Question

For what values of m and n the following system of linear equation has infinetly many solution 3x +4y=12 : (m+n)x +2(m-n)y=5m-1

Answer 1

Answer:

m=5. n=1

Step-by-step explanation:

Find the values of m and n for wch the following system of linear equations has infinitely many solutions 3x+4y=12 (m+n)x + 2(m-n)y = (5m-1)

to have infinite solutions

both equation should be same

(m+n)x + 2(m-n)y = 5m-1

should be equivalent to 3x + 4y = 12

=> (m+n)/3 = 2(m-n)/4 = (5m-1)/12

Multiplying each with 12

=> 4m + 4n = 6m - 6n = 5m -1

taking 1st 2

4m + 4n = 6m - 6n

=> 10n = 2m

=> m = 5n

taking 1st & 3rd

4m + 4n = 5m-1

=> m = 4n + 1

equating value of m

5n = 4n + 1

=> n = 1

m = 5n = 5 * 1 = 5

Hence m = 5 & n = 1

Verification of answer:

putting these value we get

6x + 8y = 24

dividing by 2

3x + 4y = 12

Hence verified

Answer 2

Given :-

• $\rm\:3x+4y=12$

• $\rm\:(m+n)x+2(m-n)=(5m-1)$

To find :-

• Value of m and n

Solution :-

The equations is written as:-

$\rm\:3x+4y-12=0\:.........................(i)$

$\rm\:(m+n)x+2(m-n)y-(5m-1)=0\:......(ii)$

These equations are of the form

$\rm\:{a_1}x+{b_1}y+{c_1}=0,$

$\rm\:{a_2}x+{b_2}y+{c_2}=0$

where $\rm\:a_1=3\:,b_1=4\:,c_1=-12$

and $\rm\:a_2=(m+n)\:,b_2=2(m-n)\:,c_1=-(5m-1)$

For infinitely many Solutions,we must have,

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

This holds only when,

$\rm\:\dfrac{3}{(m+n)}=\dfrac{4}{2(m-n)}=\dfrac{-12}{-(5m-1)}$

$\rm\longrightarrow\:\dfrac{3}{(m+n)}=\dfrac{2}{(m-n)}=\dfrac{12}{(5m-1)}$

$\rm\longrightarrow\:\dfrac{3}{(m+n)}=\dfrac{12}{(5m-1)}\:and\:\:\dfrac{2}{(m-n)}=\dfrac{12}{(5m-1)}$

$\rm\longrightarrow\:\dfrac{1}{(m+n)}=\dfrac{4}{(5m-1)}\:and\:\:\dfrac{1}{(m-n)}=\dfrac{6}{(5m-1)}$

$\rm\longrightarrow\:4(m+n)=(5m-1)\:and\:\:(5m-1)=6(m-n)$

$\rm\longrightarrow\:m-4n=1\:and\:\:m-6n=-1$

$\rm\longrightarrow\:m=5\:and\:\:n=1$[on solving the above equation]

Hence,for infinitely many solutions the value of m is 5 and n is 1.