if AB=3x-5, BC=x+2, and AC= 13, then x=
2007-11-30 03:01:34 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The figure is a line
thanks! and points :)
2007-11-30 03:03:10 · update #1
[04]
AB+BC=AC
3x-5+x+2=13
4x-3=13
4x=13+3=16
x=16/4=4
2007-11-30 03:05:44 · answer #1 · answered by alpha 7 · 2⤊ 1⤋
AB+BC=AC
3x-5+x+2=13
4x-3=13
4x=16
x=4
2007-11-30 11:33:20 · answer #2 · answered by Anonymous · 0⤊ 1⤋
I would say but not sure AC = AB+BC
so AC= 3x-5+x+2 = 4x-3=13
4x=16 and x=4
2007-11-30 11:06:06 · answer #3 · answered by maussy 7 · 1⤊ 0⤋
ok, you have to see this as on a line
A----B-----C
Soooo
AC is the biggest 1, and AB and BC together will become AC because they are part of the whole.
Now that you know that, set up the equation
3x-5 (AB) + x+2 (BC)= 13 (AC)
Combine x's and numbers on the first
4x-3=13 now it's a simple equation
Add 3 on both sides
4x=16
Divide by 4 on both sides
x= 4
2007-11-30 11:07:09 · answer #4 · answered by George 3 · 1⤊ 1⤋
4
2007-11-30 14:48:55 · answer #5 · answered by Tiny R 1 · 1⤊ 1⤋
I assume by "the figure is a line" you meant something like this:
A__________B__________C
In that case, AB + BC = AC. So 3x-5 + x+2 = 13. Combine like terms to get 4x -3 = 13. Add 3 to both sides to get 4x = 16. Divide by 4 to get x = 4.
2007-11-30 11:05:50 · answer #6 · answered by Amy F 5 · 1⤊ 1⤋
A , B and c are collinear
AB + BC = AC
3x - 5 + x + 2 = 13
4x = 16
x = 4
2007-12-03 11:16:30 · answer #7 · answered by Como 7 · 1⤊ 1⤋
ac = ab+bc
13 = (3x-5) + (x+2)
13 = 4x-3
16 = 4x
x = 4
2007-11-30 11:06:03 · answer #8 · answered by English H 2 · 1⤊ 0⤋