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How do you graph #3q+7>=13#?

Answer 1

First graph #y=3q+7# by choosing two conventient #q#'s and calculating the corresponding #y#'s (#q=0# is allways a good choice). Then draw a line through them. Also draw the horizontal line at #y=13#. Everywhere the first line is on or above the #13#-mark is a solution (answer: #q>=2#) graph{3x+7 [-31.4, 33.57, -6.62, 25.85]}

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Answer 2

Jace Anderson

To graph (3q + 7 \geq 13), you first need to solve the inequality for (q). Then, you graph the solution on a number line.

(3q + 7 \geq 13)
(3q \geq 13 - 7)
(3q \geq 6)
(q \geq \frac{6}{3})
(q \geq 2)

So, the solution to the inequality is (q \geq 2). To graph this on a number line, you draw a closed circle at (q = 2) (since (q) can be equal to 2) and shade to the right, indicating all values of (q) greater than or equal to 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.

How do you graph #3q+7&gt;=13#?

How do you graph #3q+7>=13#?