Question

If pair of linear equations x+my+4=0 ,2x+5y-3=0 are consistent then find the calue m.

Answer 1

GIVEN :

If pair of linear equations x+my+4=0 , 2x+5y-3=0 are consistent .

TO FIND :

The value of m in the given linear equations

SOLUTION:

Given pair of linear equations  x+my+4=0 , 2x+5y-3=0 are consistent

$x+my+4=0\hfill (1)$ and  

$2x+5y-3=0\hfill (2)$

We have that the  given linear equations is of the form,

$a_1x+b_1y+c_1=0$

and  $a_2x+b_2y+c_2=0$

Here $a_1=1$, $b_1=m$ and $c_1=4$

$a_2=2$, $b_2=5$ and $c_2=-3$

Since the given linear equations is consistent, it must have unique solution or infinitely many solutions.

For the system of linear equations having unique solution is given by,

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$

$\frac{1}{2}\neq \frac{m}{5}$

$\frac{1}{2}\times 5\neq m$

Rewritting we get,

$m\neq \frac{1}{2}\times 5$

$m\neq \frac{5}{2}$

For the system of linear equations having unique solution if $m\neq \frac{5}{2}$

For the system of linear equations having infinitely many solutions is given by

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{1}{2}=\frac{m}{5}=\frac{4}{-3}$

$\frac{1}{2}\times 5=m$ or $m=\frac{4}{-3}\times 5$

Rewritting we get,

$m=\frac{1}{2}\times 5$ or $m=-\frac{20}{3}$ we take only positive value of m.

∴  the value is $m=\frac{5}{2}$