Triangle TUV has sides TU, UV, and VT. TU=6x - 11, UV=3x - 1, and VT=2x + 3. Describe the possible values of x.
Any help is appreciated. I've been trying to figure this one out for a while.
All three equations represent straight lines. Those lines intersect each other at the points T, U and V.
For example, your equation for TU and your equation for UV intersect at the point U(a,b).
So x = a, and y = b are solutions to BOTH equations, since that point lies on BOTH lines.
If you think of the equations being y = 6x-11 and y = 3x-1. The point at which these two lines intersect lies on both lines. And hence the y-coordinate of that point is equal for both lines.
Hence y = y
Hence 6x-11 = 3x-1
3x = 10
x = 10/3
This is the x-coordinate of the intersection (and hence the x coordinate of point U!). If you plug that value of x into either of the two equations, you'll get the y coordinated for U.
Good.
Now, if you find the three points T U and V using the method above, then you should find which one has the LOWEST x coordinate, and which one has the HIGHEST x coordinate. The range of possible x values for this triangle is the values in between those two coordinates, including the coordinates themselves!
Let me know what answer you get, I'll let you know if you're right or not.
If and only if the given terms refer to lengths of the triangles sides you have to use the triangle inequality:
The triangle ABC has the sides a, b, c. Then
$\displaystyle a+b>c\ \wedge\ a+c>b\ \wedge\ b+c>a$
So you get:
$\displaystyle 6x-11+3x-1>2x+3\ \wedge\ 6x-11+2x+3>3x-1\ \wedge\ 3x-1+2x+3>6x-11$
$\displaystyle x>\dfrac{15}7\ \wedge\ x>\dfrac95\ \wedge\ 13>x$
Therefore you know:
$\displaystyle \boxed{\dfrac{15}7 < x< 13}$