Question

Find m and n for following pair of linear equation having infinite number of solutions (2m-1)x + 3y = 5. 3x + (n-1) y = 2.
m = 17 / 2 , n = 11/5
m= 17/4 , n = 11/5
m = 17 , n. = 11
m = 2. n = 5

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The following pair of linear equations have an infinite number of solution.

(2m-1)x + 3y = 5

3x + (n-1)y = 2

$\bf{\red{\underline{\bf{To\;find\::}}}}$

The value of m and n.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula of the infinite many solution :

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

Therefore;

$\bullet\sf{(2m-1)x+3y-5=0.....................(1)}\\\bullet\sf{3x+(n-1)y-2=0..............(2)}$

As we know that given equation compared with;

$\bullet\sf{a_1x+by_1+c_1=0}\\\bullet\sf{a_2x+by_2+c_2=0}$

$\tt{\dfrac{a_1}{a_2}} =\dfrac{(2m-1)}{3} }}\:,\:\dfrac{b_1}{b_2} =\dfrac{3}{(n-1)}} \:,\:\dfrac{c_1}{c_2}} =\dfrac{-5}{-2}}} }$

Now;

$\longrightarrow\sf{\dfrac{a_1}{a_2} =\dfrac{c_1}{c_2} }\\\\\\\longrightarrow\sf{\dfrac{(2m-1)}{3} =\dfrac{-5}{-2} }\\\\\\\longrightarrow\sf{-2(2m-1)=-5\times 3}\\\\\\\longrightarrow\sf{-4m+2=-15}\\\\\\\longrightarrow\sf{-4m=-15-2}\\\\\\\longrightarrow\sf{\cancel{-}4m=\cancel{-}17}\\\\\\\longrightarrow\sf{\pink{m=\dfrac{17}{4} }}$

&

$\longrightarrow\sf{\dfrac{a_1}{a_2} =\dfrac{b_1}{b_2} }\\\\\\\longrightarrow\sf{\dfrac{(2m-1)}{3} =\dfrac{3}{(n-1)} }\\\\\\\longrightarrow\sf{(n-1)(2m-1)=3\times 3}\\\\\\\longrightarrow\sf{2mn-n-2m+1=9}\\\\\\\longrightarrow\sf{2\bigg(\dfrac{17}{4} \bigg)n-n-2\bigg(\dfrac{17}{4} \bigg)+1=9\:\:[m=17/4]}\\\\\\\longrightarrow\sf{\dfrac{34n}{4} -n-\dfrac{34}{4} =9-1}\\\\\\\longrightarrow\sf{\dfrac{34n-4n-34}{4} =8}\\\\\\\longrightarrow\sf{34n-4n-34=32}\\\\\\\longrightarrow\sf{30n=32+34}$

$\longrightarrow\sf{30n=66}\\\\\\\longrightarrow\sf{n=\cancel{\dfrac{66}{30} }}\\\\\\\longrightarrow\sf{\pink{n=\dfrac{11}{5}}}$

Thus;

The value of m is 17/4 and n is 11/5 .

Answer 2

Answer:

Step-by-step explanation:

=17/4,n=-1/5

Step-by-step explanation:

If the condition is infinitely many solutions it is a1/a2=b1/b2=c1/c2

2m-1/3=3/n-1=-5/-2

taking a1/a2=c1/c2

2m-1/3=-5/-2

-2[2m-1]=-5[3]

=-4m+2=-15

-4m=-15-2

=-17

m=-17/-4=17/4

m=17/4

taking b1/b2=c1/c2

=3/n-1=-5/-2

5[n-1]=-2[3]

=5n-5=-6

5n=-6+5

=-1

n=-1/5

the

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