Question

2x-(a-4)y=2b+1, 4x-(a-1)y=5b-1 determine the value of a and b for which the following system of linear equations have infinite solutions

Answer 1

for infinite solution
$\frac{a1}{a2} = \frac{b1}{b2} \: = \frac{c1}{c2} \\ \frac{2}{4} = \frac{a - 4}{a - 1} = \frac{ - 2b - 1}{ - 5b + 1} \\ 2(a - 1) = 4(a - 4) \\ 2a - 2 - 4a + 16 = 0 \\ - 2a = - 14 \\ a = 7 \\ 2( - 5b + 1) = 4( - 2b - 1) \\ - 10b + 8b + 2 + 4 = 0 \\ - 2b = - 6 \\ b = 3$

Answer 2

Answer:

$a=7$ and $b=3$

Step-by-step explanation:

The given equations are:

$2x-(a-4)y=2b+1$ and $4x-(a-1)y=5b-1$

Therefore, $a_{1}=2$, $a_{2}=4$,$b_{1}=-(a-4)$, $b_{2}=-(a-1)$, $c_{1}=2b+1$, $c_{2}=5b-1$

Since, it is given that the system of linear equations have infinite many solutions, therefore

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\frac{2}{4}=\frac{-(a-4)}{-(a-1)}=\frac{2b+1}{5b-1}$

$\frac{a-4}{a-1}=\frac{2}{4}$, $\frac{2b+1}{5b-1}=\frac{2}{4}$

$2a-8=a-1$, $4b+2=5b-1$

$a=7$ and $b=3$