Answer:
See attachment for the figures drawn for each question.
10. x = 3; AB = 11
11. x = 2; AB = 6
Explanation:
10. AC = 10x + 2
AB = 2x + 5
BC = 4x + 9
AB + BC = AC (segment addition postulate)
(2x + 5) + (4x + 9) = 10x + 2 (substitution)
The equation above is what we would use to solve the problem as follows:
2x + 5 + 4x + 9 = 10x + 2
Combine like terms
6x + 14 = 10x + 2
6x - 10x = -14 + 2
-4x = -12
Divide both sides by -4
x = 3
AB = 2x + 5
Plug in the value of x
AB = 2(3) + 5 = 6 + 5
AB = 11
11. AB = 3x
BC = x - 6
AC = 8x - 14
AB + BC = AC (segment addition theorem)
3x + (x - 6) = 8x - 14 (substitution)
Use the equation to solve for x as follows:
3x + x - 6 = 8x - 14
Collect like terms
4x - 6 = 8x - 14
4x - 8x = 6 - 14
-4x = -8
x = 2 (dividing both sides by -4)
AB = 3x
AB = 3(2)
AB = 6