=-24$

Finally, divide both sides by $4$ to find $x$

$(cancel4x)/cancel4<=(-24)/4$

$x<=-6$

How do you solve 7x+8<=3x-167x+8<=3x-16?