Question

find the value of m which the pair of linear equations 2x+3y-7=0 and (m-1)x+(m+1)y=(3 m-1) has infinitely many solutions 

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Answer 1

Step-by-step explanation:

For infinitely many solutions,

2x+37−7=0

⇒2x+30 ………(1)

(m−1)x+(m+1)y−(3m−1)=0 ………..(2)

Condition is,

a

2

a

1

=

b

2

b

1

=

c

2

c

1

m−1

2

=

m+1

0

=

(3m−1)

−30

m−1

2

=

3m−1

−30

⇒6m−2=−30m+30

⇒36m=32

m= 9/8

Answer 2

Answer:

2x+3y-7=0 and (m-1)x+(m+1)y=(3 m-1)

As equations has infinite solutions so the equations are parallel.

If equations are parallel then

• A1/A2 = B1/B2 = C1/C2

NOW rearranging above equations

2x+3y-7=0 ( A1 = 2, B1 = 3 , C1 = -7 )

(m-1)x+(m+1)y - (3 m-1) =0

{HERE A2 = (m-1) , B2=(m+1) ,C2= -(3m-1)}

Now

A1 / A2 = B1/B2

2 / (m-1) = 3 / (m+1)

2(m+1) = 3(m-1)

2m+2 = 3m-3

2+3 = 3m-2m

5 = m

So m is 5