Dear Cutecumber, the answer is as follows. We will use concept of the triangle midsegment theorem: For the first problem 3x +8 is the midsegment and 2x +24 is the line the midsegment is parallel to. We know from the triangle midsegment formula that if you double the value of the midsegment it equals to line that its parallel to. Therefore, my equation will be 2(3x+8) = 2x +24. When I solve for x, I get x = 2 and the length of AB = 14. In the second question, the midsegment is AC = 3y -5 and the line its parallel to is 4y+2. I setup the following equation: 2(3y-5) = 4y +2. Solving for Y, I get y = 6. The length of HB is 13, which is half the value of HJ. In the 3rd problem, GH is the midsegment, 4z-3, and GH is the line that is parallel to the midsegment, 7z-1. I setup the following equation: 2(4z-3) = 7(z-1). Solving for z, solving for z I get z = 5. GH = 34. I hope this helps. Good day.
