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Sia ? 4 years, 9 months ago
The given system of equations is
(2a - 1)x - 3y = 5
3x + (b - 2)y = 3
{tex}\Rightarrow{/tex} (2a - 1)x - 3y - 5 = 0
{tex}\Rightarrow{/tex} 3x + (b - 2)y - 3 = 0
Here, a1 = 2a - 1, b1 = -3, c1 = -5
a2 = 3, b2 = b - 2, c2 = -3
for having infinitely many solutions, we must have
{tex}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}{/tex}
{tex}\Rightarrow \;\frac{{2a - 1}}{3} = \frac{{ - 3}}{{b - 2}} = \frac{{ - 5}}{{ - 3}}{/tex}
First and last give {tex}{\frac{{2a - 1}}{3} = \frac{{ - 5}}{{ - 3}}}{/tex}
{tex}{ \Rightarrow \;\frac{{2a - 1}}{3} = \frac{{ - 5}}{{ - 3}}}{/tex}
{tex}\Rightarrow{/tex} 2a - 1 = 5
{tex}\Rightarrow{/tex} 2a = 5 + 1 = 6
{tex}\Rightarrow{/tex} {tex}a = \frac{6}{2} = 3{/tex}
Second and last give {tex} \Rightarrow \;\frac{{ - 3}}{{b - 2}} = \frac{{ - 5}}{{ - 3}}{/tex}
{tex}\Rightarrow \;\frac{{ - 3}}{{b - 2}} = \frac{{ - 5}}{{ - 3}}{/tex}
{tex}\Rightarrow{/tex} 5(b - 2) = -9
{tex}\Rightarrow{/tex} 5b - 10 = - 9
{tex}\Rightarrow{/tex} 5b = 10 - 9 = 1
{tex}\Rightarrow \;b = \frac{1}{5}{/tex}
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