# y=3x/h-2+5 hy-8x=5 for what values of h does the system of equations above have no solutions?

a. 16/5

b. 13/8

c. 11/15

d. 5/8

### 2 Answers

- 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.

a. 16/5