## College Algebra (6th Edition)

(a) $|x -4| \lt 3$ (b) $|x - 4| \geq 3$
(a) Since the interval shown has open limits, it means that the solution presented lies between 1 and 7 without including these values. We can write this in the following manner: $$x \lt 7$$ $$and$$ $$x\gt 1$$. To write it in absolute value form, we need a constant (arbitrarily designated $b$) that can satisfy the following condition: $$x - b \lt 7$$ $$and$$ $$x - b \gt 1$$ which we can re-write as: $$x\lt 7 + b$$ $$and$$ $$x \gt 1 + b$$ Since absolute value inequalities are made from two components, one positive and one negative. we can now state the following: $$7 +b = -(1+b)$$ Solving for $b$, we get: $$7 + b= -1 -b$$ $$2b = -8$$ $$b = -4$$ The absolute value form is, therefore: $$|x - 4| \lt 3$$ (b) The solution presented must be less than or equal to 1, OR, greater than or equal to 7. We can write this in the following manner: $$x\leq 1$$ $$OR$$ $$x\geq 7$$ Since this represents the opposite situation as part (a) of this exercise, we can deduce that the absolute value form is merely the same as above with the inequality sign inverted: $$|x - 4| \geq 3$$.