SOLUTION: What is the solution (in interval notation) to the following inequality below?
-10 ≤ x + 6
Algebra.Com
Question 557100: What is the solution (in interval notation) to the following inequality below?
-10 ≤ x + 6
Found 2 solutions by Theo, bucky:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the solution to this equation is x >= -16
in interval notation, that would look like x = [ -16, infinity )
if you're looking for a good reference on interval notation and set builder notation, etc., then here's one that looks pretty good.
http://regentsprep.org/REgents/math/ALGEBRA/AP1/IntervalNot.htm
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Given to solve:
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You can generally work problems such as this one using the same steps and procedures that you would for solving an equation EXCEPT that whenever you MULTIPLY OR DIVIDE BOTH SIDES by a NEGATIVE quantity, you REVERSE THE DIRECTION of the inequality arrow. This exception does not apply for this problem.
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For the given problem, the goal is to get the unknown x by itself on one side of the inequality sign and everything else (in this case everything else is the constants) on the other side.
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Just as we would do for an equation, we want to get the +6 to be on the other side of the inequality sign, so we subtract 6 from both sides as follows:
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On the left side, the -10 and -6 combine to give -16 and on the right side the +6 and the -6 combine to zero and therefore, they no longer appear. You are left with the answer:
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This reads as "x is equal to or greater than -16" meaning that on a number line the solution will show that x must start at -16 and it can be any number along the number line at -16 and to the right of -16.
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Just as you would with an equation, you normally show the answer as having the unknown x on the left side and the constant -16 on the right side. The important thing is that the inequality arrow still shows as pointing away from the x and toward the -16. So the answer normally would be written as:
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and this would still be read as "x is equal to or greater than -16."
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I hope this helps you to understand how a problem such as this one can be worked using procedures that you already know from your work with equations.
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