# Linear equations

• Dec 18th 2005, 04:47 PM
Boban
Linear equations
I have to determine for which value of parameter m the system has
a) indefinite solutions
b) no solution.

System is:

2x + (m-1)y = 3
(m+1)x + 4y = -3

From first equation x = (3 - y(m-1)) / 2x and
pluging that x into second equation I have get that (if I am right)

y = (3(3+m)) / (m-9)(m+1)

In book stands that a) m = -3 and b) m = 3.

Can someone explain me is that right solution and if it is, why it is?
• Dec 18th 2005, 06:39 PM
ThePerfectHacker
Consider this system:
2x + (m-1)y = 3
(m+1)x + 4y = -3
If the determinant is not zero then by Cramer's Rule there exist a unique solution, thus the determinant has to be zero:
(m^2-1)-8=0 thus m^2-9=0 thus m=-3 or 3.
When, m=3 we have:
2x+2y=3
4x+4y=-3
Which is equivalent to:
4x+4y=6
4x+4y=-3
An impossibility thus at m=3 there are no solutions.
If m=-3
2x - 4y = 3
-2x + 4y = -3
Which is equivalent to:
2x - 4y = 3
2x - 4y = 3
Which has infinitely many solutions.
Thus at m=-3 there are infinitely mant solutions.
• Dec 19th 2005, 03:43 AM
CaptainBlack
Quote:

Originally Posted by Boban
I have to determine for which value of parameter m the system has
a) indefinite solutions
b) no solution.

System is:

2x + (m-1)y = 3
(m+1)x + 4y = -3

From first equation x = (3 - y.(m-1)) / 2x

This looks OK except for that extraneous trailing x.

Quote:

..and pluging that x into second equation I have get that (if I am right)

y = (3(3+m)) / (m-9)(m+1)

In book stands that a) m = -3 and b) m = 3.

Can someone explain me is that right solution and if it is, why it is?

Plugging your expression for x into the second equation gives:

$(m+1) \frac{3-y(m-1)}{2}+4.y=-3$,

expanding a bit and replacing $(m+1)(m-1)$ by $m^2-1$:

$\frac{3.(m+1)}{2}-\frac{y.(m^2-1)}{2}+4y=-3$.

Which can be rearranged to:

$y.(4-\frac{m^2-1}{2})=-3+\frac{3.(m+1)}{2}$,

or:

$y.(9-m^2)=3m-3$.

and finally:

$y=\frac{3.m-3}{9-m^2}$.

This last expression is well behaved except at $m=\pm 3$ and
will give a sensible solution. So we need only investigate
the cases $m=3$ and $m=-3$.

In the original equations put $m=3$, and they become:

$2.x+2.y=3$
$4.x+4.y=-3$,

which are clearly inconsistent and so there is no solution.

In the original equations put $m=-3$, and they become:

$2.x-4.y=3$
$-2.x+4.y=-3$,

which are clearly just multiples of one another and so any point
satisfying $2.x-4.y=3$ is a solution to the equations and so
there are an infinite number of solutions (choose any value of x, and the
equation gives a value of y which satisfies the system)

(if you know about Cramer's rule and determinants ThePerfectHackers
solution should be prefered)

RonL.