Question

Hey guys please answer the QuEStiON
It's urgent !!​

Answer 1

$\large\bf\underline{Question:-}$

If the pair of linear equations 2x + 3y = 12 and (m+n)x + (2m-n)y = 21 have infinitely many solutions ,then find the value of m and n

options :-

A) m = 1 , n = 5

B) m = 5 , n = 1

C) m = -1 ,n = 5

D) m = 5, n = -1

━━━━━━━━━━━━━

$\large\bf\underline{Given:-}$

pair of linear equations is given as

• 2x + 3y = 12 and
• (m+n)x + (2m-n)y = 21

$\large\bf\underline {To \: find:-}$

• value of m and n

$\huge\bf\underline{Solution:-}$

Given pair of linear equations is:-

• 2x +3y -7 = 0
• (m + n)x +(2m - n)y - 21 = 0

The equations are of the form

$\rm \: a_1x+b_1y + c_1 = 0 \\ \rm \: and \: a_2x+b_2y + c_2 = 0$

where,

• a1 = 2 , a2 = (m+n)
• b1 = 3 , b2 = (2m - n)
• c1 = -7 ,c2 = -21

$\rm \therefore \: \frac{a_1}{a_1} = \frac{2}{(m + n)} \: , \: \frac{b_1}{b_2} = \frac{3}{(2m - n)} \: , \: \frac{c_1}{c_2} = \frac{ - 7}{( - 21)}$

It is given that the equations have infinitely many solutions.

Then

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

$: \implies \rm \: \frac{2}{(m + n)} = \frac{3}{(2m - n)} = \frac{ - 7}{ - 21}$

1st case :-

when,

$\small: \implies \rm \: \frac{2}{(m + n)} = \frac{3}{(2m - n)} \: \\ \\ \small: \implies \rm \: 2(2m - n) = 3(m + n) \\ \\ \small: \implies \rm \: 4m - 2n = 3m + 3n \\ \\ \small: \implies \rm \: m - 5n = 0.......(i)$

Case 2nd :-

when,

$\small: \implies \rm \: \frac{3}{(2m - n)} = \frac{ - 7}{ - 21} \\ \\ \small: \implies \rm \: \frac{3}{2m - n} = \frac{1}{3} \\ \\ \small: \implies \rm \:9 = 2m - n \\ \\ \small: \implies \rm \:2m - n - 9 = 0..........(ii)$

from eq. (i)

m = 5n .........(iii)

Substituting value of m in eq.(ii)

➝ 2m - n -9 = 0

➝ 2(5n) - n - 9 = 0

➝ 10n -n -9 = 0

➝ 9n = 9

➝ n = 9/9

➝ n = 1

Substituting value of n in eq. (iii)

➝ m = 5n

➝ m = 5

So, value of m = 5 and n = 1

Option B) is correct.