for what value ofais there no unique solution for the following ?

3x - 2y = 5

5x +ay = 4

Printable View

- Jan 25th 2010, 04:18 PMsillylillysimultaneous equations problem
for what value of

*a*is there no unique solution for the following ?

3x - 2y = 5

5x +*a*y = 4 - Jan 25th 2010, 08:09 PMTKHunny
There are two answers, here.

1) No solution AT ALL, certainly would be "No Unique Solution"

2) More than one solution also would be "No Unique Solution"

#1 is trivial: 3*a - (5)*(-2) = 3a + 10 = 0 ==> a = -10/3

Another way to do this would be to put both linear equations into Slope-Intercept form. Setting the slopes equal and solving for 'a' gives the desired result. If the slopes are equal, the lines must be parallel and there is no common solution, unless...

#2 is a bit trickier, since we must show the two linear equations to represent exactly the same line. Unfortunately, we have a Degrees of Freedom problem. We can make EITHER the slopes equal OR the y-intercepts equal. We can't do both simultaneously.

Having said that, I sincerely hope the problem statement meant #1. - Jan 25th 2010, 09:16 PMArchie Meade
$\displaystyle 2y=3x-5$

$\displaystyle ay=-5x+4$

$\displaystyle y=\frac{3}{2}x-\frac{5}{2}$

$\displaystyle y=-\frac{5}{a}x+\frac{4}{a}$

There is a unique solution if the lines have a single point of intersection,

in which case the gradients must differ.

$\displaystyle \frac{3}{2} \ne -\frac{5}{a}$, for non-parallel lines

$\displaystyle a \ne -\frac{10}{3}$

If $\displaystyle a=-\frac{10}{3}$, there is no unique solution.

If the lines are parallel, there is no unique solution,

as there is no solution, since a solution gives the point of intersection.

A unique solution means the point of intersection of the two lines. - Jan 26th 2010, 04:24 AMHallsofIvy
No, this can't happen. If we multiply the second equation by $\displaystyle \frac{3}{5}$, we have

$\displaystyle 3x- 2y= 5$ and

$\displaystyle 3x- \frac{3a}{5} y= \frac{12}{5}$

There is NO value of a which will make the second equation the same as the first.

Quote:

Having said that, I sincerely hope the problem statement meant #1.

- Jan 26th 2010, 01:01 PMTKHunny
- Jan 27th 2010, 03:34 AMHallsofIvy