# Thread: System of linear equattions

1. ## System of linear equattions

I have to determine for which values of parameters p and q the system has indefinite solutions.

System is:

(p-q)x + (3p-5)y = 2pq
(p+q)x + (q-7)y = 6pq

I have solved equation for x:
$x=\frac{18p^2q-24pq-2pq^2}{2pq+q^2+2p-12q+3p^2}$

What next?

2. Originally Posted by Boban
I have to determine for which values of parameters p and q the system has indefinite solutions.

System is:

(p-q)x + (3p-5)y = 2pq
(p+q)x + (q-7)y = 6pq

I have solved equation for x:
$x=\frac{18p^2q-24pq-2pq^2}{2pq+q^2+2p-12q+3p^2}$

What next?
It's simpler than that.

These equations represent a pair if lines in the $x\ y$ plane.
They have no solutions when the lines are parallel, but not the same line;
they have a single unique solution if they are not parallel, and an infinite
number of solutions if they represent the same line.

These equations represent the same line when they are a multiple of
one another.

If $p \cdot q \neq 0$
Then the RHS of the second equation is $3$ times the RHS of the
first equation, so the same must be true of the LHS. So multiplying
the first equation by 3 and then equating coefficients of $x$ and
$y$, which gives:

$p+q=3(p-q)$,
$q-7=3(3p-5)$.

Which is a pair of linear equations which can be solved for $p$ and $q$
to give $p$ and $q$ which give an infinite number of solutions
to the original equations.

If $p=q=0$
Both equations reduce to $y=0$, so there are an infinite
number of solutions (that is any point on the x-axis is a solution).

If $p=0\ q\neq 0$
The equations become:

$-q\cdot x-5\cdot y=0$
$\ q\cdot x+(q-7)y=0$,

which implies that $q=12$.

If $p \neq 0\ q= 0$
Left as an exercise to the reader.

RonL

3. The solution in the book is
$p=\frac{16}{17}$ and $q=\frac{8}{17}$

If $p \cdot q \neq 0$
Then the RHS of the second equation is $3$ times the RHS of the
first equation, so the same must be true of the LHS. So multiplying
the first equation by 3 and then equating coefficients of $x$ and
$y$, which gives:

$p+q=3(p-q)$,
$q-7=3(3p-5)$.
I didn't understand this part.
You have multiple first equation by 3 and how did you get this? What happened to x and y?

4. Originally Posted by Boban
The solution in the book is
$p=\frac{16}{17}$ and $q=\frac{8}{17}$
Clearly $p=0$ and $q=0$ give rise to the
solutions $y=0$, $x=\mbox{anything}$. so
the answer in the book is incomplete.

RonL

5. Originally Posted by Boban
The solution in the book is
$p=\frac{16}{17}$ and $q=\frac{8}{17}$

If $p \cdot q \neq 0$
Then the RHS of the second equation is $3$ times the RHS of the
first equation, so the same must be true of the LHS. So multiplying
the first equation by 3 and then equating coefficients of $x$ and
$y$, which gives:

$p+q=3(p-q)$,
$q-7=3(3p-5)$.

Which is a pair of linear equations which can be solved for $p$ and $q$
to give $p$ and $q$ which give an infinite number of solutions
to the original equations.

I didn't understand this part.
You have multiple first equation by 3 and how did you get this? What happened to x and y?
As I said you multiply the first equation by 3, then the RHS of the
two equations are equal. Then as these are to represent the same
line the the coefficients of $x$ and $y$ on the LHS must be the same
in each equation.

For example if

$a\cdot x+b\cdot y=D$
$c\cdot x+d\cdot y=D$

with $D\neq 0$ represent the same line then $a=c$ and
$b=d$.

Note the book's solution is in fact the solution of the equations:

$p+q=3(p-q)$,
$q-7=3(3p-5)$.

RonL

6. Originally Posted by Boban
I have to determine for which values of parameters p and q the system has indefinite solutions.

System is:

(p-q)x + (3p-5)y = 2pq
(p+q)x + (q-7)y = 6pq

I have solved equation for x:
$x=\frac{18p^2q-24pq-2pq^2}{2pq+q^2+2p-12q+3p^2}$

What next?
Alternative method:

Assume what you have done so far is correct. Then for any values of
$p$ and $q$ you have a well defined value of $x$, except when:

$2pq+q^2+2p-12q+3p^2=0$

you solve this equation for $p$ and $q$ to give candidate values for
the original equations to have an infinite number of solutions. These need to
be substituted back into the original equations to check that they give the
required type of solutions.

Then repeat the procedure solving for $y$ in terms of $p$ and $q$.

You can try this method but I know I'd rather use the other.

RonL

7. Thanks for help!