1. ## linear solutions

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

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

thanks

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

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

thanks
Hello,

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

If you have the 2 linear equations

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

and

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

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

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

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 $\displaystyle 3x+my=5$ and $\displaystyle (m+2)x+5y=m$ have:

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

$\displaystyle 3x+my = 5$
$\displaystyle (m+2)x+5y = m$
$\displaystyle \left| \begin{array}{cc}3&m\\m+2&5 \end{array} \right| = 15 - m(m+2)$
$\displaystyle m(m+2) = 15 \implies m=-5,3$.