## Algebra 1

a) $20 < 2x +16 < 30$ b) $2 < x < 7$, please see the first graph. c) $2/3 < x < 10$, please see the second graph.
a) Two of the sides of the rectangle have lengths of (x+2) and 6. Since these sides are not parallel, the other two sides have sides of the same length. $20< (x+2) + (x+2) + 6 + 6 < 30$ $20 < 2x +16 < 30$ b) $20 < 2x +16 < 30$ $20 -16 < 2x + 16 - 16 < 30-16$ $4 < 2x < 14$ $4/2 < 2x/2 < 14/2$ $2 < x < 7$ c) possible pairs of triangle legs: (6, x+4), (6, 2x), (2x,x+4) $6 + x+4 > 2x$ $10 + x > 2x$ $10 + x - x > 2x - x$ $10 > x$6 + 2x > x + 46 + 2x - x - 4 > x + 4 - x - 42 + x > 02 + x - 2 > 0 - 2x > -2 x+4 + 2x > 63x + 4 > 63x + 4 - 4 > 6 - 4 3x > 2 3x/3 > 2/3x > 2/3\$ If we let x = 11, we would not have a triangle as the lengths would be 6, 15, and 22 (and 6 + 15 < 22). Thus, the solution set is 2/3 < x < 10.