# simultaneous equations problem

• Jan 25th 2010, 04:18 PM
sillylilly
simultaneous equations problem
for what value of a is there no unique solution for the following ?

3x - 2y = 5
5x + ay = 4
• Jan 25th 2010, 08:09 PM
TKHunny

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 PM
Quote:

Originally Posted by sillylilly
for what value of a is there no unique solution for the following ?

3x - 2y = 5
5x + ay = 4

$2y=3x-5$
$ay=-5x+4$

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

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

There is a unique solution if the lines have a single point of intersection,
in which case the gradients must differ.

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

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

If $a=-\frac{10}{3}$, 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 AM
HallsofIvy
Quote:

Originally Posted by TKHunny

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.

No, this can't happen. If we multiply the second equation by $\frac{3}{5}$, we have
$3x- 2y= 5$ and
$3x- \frac{3a}{5} y= \frac{12}{5}$
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 PM
TKHunny
Quote:

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

?? I do have trouble speaking english, but I thought I said that.
• Jan 27th 2010, 03:34 AM
HallsofIvy
Quote:

Originally Posted by TKHunny
?? I do have trouble speaking english, but I thought I said that.

Or I simply misunderstood what you were saying.