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Identify the solution set of |x + 3| + 7 less than or equal to 10 and describe the graph in a complete sentence.
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\(\bf |x + 3| + 7 \le 10\implies |x + 3| \le 10-7\implies |x + 3| \le 3\\ \quad \\ |x + 3| \le 3\implies \begin{cases} +(x + 3) \le 3\\ \quad \\ \bf -(x + 3) \le 3 \end{cases}\)
solve the inequalties, so you'd get 2 values for "x", 1 for each scenario as far as graphing them http://www.regentsprep.org/regents/math/algebra/ae85/grineqa.htm so you'd use a SOLID LINE, and you'd graph the x = SOME VALUE, line
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