Question

The system of linear equations $5x+my=10$ and $4x+ny=8$ have infinitely many solutions, where m and n are positive integers. Then the minimum possible value of $(m+n)$ is equal to

• A. $9$
• B. $5$
• C. $6$
• D. $10$

Answer 1

Answer: (A)

$9$

We know, if two linear equations are $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$

For infinity many solutions :

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

Here, the two linear equations are,

$5x+my=10$

$\Rightarrow$ $5x+my-10=0$

$4x+ny=8$

$\Rightarrow$ $4x+ny-8=0$

For, infinity solutions,

$\Rightarrow$ $\frac{5}{4}=\frac{m}{n}=\frac{-10}{-8}$

$\Rightarrow$ $\frac{5}{4}=\frac{m}{n}=\frac{5}{4}$

So, the least value of $m$ is $5$ and $n$ is $4$.

$\Rightarrow$ Minimum possible value of $(m+n)=5+4=9$