y=3x/h-2+5 hy-8x=5 for what values of h does the system of equations above have no solutions?
- PhilipLv 63 weeks ago
I assume your 1st eqn is y = [3/(h-2)]x + 5...(1).;
2nd eqn is hy -8x = 5, ie, y = (8/h)x + (5/h)...(2).;
Now we have both eqns in slope, y-intercept form.;
The system of (1) & (2) will have no solutions if slopes are equal and y-intercepts
are not equal.;
Suppose slopes are equal. Then 3/(h-2) = 8/h, ie., 3h = 8h-16, ie., h = (16/5).
For y-intercepts to be equal we require 5 = 5/h, ie., require h = 1. But h =/= 1. So
y-intercepts are not equal and slopes are equal for h = (16/5). Your option a. gives
the correct answer.
- llafferLv 73 weeks ago
Two linear equations will have no solutions if the slopes are the same and the intercepts are different.
If your equations are:
y = [3 / (h - 2)]x + 5 and hy - 8x = 5
Then we'll get the second equation into slope-intercept form to get its slope:
hy - 8x = 5
hy = 8x + 5
y = (8/h)x + (5/h)
So if we want the slopes to be the same, set the two slopes equal and solve for h:
3 / (h - 2) = 8/h
3h = 8(h - 2)
3h = 8h - 16
-5h = -16
h = 16/5
With h = 16/5, the intercept of the second will be different than the intercept of the first, so this system of equations will have no solutions.