# Thread: linear solutions

1. ## linear solutions

Find the value of m for which the simultaneous equations $3x+my=5$ and $(m+2)x+5y=m$ have:

a) an infinite number of solutions
b) no solutions.

mostly having trouble with a), please explain how you got your answer,
thanks

2. Originally Posted by DivideBy0
Find the value of m for which the simultaneous equations $3x+my=5$ and $(m+2)x+5y=m$ have:

a) an infinite number of solutions
b) no solutions.

mostly having trouble with a), please explain how you got your answer,
thanks
Hello,

you know that a linear equation has a straight line as it's graph.

If you have the 2 linear equations

$l_1:y=m_1x + b_1~\text{ and }l_2:y=m_2x + b_2$ then you know that there are infinite many common points if $l_1=l_2~\iff~m_1=m_2\ \wedge \ b_1=b_2$

and

There is no solution (or there aren't any common points) if

$l_1\neq l_2~\iff~m_1=m_2\ \wedge \ b_1\neq b_2$

1. Rearrange your equations ( $y=-\frac3m x+\frac5m$ , $y=-\frac{m+2}{5} x + \frac m5$)

2. Use the general equations which I mentioned above

3. For confirmation only: Infinite many solutions with m = -5 and no solution with m = 3.

3. Thanks

4. Originally Posted by DivideBy0
Find the value of m for which the simultaneous equations $3x+my=5$ and $(m+2)x+5y=m$ have:

a) an infinite number of solutions
b) no solutions.

mostly having trouble with a), please explain how you got your answer,
thanks
$3x+my = 5$
$(m+2)x+5y = m$
The determinant of the coefficients is:
$\left| \begin{array}{cc}3&m\\m+2&5 \end{array} \right| = 15 - m(m+2)$
To have non-unique solution the determinant needs to be zero:
$m(m+2) = 15 \implies m=-5,3$.
Now check which one gives infinite solutions and which one gives no solutions.*

*)Or you can use determinants again, but there is no point. Because you just need to check two values.