Question

For what walue of k will the following pair of linear equations have infinitely many.
solutions 2x-3y=7(k+1)x+(1-2k)y=5k-4​

Answer 1

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Answer 2

Given :-

• 2x - 3y = 7

• (k + 1)x + (1 - 2k)y + (4 - 5k) = 0

To find :-

• Value of k = ?

Solution :-

This equations are of the form,

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

where,

$\rm\:a_1=2,b_1=-3,c_1=-7\\\\\rm\:a_2=(k+1),b_2=(1-2k),c_2=(4-5k)$

The given system has infinity many Solutions.

Now we have,

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

$\rm\:\dfrac{2}{k+1}=\dfrac{-3}{(1-2k)}=\dfrac{-7}{(4-5k)}$

$\rm\longrightarrow\:\dfrac{2}{k+1}=\dfrac{3}{(2k-1)}$

$\rm\longrightarrow\:2(2k-1)=3(k+1)$

$\rm\longrightarrow\:4k-2=3k+3$

$\rm\longrightarrow\:4k-3k=3+2$

$\rm\longrightarrow\:k=5$

And,

$\rm\longrightarrow\:\dfrac{3}{(2k-1)}=\dfrac{7}{(5k-4)}$

$\rm\longrightarrow\:3(5k-4)=7(2k-1)$

$\rm\longrightarrow\:15k-12=14k-7$

$\rm\longrightarrow\:15k-14k=-7+12$

$\rm\longrightarrow\:k=5$

Hence,the value of k will be 5,