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solve |x+3| +7 (is less than or equal to) 10 Describe the graph of the solution in a complete sentence
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The interpretaton of \[\left| x-(-3) \right|\le3\]is "the x-values that are at most 3 units away from -3." Exactly 3 units to the right of -3 is -3+3=0 and exactly 3 units to the left of -3 is -6. Less than 3 units from -3 would be\[-6< x <0\] If you put the two sets of logic together you have the answer:\[-6 \le x \le0\] Now, of course you want to be able to solve this algebraically, ie, using the definiton of absolute value. Here's how: \[x+3\le3and -(x+3)\le3\]solving each of these\[x \le0 \ and\ x+3\ge-3\] \[x \le0 \ and\ x \ge-6\]which is \[-6 \le x \le 0\]
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