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How do you solve #5b-1>=-11#?

Answer 1

Abigail Allen

#b>=-2#

#5b-1>=-11#

Add #1# to both sides

#5b\cancelcolor(red)(-1+1)>=-11+1#

Simplify what you can

#5b>=-10#

Divide both sides by #5# so that #b# stands alone

#(\cancel(5)b)/\cancel(5)>=-10/5#

#b>=-(2xxcancel5)/cancel5#

#rArrb>=-2#

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Answer 2

Genesis Allen

To solve (5b - 1 \geq -11), you would add 1 to both sides to isolate the variable (b), then divide both sides by 5 to solve for (b).

(5b - 1 + 1 \geq -11 + 1)

(5b \geq -10)

(b \geq -2)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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How do you solve #5b-1&gt;=-11#?

How do you solve #5b-1>=-11#?