Find the value of K for …
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Sia ? 6 years, 6 months ago
for infinitely many solution
{tex}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}{/tex}
i.e., {tex}\frac{2}{6} = \frac{{k - 2}}{{2k - 1}} = \frac{k}{{2k + 5}}{/tex}
if {tex}\frac{1}{3} = \frac{{k - 2}}{{2k - 1}}{/tex}
2k - 1 = 3k - 6
k = 5
or if {tex}\frac{k}{{2k + 5}} = \frac{1}{3}{/tex}
or 3k = 2k + 5
k = 5
if {tex}\frac{{k - 2}}{{2k - 1}} = \frac{k}{{2k + 5}}{/tex}
{tex}\Rightarrow{/tex} 2k2 + 5k - 4k - 10 = 2k2 - k
{tex}\Rightarrow{/tex} k - 10 = - k
{tex}\Rightarrow{/tex} k + k = 10
{tex}\Rightarrow{/tex} 2k = 10
{tex}\Rightarrow{/tex}k = 5
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